x-intercept OQ = x1 − dy y1
dx
x1 , y1 )
(
Similarly solving (i) and (iii) we get, y-intercept OR = y1 − x1 dy
dx
(x1,y1 )
Example: 16 The sum of intercepts on co-ordinate axes made by tangent to the curve x + y = a is [Rajasthan PET 1999]
(a) a (b) 2a (c) 2 a (d) None of these
Solution: (a) x+ y= a⇒ 1 +1 dy = 0 ⇒ ∴ dy y
2x 2y dx dx = − x
Hence tangent at (x, y) is Y − y = − y (X − x) or X y + Y x = xy( x + y) = axy or X + Y = 1 .
x ax ay
Clearly its intercepts on the axes are a x and a y .
Sum of the intercepts = a( x + y ) = a . a = a .
Length of Perpendicular from Origin to the Tangent.
Length of perpendicular from origin (0, 0) to the tangent drawn at point P(x1,y1) of the curve y = f(x)
y1 − x 1 dy
dx
p= ( x1 , y1 )
dy 2
dx
1 +
Example: 17 The length of perpendicular from (0, 0) to the tangent drawn to the curve y 2 = 4(x + 2) at point (2, 4) is
Solution: (c)
(a) 1 (b) 3 (c) 6 (d) 1
2 5 5
Differentiating the given equation w.r.t. x , 2y dy = 4 at point (2, 4) dy 1
dx dx = 2
y1 − x1 dy 4 − 2 1 6.
dx 2 5
P= = =
2 1
dy 1 + 4
1 + dx
Differentiation and application of derivatives
Introduction.
In this chapter we shall study those points of the domain of a function where its graph changes its direction from
upwards to downwards or from downwards to upwards. At such points the derivative of the function, if it exists, is
necessarily zero. y
x
O
Maximum and Minimum Values of a Function.
By the maximum / minimum value of function f(x) we should mean local or regional maximum/minimum and
not the greatest / least value attainable by the function. It is also possible in Y
a function that local maximum at one point is smaller than local minimum
at another point. Sometimes we use the word extreme for maxima and f(a–h) f(a)
minima. f(a+h)
Definition: A function f(x) is said to have a maximum at x = a if decreasing
f(a) is greatest of all values in the suitably small neighbourhood of a where
x = a is an interior point in the domain of f(x) . Analytically this means Increasing f(b) Increasing
f(b–h) f(b+h) X
O a–h a a+h b b+h
(b–h)
f(a) ≥ f(a + h) and f(a) ≥ f(a − h) where h ≥ 0 . (very small quantity).
Similarly, a function y = f(x) is said to have a minimum at x = b . If f(b) is smallest of all values in the suitably
small neighbourhood of b where x = b is an interior point in the domain of f(x) . Analytically, f(b) ≤ f(b + h) and
f(b) ≤ f(b − h) where h ≥ 0 . (very small quantity).
Y
f(a)
f(a–h) f(a+h)
increasing decreasing
f(b) f(b+h)
f(b–h) X
O a–h a a+h b–h b b–h
Hence we find that, y
(i) x=a is a maximum point of f(x) f(a) − f(a + h) > 0
f (a) − f (a − h) > 0
(ii) x = b is a minimum point of f(x) f(b) − f (b + h) < 0
− f (b − h) < 0
f (b)
(iii) x = c is neither a maximum point nor a minimum point, ab cx
f (c) − f (c + h) and have opposite signs .
f (c) − f (c −
h)
Local Maxima and Local Minima.
(1) Local maximum : A function f(x) is said to attain a local maximum at x = a if there exists a
neighbourhood (a − δ ,a + δ ) of a such that f(x) < f(a) for all x ∈(a − δ ,a + δ ), x ≠ a
or f(x) − f(a) < 0 for all x ∈(a − δ ,a + δ ), x ≠ a .
In such a case f(a) is called the local maximum value of f(x) at x = a .
(2) Local minimum: A function f(x) is said to attain a local minimum at x = a if there exists a
neighbourhood (a − δ ,a + δ ) of a such that
f(x) > f(a) for all x ∈(a − δ ,a + δ ), x ≠ a
or f(x) − f(a) > 0 for all x ∈(a − δ ,a + δ ), x ≠ a
The value of function at x = a i.e., f(a) is called the local minimum value of f(x) at x = a .
The points at which a function attains either the local maximum values or local minimum values are known as
the extreme points or turning points and both local maximum and local minimum values are called the extreme
values of f(x) . Thus, a function attains an extreme value at x = a if f(a) is either a
local maximum value or a local minimum value. Consequently at an extreme point y
' a' f(x) − f(a) keeps the same sign for all values of x in a deleted nbd of a.
y = f(x) CE
In fig. we observe that the x-coordinates of the points A, C, E are points of local
maximum and the values at these points i.e., their y-coordinates are the local A D
maximum values of f(x) . The x-coordinates of points B and D are points of local B
minimum and their y-coordinates are the local minimum values of f(x) . O x
Note : By a local maximum (or local minimum) value of a function at a
point x = a we mean the greatest (or the least) value in the neighbourhood of point x = a and not
the absolute maximum (or the absolute minimum). In fact a function may have any number of points
of local maximum (or local minimum) and even a local minimum value may be greater than a local
maximum value. In fig. the minimum value at D is greater than the maximum value at A. Thus, a local
maximum value may not be the greatest value and a local minimum value may not be the least value
of the function in its domain.
• The maximum and minimum points are also known as extreme points.
• A function may have more than one maximum and minimum points.
• A maximum value of a function f(x) in an interval [a, b] is not necessarily its greatest value in that
interval. Similarly, a minimum value may not be the least value of the function. A minimum value
may be greater than some maximum value for a function.
• If a continuous function has only one maximum (minimum) point, then at this point function has its
greatest (least) value.
• Monotonic functions do not have extreme points.
Conditions for Maxima and Minima of a Function.
(1) Necessary condition: A point x = a is an extreme point of a function f(x) if f ′(a) = 0, provided f ′(a)
exists. Thus, if f ′(a) exists, then
x = a is an extreme point ⇒ f ′(a) = 0
or
f ′(a) ≠ 0 ⇒ x = a is not an extreme point
But its converse is not true i.e., f ′(a) = 0, x = a is not an extreme point.
For example if f(x) = x 3 , then f ′(0) = 0 but x = 0 is not an extreme point.
(2) Sufficient condition:
(i) The value of the function f(x) at x = a is maximum, if f ′(a) = 0 and f ′′(a) < 0 .
(ii) The value of the function f(x) at x = a is minimum if f ′(a) = 0 and f ′′(a) > 0 .
Note : If f ′(a) = 0 , f ′′(a) = 0, f ′′′(a) ≠ 0 then x = a is not an extreme point for the function f(x) .
If f ′(a) = 0, f ′′(a) = 0, f ′′′(a) = 0 then the sign of f (iv) (a) will determine the maximum and minimum
value of function i.e., f(x) is maximum, if f (iv)(a) < 0 and minimum if f (iv)(a) > 0 .
Working rule for Finding Maxima and Minima.
(1) Find the differential coefficient of f(x) with respect to x, i.e., f ′(x) and equate it to zero.
(2) Find differential real values of x by solving the equation f ′(x) = 0 . Let its roots be a, b, c......
(3) Find the value of f ′′(x) and substitute the value of a1,a2,a3...... in it and get the sign of f′′(x) for each value
of x.
(4) If f ′′(a) < 0 then the value of f(x) is maximum at x = a and if f ′′(a) > 0 then value of f(x) will be
minimum at x = a . Similarly by getting the signs of f ′′(x) at other points b, c.....we can find the points of maxima
and minima.
Example: 1 What are the minimum and maximum values of the function x5 − 5x 4 + 5x 3 − 10 [DCE 1999; Rajasthan PET 1995]
Solution: (a)
(a) −37, − 9 (b) 10, 0
(c) It has 2 minimum and 1 maximum values (d) It has 2 maximum and 1 minimum values
y = x 5 − 5x 4 + 5x 3 − 10
dy = 5x4 − 20x3 + 15x 2 = 5 x 2(x 2 − 4x + 3) = 5x 2(x − 3)(x − 1)
∴ dx
dy = 0 , gives x = 0,1, 3 ......(i)
dx
Now, d2y = 20x 3 − 60x 2 + 30x = 10x(2x 2 − 6x + 3) and d3y = 10(6x 2 − 12x + 3)
dx 2 dx 3
For x = 0 : dy = 0, d2y = 0, d3y ≠ 0 , ∴ Neither minimum nor maximum
dx dx 2 dx 3
For x = 1 , d2y = −10 =negative, ∴ Maximum value ymax . = −9
dx 2
For x = 3 , d2y = 90 =positive, ∴ Minimum value ymin. = −37 .
dx 2
Example: 2 The maximum value of sin x(1 + cos x) will be at [UPSEAT 1999]
Solution: (c)
(a) x = π (b) x = π (c) x = π (d) x = π
Example: 3 2 6 3
Solution: (d)
Example: 4 y = sin x(1 + cos x) = sin x + 1 sin 2x
Solution: (b) 2
Example: 5
Solution: (b) ∴ dy = cos x + cos 2x and d2y = − sin x − 2 sin 2x
dx dx 2
On putting dy = 0, cos x + cos 2x = 0 ⇒ cos x = − cos 2x = cos(π − 2x) ⇒ x = π − 2x
dx
∴ x = π , ∴ d2y = − sin 1 π − 2 sin 2 π = −3 − 2. 3 = −3 3 which is negative.
3 dx 2 3 3 2 2 2
x =π / 3
∴ at x = π the function is maximum.
3
If y = a log x + bx 2 + x has its extremum value at x = 1 and x = 2 , then (a, b) = [UPSEAT 2002]
(a) 1, 1 (b) 1 , 2 (c) 2, − 1 (d) − 2 ,− 1
2 2 2 3 6
dy = a + 2bx + 1 ⇒ dy = a + 2b + 1 = 0 ⇒ a = −2b − 1
dx x dx
x =1
and dy = a + 4b + 1 = 0 ⇒ −2b − 1 + 4b + 1 = 0 ⇒ − b + 4b + 1 =0 ⇒ 3b = −1 ⇒ b= −1 and a = 1 −1 = −2 .
dx 2 2 2 2 6 3 3
x = 2
1 x
x
Maximum value of is [DCE 1999; Karnataka CET 1999; UPSEAT 2003]
(a) (e)e (b) (e)1 / e (c) (e)−e (d) 1 e
e
1 x 1 x log 1 − 1
x x x
f (x) = ⇒ f ′(x) =
f ′(x) = 0 ⇒ log 1 = 1 = log e ⇒ 1 =e ⇒ x = 1 . Therefore, maximum value of function is e1/ e .
x x e
Maximum slope of the curve y = −x 3 + 3x 2 + 9x − 27 is [MP PET 2001]
(a) 0 (b) 12 (c) 16 (d) 32
y = f(x) = −x 3 + 3x 2 + 9x − 27
The slope of this curve f ′(x) = −3x 2 + 6x + 9
Let g(x) = f ′(x) = −3x 2 + 6x + 9
Differentiate with respect to x, g′(x) = −6x + 6
Put g′(x) = 0 ⇒ x = 1
Now, g′′(x) = −6 < 0 and hence at x = 1, g(x)
(Slope) will have maximum value.
∴ [g(1)]max . = −3 × 1 + 6 + 9 = 12 .
Example: 6 ∫The function f(x) = x t − 1)(t − 1)(t − 2)3 (t − 3)5dt has a local minimum at x= [IIT1999]
t(e
1
(a) 0 (b) 1 (c) 2 (d) 3
Solution: (b, d) ∫f(x) = x t − 1)(t − 1)(t − 2)3 (t − 3)5 dt , ∴ f ′(x) = x(e x − 1)(x − 1)(x − 2)3(x − 3)5
t(e
−1
For local minima, slope i.e., f ′(x) should change sign from – ve to +ve
f ′(x) = 0 ⇒ x = 0,1, 2, 3
If x = 0 − h, where h is a very small number, then f ′(x) = (−)(−)(−1)(−1)(−1) = −ve
If x = 0 + h , f ′(x) = (+)(+)(−)(−1)(−1) = −ve
Hence at x = 0 neither maxima nor minima.
If x = 1 − h , f ′(x) = (+)(+)(−)(−1)(−1) = −ve
If x = 1 + h , f ′(x) = (+)(+)(+)(−1)(−1) = +ve
Hence, at x = 1 there is a local minima.
If x = 2 − h , f ′(x) = (+)(+1)(+)(−)(−) = +ve
If x = 2 + h, f ′(x) = (+)(+)(+)(+)(−1) = −ve
Hence at x = 2 there is a local maxima.
If x = 3 − h , f ′(x) = (+)(+)(+)(+)(−) = −ve
If x = 3 + h , f ′(x) = (+)(+)(+)(+)(+) = +ve
Hence at x = 3 there is a local minima.
Example: 7 If the function f(x) = 2x 3 − 9ax 2 + 12a2x + 1 , where a > 0 attains its maximum and minimum at p and q respectively
such that p2 = q , then a equals [AIEEE 2003]
(a) 3 (b) 1 (c) 2 (d) 1
2
Solution: (c) f(x) = 2x 3 − 9ax 2 + 12a2 x + 1
f ′(x) = 6x 2 − 18ax + 12a 2
f ′′(x) = 12x − 18a
For maximum and minimum, 6x 2 − 18ax + 12a2 = 0 ⇒ x 2 − 3ax + 2a2 = 0
x = a or x = 2a at x = a maximum and at x = 2a minimum
p2 = q
a2 = 2a ⇒ a = 2 or a = 0 but a > 0, therefore a = 2 .
Example: 8 ∫The points of extremum of the function φ(x) = x e−t2 / 2(1 − t 2)dt are
1
(a) x = 0 (b) x = 1 (c) x = 1 (d) x = −1
2
Solution: (b,d) ∫φ(x) = x e−t2 / 2(1 − t 2)dt ⇒ φ ′(x) = e−x2 / 2(1 − x 2)
1
Now φ ′(x) = 0 ⇒ 1 − x 2 = 0 ⇒ x = ±1
Hence, x = ±1 are points of extremum of φ(x) .
Point of Inflection.
A point of inflection is a point at which a curve is changing concave upward to concave downward or vice-
versa. A curve y = f(x) has one of its points x = c as an inflection point, if y
f ′′(c) = 0 or is not defined and if f ′′(x) changes sign as x increases through x = c .
The later condition may be replaced by f′′′(c) ≠ 0 , when f′′′(c) exists. y = f(x)
Thus, x = c is a point of inflection if f ′′(c) = 0 and f′′′(c) ≠ 0 . f(c)
Properties of maxima and minima
(i) If f(x) is continuous function in its domain, then at least one maxima and O c x
one minima must lie between two equal values of x.
(ii) Maxima and minima occur alternately, that is, between two maxima there is one minimum and vice-versa.
(iii) If f(x) → ∞ as x → a or b and f ′(x) = 0 only for one value of x (say c) between a and b, then f(c) is
necessarily the minimum and the least value.
If f(x) → −∞ as x → a or b, then f(c) is necessarily the maximum and the greatest value.
Greatest and Least Values of a Function in a given Interval.
If a function f(x) is defined in an interval [a, b], then greatest or least values of this function occurs either at
x = a or x = b or at those values of x where f′(x) = 0 .
Remember that a maximum value of the function f(x) in any interval [a, b] is not necessarily its greatest value
in that interval. Thus greatest value of f(x) in interval [a, b] = max. [ f(a), f(b), f(c)]
Least value of f(x) interval [a, b]= min. [ f(a), f(b), f(c)]
Where x = c is a point such that f′(c) = 0
Example: 9 The maximum and minimum values of x3 − 18x2 + 96 in interval (0, 9) are [RPET 1999]
Solution: (c)
(a) 160, 0 (b) 60, 0 (c) 160, 128 (d) 120, 28
Example: 10
Solution: (d) Let y= x3 − 18x2 + 96x ⇒ dy = 3x2 − 36x + 96 = 0
Example: 11 dx
∴ x2 − 12x + 32 = 0 ⇒ (x − 4)(x − 8) = 0, x = 4,8
Now, d2y = 6x − 36 at x = 4, d2y = 24 − 36 = −12 < 0
dx 2 dx 2
∴ at x=4 function will be maximum and [ f(x)]max . = 64 − 288 + 384 = 160 at x = 8 d2y = 48 − 36 = 12 > 0
dx 2
∴ at x = 8 function will be minimum and [ f(x)]min . = 128 .
The minimum value of the function 2cos 2x − cos 4 x in 0 ≤ x ≤ π is
(a) 0 (b) 1 (c) 3 (d) – 3
2
y = 2cos 2x − cos 4 x = 2cos 2x(1 − cos 2x) + 1 = 4 cos 2x sin2 x + 1
Obviously, sin2 x ≥ 0 [AMU 2002]
Therefore, to be least value of y, cos 2x should be least i.e., – 1. Hence least value of y is – 4 + 1 = –3.
On [1, e] the greatest value of x2 log x
(a) e2 (b) 1 log 1 (c) e2 log e (d) None of these
e e
Solution: (a) f(x) = x2 log x ⇒ f′(x) = (2log x + 1)x
Now f ′(x) = 0 ⇒ x = e−1/ 2,0
0 < e−1/ 2 < 1 , None of these critical points lies in the interval [1,e]
∴ So we only compare the value of f(x) at the end points 1 and e. We have f(1) = 0, f(e) = e2
∴ greatest value = e2
Maxima and Minima of Functions of Two Variables.
If a function is defined in terms of two variables and if these variables are associated with a given relation then
by eliminating one variable, we convert function in terms of one variable and then find maxima and minima by
known methods.
Example: 12 x and y be two variables such that x > 0 and xy = 1 . Then the minimum value of x + y is
Solution: (a)
(a) 2 (b) 3 (c) 4 [Kurukshetra CEE 1988; MP PET 2002]
Example: 13
Solution: (c) (d) 0
Example: 14 xy =1 ⇒ y= 1 and let z = x + y
Solution: (b) x
z = x + 1 ⇒ dz =1− 1 ⇒ dz =0 ⇒ 1− 1 =0 ⇒ x = −1, + 1 and d2z 2
x dx x2 dx x2 dx 2 = x3
d2z x =1 = 2 = 2 = +ve , ∴ x =1 is point of minima.
dx 2 1
x = 1, y = 1 , ∴ minimum value = x + y = 2 .
The sum of two non-zero numbers is 4. The minimum value of the sum of their reciprocals is [Kurukshetra CEE 1998]
(a) 3 (b) 6 (c) 1 (d) None of these
4 5
Let x + y = 4 or y = 4 − x
1 + 1 = x+y or f(x) = 4 = 4
x y xy xy x(4 − x)
f(x) = 4 x2 , f ′(x) = −4 .(4 − 2x)
4x − (4 x − x 2)2
Put f′(x) = 0 ⇒ 4 − 2x = 0 ⇒ x = 2 and y = 2
∴ min. 1 + 1 = 1 + 1 =1.
x y 2 2
The real number which most exceeds its cube is [MP PET 2000]
(a) 1 (b) 1 (c) 1 (d) None of these
2 3 2
Let number = x, then cube = x 3
Now f(x) = x − x 3 (Maximum) ⇒ f ′(x) = 1 − 3x 2
Put f ′(x) = 0 ⇒ 1 − 3x 2 = 0 ⇒ x = ± 1
3
Because f ′′(x) = −6x = −ve . when x = + 1 .
3
Geometrical Results related to Maxima and Minima.
The following results can easily be established.
(1) The area of rectangle with given perimeter is greatest when it is a square.
(2) The perimeter of a rectangle with given area is least when it is a square.
(3) The greatest rectangle inscribed in a given circle is a square.
(4) The greatest triangle inscribed in given circle is equilateral.
(5) The semi vertical angle of a cone with given slant height and maximum volume is tan−1 2
(6) The height of a cylinder of maximum volume inscribed in a sphere of radius a is a 2a / 3 .
Important Tips
Equilateral triangle: Area = ( 3 / 4)x2, where x is its side.
• Square: Area = a 2 , perimeter = 4a , where a is its side.
• Rectangle: Area = ab, perimeter = 2(a + b) , where a, b are its sides.
• Trapezium: Area = 1 (a + b)h , where a, b are lengths of parallel sides and h be the distance between them.
2
• Circle: Area = πa 2 , perimeter = 2πa , where a is its radius.
• Sphere: Volume = 4 πa 3 , surface area = 4πa 2 , where a is its radius.
3
• Right circular cone: Volume = 1 πr 2h, curved surface = πrl , where r is the radius of its base, h is its height and l is its slant
3
height.
• Cylinder: Volume = πr2h , whole surface = 2πr(r + h) , where r is the radius of the base and h is its height.
Example: 15 The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area are [MP PET 1993]
Solution: (c)
(a) 10 cm and 40 cm (b) 20 cm and 30 cm (c) 25 cm and 25 cm (d) 15 cm and 35 cm
Example: 16
Solution: (b) 2x + 2y = 100 ⇒ x + y = 50 .....(i)
Let area of rectangle is A, ∴ A = xy ⇒ y = A
x
From (i), x+ A = 50 ⇒ A = 50x − x2 ⇒ dA = 50 − 2x
x dx
for maximum area dA =0
dx
∴ 50 − 2x = 0 ⇒ x = 25 and y = 25
∴ adjacent sides are 25 cm and 25 cm.
The radius of the cylinder of maximum volume, which can be inscribed a sphere of radius R is [AMU 1999]
(a) 2 R (b) 2 R (c) 3 R (d) 3 R
3 3 4 4
If r be the radius and h the height, the from the figure, r2 + h 2 = R2 ⇒ h2 = 4(R2 − r 2)
2
Now, V = πr 2h = 2πr 2 R2 − r 2
∴ dV = 4πr R2 − r2 + 2πr 2 . 1 (−2r)
dr 2 R2 − r2
D LC
dV
For max. or min., dr =0
⇒ 4πr R2 − r 2 = 2πr 3 ⇒ 2(R2 − r 2) = r 2 O
R2 − r2 R
2 d2V 2 θ B
3 dr 2 3 A rM
⇒ 2R2 = 3r 2 ⇒ r = R ⇒ = −ve . Hence V is max. when r= R .
Example: 17 The ratio of height of a cone having maximum volume which can be inscribed in a sphere with the diameter of sphere is
Solution: (a)
[MNR 1985]
(a) 2 (b) 1 (c) 3 (d) 1
3 3 4 4
Let OM = x
Then height of cone i.e., h = x + a (where a is radius of sphere)
Radius of base of cone = a2 − x2
Therefore, volume V = 1 π (a 2 − x 2 )(x + a) ⇒ dV = π (a + x)(a − 3x)
3 dx 3
A
Now, dV =0 ⇒ x = −a, a
dx 3 a
But x ≠ −a, So, x = a O
3 aa
The volume is maximum at x = a x
3 B MC
Height of a cone h = a+ a = 4 a
3 3
Therefore ratio of height and diameter = 4 a 2 .
3 3
2a =
Differentiation and application of derivatives
Definition.
(1) Strictly increasing function : A function f(x) is said to be a strictly y
increasing function on (a, b), if x1 < x2 ⇒ f(x1) < f(x2) for all x1, x2 ∈ (a, b).
y = f(x)
Thus, f(x) is strictly increasing on (a, b), if the values of f(x) increase with the f(x1) f(x2)
increase in the values of x. O a x1 x2 b
(2) Strictly decreasing function : A function f(x) is said to be a strictly
decreasing function on (a,b) , if x1 < x2 ⇒ f(x1) > f(x2) for all x1, x2 ∈ (a, b) . Thus, x
f(x) is strictly decreasing on (a, b), if the values of f(x) decrease with the increase in
the values of x. y
x′ a f(x1) f(x2)
O x1 x2 b x
Example: 1 On the interval (1, 3) the function f(x) = 3x + 2 is [AMU 1999]
Solution: (b) x
Example: 2
Solution: (c) (a) Strictly decreasing (b) Strictly increasing
(d) Neither increasing nor decreasing
Example: 3 (c) Decreasing in (2, 3) only
Solution: (a)
Example: 4 f(x) = 3x + 2 ⇒ f ′(x) = 3 − 2
x x2
Clearly f ′(x) > 0 on the interval (1, 3)
∴ f(x) is strictly increasing.
For which value of x, the function f(x) = x 2 − 2x is decreasing [Ranchi BIT 1990]
(a) x > 1 (b) x > 2 y
(c) x < 1 (d) x < 2
f(x) = (x − 1)2 − 1 (1, y=
Hence decreasing in x < 1 O x
Alternative method: f ′(x) = 2x − 2 = 2(x − 1)
To be decreasing, 2(x − 1) < 0 ⇒ (x − 1) < 0 ⇒ x < 1 . (1, –
2x 3 + 18x 2 − 96x + 45 = 0 is an increasing function when [Rajasthan PET 1997]
(a) x ≤ −8, x ≥ 2 (b) x < −2, x ≥ 8 (c) x ≤ −2, x ≥ 8 (d) 0 < x ≤ −2
f ′(x) = 6x 2 + 36x − 96 > 0 , for increasing
⇒ f′(x) = 6(x + 8)(x − 2) ≥ 0 ⇒ x ≥ 2, x ≤ −8 .
The function x x is increasing, when [MP PET 2003]
(a) x > 1 (b) x < 1 (c) x < 0 (d) For all real x
e e
Solution: (a) Let y = xx ⇒ dy = x x (1 + log x) ; For dy >0
dx dx
x x (1 + log x) > 0 ⇒ 1 + log x >0 ⇒ log e x > log e 1
e
For this to be positive, x should be greater than 1 .
e
Monotonic Function.
A function f(x) is said to be monotonic on an interval (a, b) if it is either increasing or decreasing on (a, b).
(1) Monotonic increasing function : A function is said to be a monotonic increasing function in defined
interval if, yy
x1 > x2 ⇒ f(x1) ≥ f(x2)
or x1 > x2 ⇒ f(x1) < f(x2)
or x1 < x2 ⇒ f(x1) ≤ f(x2)
or x1 < x2 ⇒ f(x1)> f(x2) O x x
O
(2) Monotonic decreasing function: A function is said to be a monotonic decreasing function in defined
interval, if x1 > x2 ⇒ f(x1) ≤ f(x2) y y
or x1 > x2 ⇒ f(x1) > f(x2)
or x1 < x 2 ⇒ f(x1) ≥ f(x 2 )
or x1 < x2 ⇒ f(x1) < f(x2)
O xO x
Example: 5 The function f(x) = cos x − 2px is monotonically decreasing for [MP PET 2002]
Solution: (b)
Example: 6 (a) p < 1 (b) p > 1 (c) p < 2 (d) p > 2
Solution: (c) 2 2
Example: 7
f(x) will be monotonically decreasing, if f ′(x) < 0 .
⇒ f ′(x) = − sin x − 2p < 0 ⇒ 1 sin x + p > 0 ⇒ p > 1 [ − 1 ≤ sin x ≤ 1]
2 2
If f(x) = x5 − 20x 3 + 240x , then f(x) satisfies which of the following [Kurukshetra CEE 1996]
(a) It is monotonically decreasing everywhere (b) It is monotonically decreasing only in (0,∞)
(c) It is monotonically increasing every where (d) It is monotonically increasing only in (−∞,0)
f ′(x) = 5x 4 − 60x 2 + 240 = 5(x 4 − 12x 2 + 48) = 5[(x 2 − 6)2 + 12]
⇒ f ′(x) > 0 ∨ x ∈ R
i.e., f(x) is monotonically increasing everywhere.
The value of a for which the function (a + 2)x 3 − 3ax 2 + 9ax − 1 decrease monotonically throughout for all real x, are
[Kurukshetra CEE 2002]
(a) a < −2 (b) a > −2 (c) −3 < a < 0 (d) −∞ < a ≤ −3
Solution: (d) If f(x) = (a + 2)x 3 − 3ax 2 + 9ax − 1 decreases monotonically for all x ∈ R, then f ′(x) ≤ 0 for all x ∈ R
Example: 8 ⇒ 3(a + 2)x 2 − 6ax + 9a ≤ 0 for all x ∈ R ⇒ (a + 2)x 2 − 2ax + 3a ≤ 0 for all x ∈ R
Solution: (d)
⇒ a + 2 < 0 and discriminant ≤ 0 ⇒ a < −2 and − 8a2 − 24a ≤ 0
⇒ a < −2 and a(a + 3) ≥ 0 ⇒ a < −2 and a ≤ −3 or a ≥ 0 ⇒ a ≤ −3 ⇒ −∞ < a ≤ −3
Function f(x) = λ sin x + 6 cos x is monotonic increasing if [MP PET 2001]
2 sin x + 3 cos x
(a) λ > 1 (b) λ < 1 (c) λ<4 (d) λ > 4
The function is monotonic increasing if, f ′(x) > 0
⇒ (2 sin x + 3 cos x)(λ cos x − 6 sin x) – (λ sin x + 6 cos x)(2 cos x − 3 sin x) > 0
(2 sin x + 3 cos x)2 (2 sin x + 3 cos x)2
⇒ 3λ(sin2 x + cos2 x) − 12(sin2 x + cos2 x) > 0 ⇒ 3λ − 12 > 0 ⇒ λ > 4 .
Necessary and Sufficient Condition for Monotonic Function.
In this section we intend to see how we can use derivative of a function to determine where it is increasing and
where it is decreasing
(1) Necessary condition : From figure we observe that if f(x) is an increasing function on (a, b), then
tangent at every point on the curve y = f(x) makes an y y
O P(x, y)
acute angle θ with the positive direction of x-axis.
∴ tan θ >0⇒ dy > 0 or f' (x) > 0 for all x ∈ (a, b) P(x,y)
dx y=f(x)
y=f(x)
It is evident from figure that if f(x) is a decreasing θ ab θ
ab x
function on (a, b), then tangent at every point on the xO
curve y = f(x) makes an obtuse angle θ with the positive
direction of x-axis.
∴ tan θ < 0 ⇒ dy < 0 or f ' (x) < 0 for all x ∈ (a, b).
dx
Thus, f ' (x) > 0(< 0) for all x ∈ (a, b) is the necessary condition for a function f (x) to be increasing (decreasing)
on a given interval (a, b). In other words, if it is given that f (x) is increasing (decreasing) on (a, b), then we can say
that f ' (x) > 0 (< 0).
(2) Sufficient condition : Theorem : Let f be a differentiable real function defined on an open interval (a, b).
(a) If f ' (x) > 0 for all x ∈ (a, b) , then f(x) is increasing on (a, b).
(b) If f ' (x) < 0 for all x ∈ (a, b) , then f(x) is decreasing on (a, b).
Corollary : Let f (x) be a function defined on (a, b).
(a) If f'(x) > 0 for all x ∈ (a, b) , except for a finite number of points, where f'(x) = 0, then f(x) is increasing on
(a, b).
(b) If f'(x) < 0 for all x ∈ (a, b) , except for a finite number of points, where f'(x) = 0, then f (x) is decreasing on
(a, b).
Example: 9 The function f(x) = ln(π + x) is [IIT 1995]
Solution: (b) ln(e + x)
(a) Increasing on [0, ∞) (b) Decreasing on [0, ∞)
(c) Decreasing on 0, π and increasing on π , ∞ (d) Increasing on 0, π and decreasing on π , ∞
e e e e
Let f(x) = ln(π + x)
ln(e + x)
f ′(x) ln(e + x)× π 1 x − ln(π + x) e 1 x = (e + x) ln(e + x) − (π + x) ln(π + x)
+ + {ln(e + x)}2 × (e + x)(π + x)
∴ = {ln(e + x)}2
⇒ f′(x) < 0 for all x ≥ 0 {π > e} . Hence, f(x) is decreasing in [0, ∞) .
Example: 10 Which of the following is not a decreasing function on the interval 0, π
Solution: (c) 2
Example: 11
(a) cos x (b) cos 2x (c) cos 3x (d) cot x
(d) (−∞, − 1)
Obviously, here cos 3x in not decreasing in 0, π because d cos 3x = −3 sin 3x .
2 dx
But at x = 75o , −3 sin 3x > 0 . Hence the result.
The interval of increase of the function f(x) = x − ex + tan 2π is
7
(a) (0, ∞) (b) (−∞, 0) (c) (1, ∞)
Solution: (b, d) We have f(x) = x − ex + tan 2π ⇒ f′(x) = 1 − ex
7
For f(x) to be increasing, we must have f′(x) > 0 ⇒ 1 − ex > 0 ⇒ ex < 1 ⇒ x < 0 ⇒ x ∈ (−∞, 0) ⇒ (−∞, − 1) ⊆ (−∞, 0)
Test for Monotonicity.
(1) At a point : (i) Function f (x) will be monotonic increasing in domain at a point if and only if, f ' (a) > 0
(ii) Function f (x) will be monotonic decreasing in domain at a point if and only if, f ' (a) < 0 .
(2) In an interval : Function f (x), defined in [a, b] is
(i) Monotonic increasing in (a, b) if, f' (x) ≥ 0 , a< x <b
(ii) Monotonic increasing in [a, b] if, f' (x) ≥ 0 , a≤ x ≤b
(iii) Strictly increasing in [a, b], if, f' (x) > 0 , a ≤ x ≤ b
(iv) Monotonic decreasing in (a, b), if, f ' (x) ≤ 0 , a < x < b
(v) Monotonic decreasing in [a, b], if, f' (x) ≤ 0 , a ≤ x ≤ b
(vi) Strictly decreasing in [a, b], if, f' (x) < 0 , a≤ x ≤b
Example: 12 f(x) = xe x (1−x) then f(x) is [IIT Screening 2001]
(a) Increasing on −1 , 1 (b) Decreasing on R (c) Increasing on R (d) Decreasing on −1 ,1
2 2
Solution: (a) f ′(x) = e x(1− x) + x . e x(1− x) .(1 − 2x) = e x(1−x){1 + x(1 − 2x)} = e x(1−x) .(−2x2 + x + 1)
Now by the sign-scheme for − 2x 2 + x + 1
– +–
– 1/2 1
f ′(x) ≥ 0, if x ∈ − 1 ,1 , because e x(1− x) is always positive. So, f(x) is increasing on − 1 ,1 .
2 2
Example: 13 x tends 0 to π then the given function f(x) = x sin x + cos x + cos2 x is
Solution: (b)
(a) Increasing (b) Decreasing
(c) Neither increasing nor decreasing (d) None of these
f(x) = x sin x + cos x + cos2 x
∴ f ′(x) = sin x + x cos x − sin x − 2 cos x sin x = cos x(x − 2 sin x)
Hence x → 0 to π , then f′(x) ≤ 0 , i.e., f(x) is decreasing function.
Properties of Monotonic Function.
(1) If f(x) is strictly increasing function on an interval [a, b], then f −1 exists and it is also a strictly increasing
function.
(2) If f(x) is strictly increasing function on an interval [a, b] such that it is continuous, then f −1 is continuous on
[ f(a), f(b)]
(3) If f(x) is continuous on [a, b] such that f ′(c) ≥ 0(f ′(c) > 0) for each c ∈ (a,b), then f(x) is monotonically
(strictly) increasing function on [a, b].
(4) If f(x) is continuous on [a, b] such that f ′(c) ≤ 0(f ′(c) < 0) for each c ∈ (a,b) , then f(x) is monotonically
(strictly) decreasing function on [a, b]
(5) If f(x) and g(x) are monotonically (or strictly) increasing (or decreasing) functions on [a, b], then gof(x) is a
monotonically (or strictly) increasing function on [a, b]
(6) If one of the two functions f(x) and g(x) is strictly (or monotonically) increasing and other a strictly
(monotonically) decreasing, then gof(x) is strictly (monotonically) decreasing on [a, b].
Example: 14 The interval in which the function x 2e−x is non decreasing, is
(a) (−∞, 2] (b) [0, 2] (c) [2, ∞) (d) None of these
Solution: (b) Let f(x) = x 2e−x
⇒ dy = 2xe − x − x2e−x = e− x (2x − x 2)
dx
Hence f ′(x) ≥ 0 for every x ∈[0, 2] , therefore it is non-decreasing in [0, 2].
Example: 15 The function sin4 x + cos4 x increase if [IIT 1999]
Solution: (b)
(a) 0 < x < π (b) π < x < 3π (c) 3π < x < 5π (d) 5π < x < 3π
8 4 8 8 8 8 4
f(x) = sin4 x + cos4 x = (sin2 x + cos2 x)2 − 2 sin2 x cos2 x
= 1− 4 sin2 x cos2 x =1− sin2 2x =1− 1 (2 sin 2 2x)
2 2 4
= 1− 1 − cos 4x = 3 + 1 cos 4x
4 4 4
Hence function f(x) is increasing when f ′(x) > 0
f ′(x) = − sin 4 x > 0 ⇒ sin 4 x < 0
Hence π < 4x < 3π or π < x < 3π .
2 4 8
Differentiation and application of derivatives
Definition.
Let f be a real valued function defined on the closed interval [a, b] such that,
(1) f(x) is continuous in the closed interval [a, b]
(2) f(x) is differentiable in the open interval ]a,b[ and
(3) f(a) = f(b)
Then there is atleast one value c of x in open interval ]a, b[ for which f ′(c) = 0 .
Analytical Interpretation.
Now, Rolle's theorem is valid for a function such that y
(1) f(x) is continuous in the closed interval [a, b]
(2) f(x) is differentiable in open interval ]a, b[ and y=c
(3) f(a) = f(b)
So, generally two cases arises in such circumstances. Oa b x
Case I: f(x) is constant in the interval [a, b] then f ′(x) = 0 for all x ∈[a,b] . Hence, Rolle's theorem follows, and
we can say, f ′(c) = 0, where a < c < b
Case II: f(x) is not constant in the interval [a, b] and since f(a) = f(b) .
y
C decrease
Increase B
A
O x=a x=c x=b x
positive
The function should either increase or decrease when x takes values slightly greater than a.
Now, let f(x) increases for x > a
Since, f(a) = f(b) , hence the function must seize to increase at some value x = c and decreasing upto x = b .
Clearly at x = c function has maximum value.
Now let h be a small positive quantity then, from definition of maximum value of the function,
f(c + h) − f(c) < 0 and f(c − h) − f(c) < 0
∴ f (c + h) − f (c) <0 and f (c − h) − f (c) > 0
h −h
So, lim f(c + h) − f(c) ≤ 0 and lim f(c − h) − f(c) ≥ 0 .....(i)
h −h
h→0 h→0
But, if lim f (c + h) − f (c) ≠ lim f (c − h) − f (c) ,
h −h
h→0 h→0
The Rolle's theorem cannot be applicable because in such case,
RHD at x = c ≠ LHD at x = c .
Hence, f(x) is not differentiable at x = c, which contradicts the condition of Rolle's theorem.
∴ Only one possible solution arises, when lim f (c + h) − f (c) = lim f (c − h) − f (c) = 0
h −h
h→0 h→0
Which implies that, f ′(c) = 0 where a < c < b
Hence, Rolle's theorem is proved. y
AB
Increase minimum
Decreas x
O x=a x=cx=b
Similarly, the case where f(x) decreases in the interval a < x < c and then increases in the interval
c < x < b, f ′(c) = 0 . But when x = c, the minimum value of f(x) exists in the interval [a, b].
Geometrical Interpretation.
Consider the portion AB of the curve y = f(x) , lying between x = a and y
x = b , such that
C1 C2
(1) It goes continuously from A to B.
(2) It has tangent at every point between A and B and A BD
(3) Ordinate of A = ordinate of B
From figure, it is clear that f(x) increases in the interval AC1 , which implies O x=a x=c x=b x
that f ′(x) > 0 in this region and decreases in the interval C1B which implies
f ′(x) < 0 in this region. Now, since there is unique tangent to be drawn on the curve lying in between A and B and
since each of them has a unique slope i.e., unique value of f ′(x) .
∴ Due to continuity and differentiability of the function f(x) in the region A to B. There is a point x = c where
f ′(c) = 0 . Hence, f ′(c) = 0 where a < c < b
Thus Rolle's theorem is proved.
Similarly the other parts of the figure given above can be explained, establishing Rolle's theorem throughout.
Note : On Rolle's theorem generally two types of problems are formulated.
To check the applicability of Rolle's theorem to a given function on a given interval.
To verify Rolle's theorem for a given function in a given interval.
In both types of problems we first check whether f(x) satisfies the condition of Rolle's theorem or not.
The following results are very helpful in doing so.
(i) A polynomial function is everywhere continuous and differentiable.
(ii) The exponential function, sine and cosine functions are everywhere continuous and differentiable.
(iii) Logarithmic functions is continuos and differentiable in its domain.
(iv) tan x is not continuous and differentiable at x = ± π , ± 3π , ± 5π ,.......
2 2 2
(v) |x| is not differentiable at x = 0 .
(vi) If f ′(x) tends to ± ∞ as x → K , then f(x) is not differentiable at x = K .
1 1 +
2x −1 2
For example, if f(x) = (2x − 1)1 / 2 , then f ′(x) = is such that as x → ⇒ f ′(x) → ∞
So, f(x) is not differentiable at x= 1 .
2
Example: 1 The function f(x) = x(x + 3)e−1/ 2x satisfies all the condition of Rolle's theorem in [– 3, 0]. The value of c is
Solution: (c)
(a) 0 (b) 1 (c) – 2 (d) – 3
Example: 2
Solution: (a) To determine 'c' in Rolle's theorem, f ′(c) = 0
{ }Here 2)x . − 1 2)x − 1 3 1
f ′(x) = (x 2 + 3x)e −(1 / 2 + (2x + 3)e −(1 / = e −(1 / 2)x 2 (x 2 + 3x) + 2x + = − 2 e −(x / 2) x2 − x −6
∴ f ′(c) = 0 ⇒ c 2 − c − 6 = 0 ⇒ c = 3, − 2 .
But c = 3 ∉[−3,0] , Hence c = –2.
If the function f(x) = x 3 − 6x 2 + ax + b satisfies Rolle's theorem in the interval [1, 3] and f ′ 2 3+ 1 = 0 then
3
[MP PET 2002]
(a) a = 11 (b) a = −6 (c) a = 6 (d) a = 1
f(x) = x 3 − 6x 2 + ax + b ⇒ f ′(x) = 3x 2 − 12x + a
⇒ f ′(c) = 0 ⇒ f ′ 2 + 1 = 0 ⇒ 3 2 + 1 2 − 12 2 + 1 + a = 0
3 3 3
⇒ 3 4 + 1 + 4 − 12 2 + 1 + a = 0 ⇒ 12 + 1 + 4 3 − 24 − 4 3 + a = 0 ⇒ a = 11 .
3 3 3
Differentiation and application of derivatives
Definition.
If a function f(x) ,
(1) Is continuous in the closed interval [a, b] and
(2) Is differentiable in the open interval (a, b)
Then there is atleast one value c ∈(a,b) , such that; f′(c) = f(b) − f(a)
b−a
Analytical Interpretation.
First form: Consider the function, φ(x) = f(x) − f(b) − f (a) x
b− a
Since, f(x) is continuous in [a, b]
∴ φ(x) is also continuous in [a,b]
since, f ′(x) exists in (a, b) hence φ ′(x) also exists in (a,b) and φ′(x) = f ′(x) − f(b) − f(a) .....(i)
b−a
Clearly, φ(x) satisfies all the condition of Rolle's theorem
∴ There is atleast one value of c of x between a and b such that φ′(c) = 0 substituting x = c in (i) we get,
f′(c) = f(b) − f(a) which proves the theorem.
b−a
Second form: If we write b = a + h then a < c < b,c = a + θh where 0 < θ < 1
Thus, the mean value theorem can be stated as follows:
If (i) f(x) is continuous in closed interval [a, a+h]
(ii) f ′(x) exists in the open interval (a,a + h) then there exists at least one number θ (0 < θ < 1)
Such that f(a + h) = f(a) + hf ′(a + θh) .
Geometrical Interpretation.
Let f(x) be a function defined on [a, b] and let APB be the curve represented by y = f(x) . Then co-ordinates of
A and B are (a, f(a)) and (b, f(b)) respectively. Suppose the chord AB makes an angle ψ with the axis of x. Then
from the triangle ARB, we have
tanψ = BR ⇒ tanψ = f(b) − f(a) y
AR b−a B[b, f(b)]
By Lagrange's Mean value theorem, we have, f′(c) = f(b) − f(a) ∴ tanψ = f ′(c) [a, f(a)]A ψ P[c, f(c)]
b−a ψR
⇒ slope of the chord AB = slope of the tangent at (c, f(c))
OM N x
Example: 1 In the mean-value theorem f(b) − f(a) = f ′(c) , if a =0, b= 1 and f(x) = x(x − 1)(x − 2), the value of c is [MP PET 2003]
Solution: (c) b−a 2
Example: 2 (a) 1− 15 (b) 1 + 15 (c) 1− 21 (d) 1 + 21
Solution: (a) 6 6
Example: 3
Solution: (a) From mean value theorem f ′(c) = f(b) − f(a)
b−a
a = 0, f(a) = 0 ⇒ b= 1 , f (b) = 3
2 8
f ′(x) = (x − 1)(x − 2) + x(x − 2) + x(x − 1) ,
f ′(c) = (c − 1)(c − 2) + c(c − 2) + c(c − 1) = c 2 − 3c + 2 + c 2 − 2c + c 2 − c , f ′(c) = 3c 2 − 6c + 2
According to mean value theorem
f(b) − f(a) 3 − 0 3 5
b−a 8 4 4
⇒ f ′(c) = ⇒ 3c 2 − 6c + 2 = = ⇒ 3c 2 − 6c + = 0
1
2 − 0
c = 6± 36 − 15 = 6± 21 =1± 21 .
2×3 6 6
From mean value theorem f(b) − f(a) = (b − a) f ′(x1),a < x1 < b if f(x) = 1 then x1 [MP PET 1996]
x
(a) ab (b) 2ab (c) a+b (d) b−a
a+b 2 b+a
11
−1 −1 b−a 1
f ′(x1) = x12 , ∴ x12 = b−a = − ab ⇒ x1 = ab .
The abscissae of the points of the curve y = x3 in the interval [–2, 2], where the slope of the tangent can be obtained by
mean value theorem for the interval [– 2, 2] are [MP PET 1993]
(a) ± 2 (b) 3 (c) ± 3 (d) 0
3 ±2
Given that equation of curve y = x3 = f(x)
So f(2) = 8 and f(−2) = −8
Now f ′(x) = 3x 2 ⇒ f ′(x) = f(2) − f(−2) ⇒ 8 − (−8) = 3x2; ∴ x = ± 2.
2 − (−2) 4 3
Definite integral and area under curves
Definition.
Let φ(x) be the primitive or anti-derivative of a function f(x) defined on [a, b] i.e., d [φ(x)] = f(x) . Then the
dx
definite integral of f(x) over [a, b] is denoted by bf(x)dx and is defined as [φ(b) − φ(a)] i.e., bf(x)dx = φ(b) − φ(a).
∫ ∫a a
This is also called Newton Leibnitz formula.
The numbers a and b are called the limits of integration, ‘a’ is called the lower limit and ‘b’ the upper limit. The
interval [a, b] is called the interval of integration. The interval [a, b] is also known as range of integration.
Important Tips
∫• In the above definition it does not matter which anti-derivative is used to evaluate the definite integral, because if f (x)dx = φ (x) + c,
then b f (x)dx = [φ(x) + c ] ba = (φ(b) + c)− (φ(a) + c) = φ(b) − φ(a).
∫
a
In other words, to evaluate the definite integral there is no need to keep the constant of integration.
• Every definite integral has a unique value.
Example: 1 ∫3 tan−1 x + tan−1 x 2+ 1 dx = [Karnataka CET 2000]
Solution: (b) x2 +1 x
−1 (d) None of these
Example: 2
Solution: (b) (a) π (b) 2π (c) 3π [MP PET 1999]
(c) 0
∫I = 3 tan−1 x + cot −1 x x 1 dx (d) None of these
−1 x2 +1 2+
I= 3 π dx I π [ x]3−1 π [3 1] 2π .
∫⇒ −1 2 ⇒ = 2 = 2 + =
∫π 2 x dx is equal to
sin
0
(a) π (b) π/2
1 π sin 2 x dx 1 π
∫ ∫I= 2 = 2
2 [1 − cos 2x]dx
0
0
⇒ I = 1 x − sin 2x π ⇒ I = 1 [π ] = π .
2 2 0 2 2
Definite Integral as the Limit of a Sum.
Let f(x) be a single valued continuous function defined in the interval a ≤ x ≤ b, where a and b are both finite.
Let this interval be divided into n equal sub-intervals, each of width h by inserting (n – 1) points
a + h, a + 2h, a + 3h......a + (n − 1)h between a and b. Then nh = b − a .
Now, we form the sum hf(a) + hf(a + h) + hf(a + 2h) + ........ + hf(a + rh) + ...... + hf[a + (n − 1)h]
= h[ f(a) + f(a + h) + f(a + 2h) + ..... + f(a + rh) + .... + f{a + (n − 1)h}]
n−1
∑= h f(a + rh)
r=0
where, a + nh = b ⇒ nh = b − a
n−1
∑The lim h f(a + rh), if it exists is called the definite integral of f(x) with respect to x between the limits a and
h→0 r =0
b and we denote it by the symbol bf(x)dx .
∫a
bf(x)dx = lim h[ f(a) + f(a + h) + f(a + 2h) + ...... + f{a + (n − 1)h}] ⇒ b n−1
h
∫ ∫ ∑Thus, f (a rh)
f (x)dx = lim +
a h→0 a h→0 r =0
where, nh = b − a, a and b being the limits of integration.
The process of evaluating a definite integral by using the above definition is called integration from the first
principle or integration as the limit of a sum.
Important Tips
• In finding the above sum, we have taken the left end points of the subintervals. We can take the right end points of the sub-intervals
throughout.
Then we have, ∫ab f(x)dx = lim h[ f(a + h) + f(a + 2h) + ..... + f(a + nh)] , ⇒ ∫ab f(x)dx = n
h→0
h ∑ f(a + rh)
r =1
where, nh = b – a.
∫• β (x dx − x) (β > α) = π ∫ β (x −α )(β − x) dx = π (β − α )2
α α 8
−α )(β
∫• b x−a dx π (b a) b f (x ) dx 1 nb
a b−x = 2 − ∫ ∫ a = n
f (x) dx
na
(b−c c) dx b or (b+c c) dx b
fx f (x) dx fx f (x) dx
• ∫ ∫ ∫ ∫a−c a a+c a+ = − =
Some useful results for evaluation of definite integrals as limit for sums
(i) 1+2 +3+ …….+ n = n(n + 1) (ii) 12 + 22 + 32 + ........ + n2 = n(n + 1)(2n + 1)
2 6
(iii) 13 + 23 + 33 + ....... + n3 = n(n + 1) 2 (iv) a + ar + ar 2 + ........ + ar n−1 = a(r n − 1) , r ≠ 1, r >1
2 r −1
(v) a + ar + ar 2 + ............... + ar n−1 a (1 − r n ) , r ≠ 1, r <1
1 − r
=
sina + n − 1 h sinn2h
2
∑(vi) n−1
sin a + sin(a + h) + ........ + sin[a + (n − 1)h] = + nh)] =
[sin(a sin h
2
r =0
cosa + n − 1 h sinn2h
2
∑(vii) n−1
cos a + cos(a + h) + cos(a + 2h) + ..... + cos[a + (n − 1)h] = + nh)] =
[cos(a sin h
2
r =0
(viii) 1− 1 + 1 − 1 + 1 − 1 + .......... .......... ∞ = π2 (ix) 1+ 1 + 1 + 1 + 1 + 1 .......... ...... ∞ = π2
22 32 42 52 62 12 22 32 42 52 62 6
(x) 1+ 1 + 1 + .................... ∞ = π2 (xi) 1 + 1 + 1 + .......... .......... ∞ = π2
32 52 8 22 42 62 24
(xii) cosθ = e iθ + e −iθ and sinθ e iθ − e −iθ (xiii) coshθ = eθ + e −θ and sinhθ eθ − e −θ
2 = 2 2 = 2
Evaluation of Definite Integral by Substitution.
When the variable in a definite integral is changed, the substitutions in terms of new variable should be effected
at three places.
(i) In the integrand (ii) In the differential say, dx (iii) In the limits
For example, if we putφ(x) = t in the integral bf{φ(x)}φ' (x)dx , then b f{φ(x)}φ' (x)dx = φ(b)f(t)dt .
∫ ∫ ∫a a φ(a)
Important Tips
π dx π / 2 dx
0 1 + sin x 0 sin x + cos x =
=2 ∫ ( )
∫• 2 log 2 +1
∫• π / 2 log (tan x)dx = 0 ∫a dx = a , where f (a − x) = − f (x)
0 0 1 + e f(x) 2
∫• a dx = π ∫a dx = π ∫ a a2 − x2 dx = πa2
0 a2 − x2 2 0 x2 + a2 2a 0 4
Example: 3 ∫If h(a) = h(b), then b f (g[h(x)])]−1 f ' (g[h(x)]) g' [h(x)] h' (x) dx is equal to [MP PET 2001]
Solution: (a)
Example: 4 [ (d) None of these
Solution: (b) a
Example: 5 (a) 0 (b) f(a) − f(b) (c) f[g(a)] − f[g(b)]
Solution: (b)
Put f(g[h(x)]) = t ⇒ f '(g[h(x)]) g'[h(x)]h'(x)dx = dt
f (g[h(b)]) −1dt f (g[h(b)])
f (g[h(a)])
t
∫ [ ]∴
f (g[h(a)]) = log(t) = 0 [ h(a) = h(b)]
∫The value of the integral e2 log e x dx is [IIT 2000]
e −1 x
(a) 3/2 (b) 5/2 (c) 3 (d) 5
Put log e x = t ⇒ e t = x
∴ dx = e t dt
and limits are adjusted as –1 to 2
I= 2 t etdt 2 I= 0 2 I − t2 0 t2 2 I = 5/2
−1 et 2 2
|t |dt ⇒ − tdt + tdt −1 0
−1 −1 0
∫ ∫ ∫ ∫∴ = ⇒ = + ⇒
∫π / 2 dx equals [MNR 1983; Rajasthan PET 1990; Kurukshetra CEE 1997]
0 1 + sin x
(a) 0 (b) 1 (c) –1 (d) 2
∫I = π/ / 2 dx
0 sin 2 x/2 + cos 2 x/2 + 2 sin x/2 cos x/2
π / 2 dx π /2 sec 2 x/2 dx
0 (sin x/2 + cos x/2)2 0 (1 + tan x/2)2
∫ ∫I = =
Put (1 + tan x/2) = t ⇒ 1 sec 2 x/2 dx = dt
2
I=2 2 dt −21t 2 −2 1 1 =1
∫∴ 1 t2 = 1 = 2 − 1
Properties of Definite Integral.
∫ ∫(1) bf(x)dx = bf(t)dt i.e., The value of a definite integral remains unchanged if its variable is replaced by any
aa
other symbol.
Example: 6 ∫6 x 1 1 dx is equal to
Solution: (c) 3 +
(a) [log(x + 1)] 6 (b) [log(t + 1]63 (c) Both (a) and (b) (d) None of these
3
∫I = 6 1 dx = [log(x + 1)]63 , ∫I = 6 t 1 dt = [log(t + 1)]63
3 x +1 3 +1
∫ ∫(2) b f(x)dx = − af(x)dx i.e., by the interchange in the limits of definite integral, the sign of the integral is
ab
changed.
Example: 7 10 [MP PET 2000]
Solution: (d)
Suppose f is such that f(–x) = –f(x) for every real x and f(x) dx = 5 , then f(t) dt =
∫ ∫0 −1
(a) 10 (b) 5 (c) 0 (d) –5
1
∫Given, f(x)dx = 5
0
Put x = −t ⇒ dx = −dt
−1 0
∫ ∫∴ I = − f(−t)dt = − f(t)dt ⇒ I = −5
0 −1
∫ ∫ ∫(3) b f(x)dx = cf(x)dx + bf(x)dx , (where a < c < b)
a ac
bf(x)dx = c1 f(x)dx + c2 f(x)dx + ..... + b f(x)dx; (where
a a c1 cn
∫ ∫ ∫ ∫or a < c1 < c2 < ........cn < b)
Generally this property is used when the integrand has two or more rules in the integration interval.
Important Tips
∫• b x−a + x−b ) dx = (b − a)2
(
a
Note : Property (3) is useful when f (x) is not continuous in [a, b] because we can break up the integral
into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.
The expression for f (x) changes at the end points of each of the sub-interval. You might at times be
puzzled about the end points as limits of integration. You may not be sure for x = 0 as the limit of the first
integral or the next one. In fact, it is immaterial, as the area of the line is always zero. Therefore, even if you
∫write 0 (1 − 2x) dx, whereas 0 is not included in its domain it is deemed to be understood that you are
−1
approaching x = 0 from the left in the first integral and from right in the second integral. Similarly for x = 1.
Example: 8 ∫2 − x 2 | dx is equal to [IIT 1989; BIT Ranchi 1996; Kurukshetra CEE 1998]
Solution: (b) |1
−2
Example: 9
Solution: (b) (a) 2 (b) 4 (c) 6 (d) 8
Example: 10
Solution: (c) 2 |1 − x 2 | dx = −1 x 2 | dx 1 x 2 2 x 2 | dx
∫ ∫ ∫ ∫I = − + − | dx + −
|1 |1 |1
−2 −2 −1 1
I =− −1 x 2 )dx 1 (1 − x2)dx − 2 x 2)dx I 4 4 4 4 .
3 3 3
(1 −1 (1
∫ ∫ ∫⇒ −2 − + 1 − ⇒ = + + =
∫1.5 2 ]dx , where [.] denotes the greatest integer function, equals [DCE 2000, 2001; IIT 1988; AMU 1998]
[x
0
(a) 2 + 2 (b) 2 − 2 (c) 1 + 2 (d) 2 − 1
1.5 2 ]dx 1 2 2 [x 2 ]dx + 1.5 2 I =0+ 2 1.5
[x [x [x 1dx + 2dx = 2 − 1 + 3 − 2 2 ⇒ I = 2 − 2
∫ ∫ ∫ ∫ ∫ ∫I = = ]dx + ]dx ⇒
0 01 2 12
ecos x. sin x, | x |≤ 2 3
2, otherwise
∫If f(x) = , then f(x)dx = [IIT 2000]
−2
(a) 0 (b) 1 (c) 2 (d) 3
| x|≤ 2 ⇒ −2 ≤ x ≤ 2 and f(x) = e cos x sin x is an odd function.
3 2 3 3 = [2x]32 2 a
f(x)dx = f(x)dx + f(x)dx ⇒ I = 0 + 2dx ∫[ f(x) = 0 if f(x) is odd and in (2, 3) f(x) is 2]
∫ ∫ ∫ ∫∴ I = 2 = −a
−2 −2 2
∫ ∫(4) a f(x)dx = af(a − x)dx : This property can be used only when lower limit is zero. It is generally used for
00
those complicated integrals whose denominators are unchanged when x is replaced by (a – x).
Following integrals can be obtained with the help of above property.
π /2 sinn x π /2 cos n x π
0 x + cosn 0 cos n x + sinn 4
∫ ∫(i) dx = dx =
sin n x x
π /2 tann x π /2 cot n x π π / 2 1 π/ 2 1 π
0 + tann 0 + cot n 4 ∫ ∫(iii) 0 tann 0 cotn 4
∫ ∫(ii)1 x dx = 1 x dx = 1 + x dx = 1 + xdx =
π/ 2 secn x π /2 cosecn x π
0 x + cosecnx 0 cosecn x + secn 4
∫ ∫(iv) dx = dx =
sec n x
∫ ∫(v) π/2 f(sin 2x) sin xdx = π / 2 f(sin 2x) cos xdx ∫ ∫(vi) π / 2 f(sin x)dx = π / 2 f(cos x)dx
00 00
∫ ∫(vii) π / 2 f(tan x)dx = π / 2 f(cot x)dx ∫ ∫(viii) 1 f(log x)dx = 1 f[log(1 − x)]dx
00 00
∫ ∫(ix) π / 2 log tan xdx = π / 2 log cot xdx ∫(x) π /4 log(1 + tan x)dx = π log 2
00 0 8
∫ ∫(xi) π / 2 −π π 1
0 2 2 2
log sin xdx = π / 2 log cos xdx = log 2 = log
0
π /2 a sin x + b cos x π /2 a sec x + b cosec x π /2 a tan x + b cot x π
0 sin x + cos x 0 sec x + cosec x 0 tan x + cot x 4
∫ ∫ ∫(xii) dx = dx = dx = (a + b)
Example: 11 π
Solution: (b) ∫ e sin2 x cos 3 xdx =
Example: 12 0
Solution: (a)
OR
Example: 13
Solution: (a) ∫For any integer n, π e sin2 x cos 3 (2n + 1)xdx = [MP PET 2002; MNR 1992, 98]
0
(d) None of these
(a) –1 (b) 0 (c) 1
Let, f1(x) = cos 3 x = − f(π − x)
and f2 (x) = cos 3 (2n + 1)x = − f(π − x)
∴ I = 0.
∫2a f(x) x) dx is equal to [IIT 1988; Kurukshetra CEE 1999; Karnataka CET 2000]
0 f(x) + f(2a −
(a) a (b) a/2 (c) 2a (d) 0
2a f(x) x) dx 2a f(2a − x)
0 f(x) + f(2a 0 (2a − x) + f(x)
∫ ∫I = − = f dx
2a f(x) + f (2a − x) 2a = [x]02a 2a
0 f(x) + f (2a − x)
∫ ∫2I = dx = dx =
0
∴I =a.
∫ ∫π / 2 tan x π /2 sin x dx is equal to [MNR 1989, 90; Rajasthan PET 1991, 95]
= (d) 1
0 1 + tan x 0 sin x + cos x
(a) π/4 (b) ∞ (c) –1
∫We know, π / 2 tann xdx = π for any value of n
0 1 + tann x 4
∴ I =π /4.
∫ ∫(5) a f(x) dx = a f(x) + f(−x) dx .
−a 0
∫ ∫In special case : a 2 a f(x)dx, if f(x) is even function or f(−x) = f(x)
−a f (x) dx = 0 0 ,if f(x)is odd is odd function or f(−x) = f(x)
−
This property is generally used when integrand is either even or odd function of x.
Example: 14 ∫The integral 1/ 2 [x] + ln 1 + x dx equal to
−1 / 2 1 − x
Solution: (a)
Example: 15 (a) −1 (b) 0 (c) 1 (d) 2 ln 1
Solution: (a) 2 2
log 1 + x is an odd function of x as f(−x) = − f(x)
1 − x
1/ 2 0 1/ 2 0 − [x]0−1 / 2 −1 .
2
[x]dx + 0 ⇒ [x]dx + [x]dx ⇒ I = − 1dx + 0
−1 / 2 −1 / 2 0 −1 / 2
∫ ∫ ∫ ∫I = I= ⇒ =
∫The value of the integral 1 log x + x 2 + 1dx is [MP PET 2001]
−1
(a) 0 (b) log 2 (c) log 1/2 (d) None of these
f(x) = log[x + x2 + 1] is a odd function i.e. f(−x) = − f(x) ⇒ f(x) + f(−x) = 0 ⇒ I = 0.
Example: 16 ∫The value of π − x 2 ) sin x cos 2 xdx is [Roorkee 1992; MNR 1998]
Solution: (a)
(1 (d) 3
−π
(a) 0 (b) 1 (c) 2
Let, f1(x) = (1 − x 2 ) , f2(x) = sin x and f3 (x) = cos 2 x
Now, f1(x) = f1(−x) , f2(x) = − f2(−x) and f3(x) = f3(−x)
ππ π
∫ ∫ ∫∴ I = f(x)dx = [ f1(x).f2 (x).f3 (x)]dx = − [ f1(−x).f2 (−x).f3 (−x)]dx
−π −π −π
∴ I=0
∫ ∫ ∫(6) 2af(x)dx = a f(x)dx + a f(2a − x) dx
00 0
∫ ∫In particular, 2a 0 , if f(2a − x) = − f(x)
0 f (x) dx = 2 a f(x)dx , if f(2a − x) = f(x)
0
It is generally used to make half the upper limit.
Example: 17 ∫If n is any integer, then π ecos2 x cos3(2n + 1)x dx is equal to [IIT 1985; Rajasthan PET 1995]
Solution: (c) 0
(d) None of these
Example: 18 (a) x (b) 1 (c) 0
Solution: (c) (d) I1 = 4I 2
∫I = π ecos2(π −x). cos3(2n + 1)(π − x) dx
0
∫⇒ I = − π e cos2 x . cos 3 (2n + 1)x dx ⇒ I = −I
0
⇒ 2I = 0 ⇒ I = 0.
3π 2 and π 2 then
f (cos f
∫ ∫If I1 =
x) dx I2 = (cos x) dx
00
(a) I1 = I 2 (b) I1 = 2I 2 (c) I1 = 3I 2
f(cos 2 x) = f(cos 2 (3π − x))
I1 =3 π (cos 2 x) dx I1 = 3I2
∫∴ ⇒
f
0
∫ ∫(7) bf(x)dx = bf(a + b − x)dx
aa
∫Note : b f (x) dx − x) = 1 (b − a)
a f(x) + f (a + b 2
Example: 19 ∫3π / 4 dx is equal to [IIT 1999]
Solution: (a)
π / 4 1 + cos x
Example: 20
(a) 2 (b) –2 (c) 1/2 (d) –1/2
∫I = 3π / 4 1 − 1 x dx [ cos π + 3π − x = − cos x
π /4 cos 4 4
2I = 3π / 4 2 dx
∫∴ π /4 cos2
1− x
∫⇒ 2I = 2 3π / 4 cosec2xdx ⇒ 2I = −2[cot x]π3π/4/4 = 4 ⇒ I = 2.
π /4
3 x dx is [IIT 1994; Kurukshetra CEE 1998]
5−x + x
∫The value of (d) 1/2
2 (b) 0 (c) –1
(a) 1
Solution: (d) ∫3 x dx
2 5−x + x
Example: 21 I=
Solution: (b)
Put x = 2 + 3 − t ⇒ dx = −dt
2 5−t 3 5−x dx and 2I = 3 x+ 5 − x dx = 3
∴ I= (−dt) = 1 dx
∫ ∫ ∫ ∫3 5 − t + t
2 5−x + x 2 5−x + x 2
⇒ 2I = [x]22 = 1 ⇒ I = 1 / 2
b [Kurukshetra CEE 1993; AIEEE 2003]
∫If f(a + b − x) = f(x) then x f(x)dx is equal to
a
∫(a)a+b b ∫(b) a+b b ∫(c) b−a b (d) None of these
2 2 2
f(b − x)dx f (x)dx f (x)dx
a a a
bb
∫ ∫I = x f(x)dx and I = (a + b − x)f(a + b − x)dx
aa
b b b 2I b (x)dx (a b) a +b b
2
(a + b − x)f(x)dx ⇒ (a + b)f(x)dx − x f(x)dx f f (x)dx
a a a a a
∫ ∫ ∫ ∫ ∫⇒ I = I= ⇒ = + ⇒ I =
∫ ∫(8)a 1 a f (x)dx if f(a − x) = f(x)
0 x f(x)dx = 2 a 0
Example: 22 π π
Solution: (b)
x f(sin x)dx = k
∫ ∫If f(sin x)dx , then the value of k will be [IIT 1982]
00
(a) π (b) π/2 (c) π/4 (d) 1
ππ
∫ ∫Given, x f(sin x)dx = k f(sin x)dx
00
π π ππ π
⇒ (π − x)f(sin(π − x))dx = k f(sin(π − x))dx ⇒ π f(sin x)dx − x f(sin x)dx = k f(sin x)dx
∫ ∫ ∫ ∫ ∫0 0 0
00
ππ π
∫ ∫ ∫⇒ π f(sin x)dx − 2k f(sin x)dx = 0 ⇒ (π − 2k) f(sin x)dx = 0
00 0
∴ π − 2k = 0 ⇒ k = π /2 .
(9) If f(x) is a periodic function with period T, then nT f(x)dx = n T f(x)dx ,
∫ ∫0 0
Deduction : If f(x) is a periodic function with period T and a ∈ R + , then a+nT f(x)dx = a f(x)dx
∫ ∫nT 0
(10) (i) If f(x) is a periodic function with period T, then
∫ ∫a+nT f(x) dx = n T f(x) dx where n ∈ I
a0
(a) In particular, if a = 0
∫ ∫nT f(x) dx = n T f(x) dx, where n ∈ I
00
∫ ∫(b) If n = 1, a+T f(x) dx = T f(x) dx,
00
∫ ∫(i) nT f(x) dx = (n − m) T f(x) dx, where n, m ∈ I
mT 0
∫ ∫(ii) b+nT f(x) dx = b f(x) dx, a where n ε I
a + nT
(11) If f(x) is a periodic function with period T, then a+T f(x) is independent of a.
∫a
∫ ∫(12) b f(x) dx = (b − a) 1 f ((b − a) x + a) dx
a0
∫(13) If f(t) is an odd function, then φ (x) = x f(t) dt is an even function
a
∫(14) If f(x) is an even function, then φ (x) = x f(t) dt is on odd function.
0
∫Note : If f(t) is an even function, then for a non zero ‘a’, x f(t)dt is not necessarily an odd function. It will
0
∫be odd function if af(t) dt = 0
0
Example: 23 ∫For 2π x sin 2n x is equal to
Solution: (a) n > 0, 0 sin 2n x + cos 2n dx [IIT 1996]
x
Example: 24
Solution: (c) (a) π 2 (b) 2π 2 (c) 3π 2 (d) 4π 2
2π x sin2n xdx 2π (2π − x) sin2n(2π − x)dx a a
0 sin2n x + cos2n x 0 sin2n(2π − x) + cos2n(2π − x)
f(x) = f (a
0 0
and I = ∫ ∫ x)
∫ ∫I = −
2I = 2π 2π sin2n π dx I =π 2π sin2n x dx
0 sin2n x + cos2n 0 sin2n x + cos2n
∫ ∫∴ x ⇒ x
nT T
∫ ∫using f(x) = n f(x)dx
00
I = 4π π /2 sin2n x dx I = 4π (π /4) = π 2 .
∫∴ 0 sin2n x + cos 2n x ⇒
a+T
If f(x) is a continuous periodic function with period T, then the integral I = f(x)dx is
∫a
(a) Equal to 2a (b) Equal to 3a (c) Independent of a (d) None of these
a+T 0 T a+T
Consider the function g(a) = f(x)dx = f(x)dx + f(x)dx + f(x)dx
∫ ∫ ∫ ∫a a 0 T
Putting x − T = y in last integral, we get a+T aa
f(x)dx = f(y + T)dy = f(y)dy
∫ ∫ ∫T 0 0
0T a T
∫ ∫ ∫ ∫⇒ g(a) = f(x)dx + f(x)dx + f(x)dx = f(x)dx
a0 0 0
Hence g(a) is independent of a.
Important Tips
• Every continuous function defined on [a, b] is integrable over [a, b].
• Every monotonic function defined on [a, b] is integrable over [a, b].
b
• If f(x) is a continuous function defined on [a, b], then there exists c ∈ (a, b) such that f(x)dx = f(c).(b − a) .
∫a
∫The number f (c) = (b 1 a) b is called the mean value of the function f(x) on the interval [a, b].
−
f (x)dx
a
x
∫• If f is continuous on [a, b], then the integral function g defined by g(x) = f(t)dt for x ∈[a, b] is derivable on [a, b] and g′(x) = f(x)
a
for all x ∈ [a, b] .
b
• If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then m(b − a) ≤ f(x)dx ≤ M(b − a)
∫a
• If the function ϕ(x) and ψ (x), are defined on [a, b] and differentiable at a point xε(a,b) and f(t) is continuous for φ (a) ≤ t ≤ ψ (b) ,
∫then ψ (x) = f(ψ (x))ψ ' (x) − f(ϕ(x))ϕ' (x)
f (t)dt
ϕ( x )
bb
∫ ∫f(x)dx ≤ | f(x)| dx
aa
b ≤ b 1 / 2 b 1 / 2
If f 2 (x) and g 2(x) are integrable on [a, b], then f(x) g(x)dx f 2 (x) dx g 2 (x) dx
a a a
∫ ∫ ∫•
• Change of variables : If the function f(x) is continuous on [a, b] and the function x = ϕ(t) is continuously differentiable on the
b t2f(ϕ(t))ϕ' (t)dt .
f(x)dx =
∫ ∫interval [t1, t2] and a = ϕ(t1),b = ϕ(t2), then
a t1
b
• Let a function f(x,α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d . Then for any α ∈[c,d] , if I(α) = f(x,α)dx, then
∫a
b
∫I'(α) = f'(x,α)dx,
a
Where I'(α) is the derivative of I(α) w.r.t. α and f '(x,α) is the derivative of f(x,α) w.r.t. α, keeping x constant.
• For a given function f(x) continuous on [a, b] if you are able to find two continuous function f1(x) and f2 (x) on [a, b] such that
b bb
∫ ∫ ∫f1(x) ≤ f(x) ≤ f2(x) ∀ x ∈ [a,b], then a f1(x)dx ≤ a f(x)dx ≤ a f2(x)dx
Summation of Series by Integration.
∫ ∑We know that bf(x)dx = lim h n f(a + rh), where nh = b − a
a n→∞ r =1
a = 0, b = 1, or 1 . Hence 1f (x)dx 1 r
n n n
0
∫ ∑Now, put
∴ nh = 1 h= = lim f
n→∞
Note : Express the given series in the form 1 f r . Replace r by x, 1 by dx and the limit of the sum is
∑ h n
n
n
∫1 f(x)dx .
0
Example: 25 If Sn = 1 n + 1 + ........ + 1 then lim Sn is equal to [Roorkee 2000]
Solution: (b) 1+ 2 + 2n n + n2
n→∞
Example: 26
(a) log 2 (b) 2 log 2 (c) 3 log 2 (d) 4 log 2
∑ ∑lim 1 = lim 1
n→∞ r + rn n→∞ n r r
n
+ n
∴ lim Sn = ∫10 1 dx
x (1 + x)
n→∞
= 2[log(1 + x )]10 = 2log 2
(n! )1 / n lim n! 1 / n
n n→∞ nn
lim or is equal to [WB JEE 1989; Kurukshetra CEE 1998]
n→∞
(a) e (b) e–1 (c) 1 (d) None of these
Solution: (b) Let A lim (n! )1 / n
n
= n→∞
log 1.2.3......n 1/n log 1 2 3 n 1/n 1 n log r
nn n n n n n r =1 n
∑⇒ log A = lim ⇒ log A = lim . . .... ⇒ log A = lim
n→∞ n→∞ n→∞
∫⇒ log A = 1 = [x log x − x]10 ⇒ log A = −1 ⇒ A = e−1
log xdx
0
Gamma Function.
Γ m + 1 Γ n + 1
2 2
∫If m and n are non-negative integers, then π / 2 sin m x cos n xdx =
0
2Γ m + n + 2
2
where Γ(n) is called gamma function which satisfied the following properties
Γ(n + 1) = nΓ(n) = n! i.e. Γ (1) = 1 and Γ(1 / 2) = π
In place of gamma function, we can also use the following formula :
∫π/ 2 sin m x cos n xdx = (m − 1)(m − 3).....(2 or 1)(n − 1)(n − 3).....(2 or 1)
0 (m + n)(m + n − 2)....(2 or 1)
It is important to note that we multiply by (π/2); when both m and n are even.
Example: 27 ∫The value of π / 2 sin 4 x cos 6 xdx = [Rajasthan PET 1999]
Solution: (c) 0
(d) 5π / 312
(a) 3π /312 (b) 5π / 512 (c) 3π / 512
I = (4 + (4 − 1).(4 − 3).(6 − 1).(6 − 3).(6 − 5) 6 − 8) . π = 3.1.5.3.1 . π = 3π
6)(4 + 6 − 2)(4 + 6 − 4)(4 + 6 − 6)(4 + 2 10.8.6.4.2. 2 512
Reduction formulae Definite Integration.
∫(1) ∞ e −ax sin bxdx = a2 b ∫(2) ∞ e −ax cos bxdx = a2 a b2 ∫(3) ∞ e −ax x ndx = n!
+ b2 + an +1
0 0 0
Example: 28 ∞ x n−1dx, ∞ −λx n−1dx is equal to
Solution: (c) ∫ ∫If In = x then x
e− e
00
(a) λIn (b) 1 In (c) In (d) λn In
λ λn
Put, λx = t, λdx = dt, we get,
∫ ∫ ∫∞ −λx 1 ∞ −t n−1dt 1 ∞ x n−1dx In
e λn λn λn
xn−1dx = e t = e − x =
0 0 0
Walli's Formula.
n−1 . n − 3 . n− 5 ...... 2 , when n is odd
. n − 2 . 4 3 when n is even
∫ ∫π / 2 sinn xdx = π / 2 cosn xdx = n n − 3 n − 5
0 0 n − n − 2 n − 4
n 1 ....... 3 . 1 . π ,
4 2 2
n−
∫0π / 2 sin m x cos n dx = (m − 1)(m − 3)..........(n − 1)(n − 3)........ [If m, n are both odd +ve integers or one odd +ve integer]
(m + n)(m + n − 2)
= (m − 1) (m − 3)............(n − 1)(n − 3) . π [If m, n are both +ve integers]
(m + n)(m + n − 2) 2
Example: 29 ∫π / 2 sin7 xdx has value [BIT Ranchi 1999]
Solution: (c) 0
(a) 37 (b) 17 (c) 16 (d) 16
184 45 35 45
Using Walli’s formula, ⇒ I = 7 − 1 . 7 − 3 . 7 − 5 = 6.4.2 = 16
7 7 − 2 7 − 4 7.5.3 35
Leibnitz’s Rule.
(1) If f(x) is continuous and u(x), v(x) are differentiable functions in the interval [a, b], then,
∫d v(x) f(t)dt = f {v(x)} d {v(x)} − f {u(x)} d {u(x)}
dx dx
dx u( x )
(2) If the function φ (x) and ψ (x) are defined on [a,b] and differentiable at a point x ∈(a, b), and f(x, t) is
d ψ(x) ψ (x) d d ψ (x) dφ (x)
dx φ(x) φ(x) dx
∫ ∫continuous, then, f (x, t) dt = f(x, t) dt + dx f(x,ψ (x)) − dx f(x, φ(x))
Example: 30 x 2 − t2 dt . Then the real roots of the equation x2 − f ' (x) = 0 are [IIT 2002]
Solution: (a)
Example: 31 ∫Let f(x) = (b) ± 1 (c) ± 1 (d) 0 and 1
Solution: (c) 1 2 2
(a) ±1
∫f(x) = x 2 − t2 dt ⇒ f ' (x) = 2 − x2 .1 − 2 − 1.0 = 2 − x2
1
∴ x2 = f ' (x) = 2 − x2 ⇒ x 4 + x 2 − 2 = 0 ⇒ (x2 + 2)(x2 − 1) = 0
∴ x = ±1 (only real).
∫Let f : (0, ∞) → R and f(x) = x f(x2) = x2(1 + x), then f (4) equals [IIT 2001]
f(t)dt . If
0
(a) 5/4 (b) 7 (c) 4 (d) 2
∫By definition of f(x) we have f(x2) = x2 = x2 + x3 (given)
f (t)dt
0
Differentiate both sides, f(x2).2x + 0 = 2x + 3x2
Put, x = 2 ⇒ 4 f(4) = 16 ⇒ f(4) = 4
Integrals with Infinite Limits (Improper Integral).
∫A definite integral bf(x)dx is called an improper integral, if
a
The range of integration is finite and the integrand is unbounded and/or the range of integration is infinite and
the integrand is bounded.
∫e.g., The integral 1 1 dx is an improper integral, because the integrand is unbounded on [0, 1]. Infact,
0 x2
∫1 as ∞ 1 is an improper integral, because the range of integration is not finite.
0 +x
x2
→∞ x → 0 . The integral 1 2 dx
There are following two kinds of improper definite integrals:
∫(1) Improper integral of first kind : A definite integral bf(x)dx is called an improper integral of first kind if
a
the range of integration is not finite (i.e., either a → ∞ or b → ∞ or a → ∞ and b → ∞ ) and the integrand f(x) is
bounded on [a, b].
1 ∞ 1 ∞1 ∞ 3x
x2 0 +x −∞ 1 + x 2 1 (1 + 2x)3
∫ ∫ ∫ ∫∞ dx, 1 2 dx, dx, dx are improper integrals of first kind.
1
Important Tips
• In an improper integral of first kind, the interval of integration is one of the following types [a, ∞), (–∞, b], (–∞, ∞).
The improper integral ∞ is said to be convergent, if lim k
f (x)dx f(x)dx exists finitely and this limit is called the value of the improper
∫ ∫•
a k→∞ a
k
∫integral. If lim f(x)dx is either +∞ or –∞, then the integral is said to be divergent.
k→ ∞ a
∞
∫• The improper integral f(x)dx is said to be convergent, if both the limits on the right-hand side exist finitely and are independent of
−∞
∞
∫each other. The improper integral f(x)dx is said to be divergent if the right hand side is +∞ or –∞
−∞
∫(2) Improper integral of second kind : A definite integral bf(x)dx is called an improper integral of second
a
kind if the range of integration [a, b] is finite and the integrand is unbounded at one or more points of [a, b].
∫If bf(x)dx is an improper integral of second kind, then a, b are finite real numbers and there exists at least one
a
point c ∈ [a, b] such that f(x) → +∞ or f(x) → −∞ as x → c i.e., f(x) has at least one point of finite discontinuity
in [a, b].
For example :
∫(i) The integral 3 1 2 dx, is an improper integral of second kind, because lim x 1 2 = ∞.
1 x− x → 2 −
∫(ii) The integral 1log xdx; is an improper integral of second kind, because log x → ∞ as x → 0 .
0
∫(iii) The integral 2π 1 + 1 x dx, is an improper integral of second kind since integrand 1 becomes
0 cos 1 + cos x
infinite at x = π ∈[0, 2π ].
∫(iv)1 sin x dx, is a proper integral since lim sin x = 1.
0 x x
x→0
Important Tips
• Let f(x) be bounded function defined on (a, b] such that a is the only point of infinite discontinuity of f(x) i.e., f(x) → ∞ as x → a. Then
b bb
f(x)dx and is defined as f(x)dx . Provided that the limit on
a a+ε
∫ ∫ ∫the improper integral of f(x) on (a, b] is denoted by
a f(x)dx = εl→im0
b
right hand side exists. If l denotes the limit on right hand side, then the improper integral f(x)dx is said to converge to l, when l is
∫a
finite. If l = + ∞ or l = –∞, then the integral is said to be a divergent integral.
• Let f(x) be bounded function defined on [a, b) such that b is the only point of infinite discontinuity of f(x) i.e. f(x) → ∞ as x → b. Then
b b b−ε
the improper integral of f(x) on [a, b) is denoted by f(x)dx and is defined as f(x)dx = lim f (x)dx
∫ ∫ ∫a a ε → 0 a
b
Provided that the limit on right hand side exists finitely. If l denotes the limit on right hand side, then the improper integral f(x)dx is
∫a
said to converge to l, when l is finite.
If l = +∞ or l = −∞ , then the integral is said to be a divergent integral.
• Let f(x) be a bounded function defined on (a, b) such that a and b are only two points of infinite discontinuity of f(x) i.e., f(a) → ∞,
f(b) → ∞.
b
Then the improper integral of f(x) on (a, b) is denoted by f(x)dx and is defined as
∫a
bc f(x)dx + lim b−δ f(x)dx, a < c < b
∫ ∫ ∫f(x)dx = lim
a ε → 0 a+ε δ →0 a
Provided that both the limits on right hand side exist.
• Let f(x) be a bounded function defined [a, b]-{c}, c∈[a, b] and c is the only point of infinite discontinuity of f(x) i.e. f(c)→∞. Then the
b b c−x b
improper integral of f(x) on [a, b] – {c} is denoted by f(x)dx and is defined as f(x)dx = lim f(x)dx + lim f (x)dx
∫ ∫ ∫ ∫a
a x→0 a δ → 0 c+δ
b
Provided that both the limits on right hand side exist finitely. The improper integral f(x)dx is said to be convergent if both the limits
∫a
on the right hand side exist finitely.
• If either of the two or both the limits on RHS are ±∞, then the integral is said to be divergent.
Example: 32 ∫The improper integral ∞ − x dx is …… and the value is….
Solution: (a)
Example: 33 e
0
Solution: (b)
(a) Convergent, 1 (b) Divergent, 1 (c) Convergent, 0 (d) Divergent, 0
Example: 34
Solution: (d) ∞ e−xdx = lim k xdx x ]k0 lim [−e−k e0] I = lim (1 − e−k ) = 1 − 0 = 1 [ lim e−k = e−∞ = 0]
0 k→∞ e k→∞ k→∞ k→∞
∫ ∫I = − ⇒ I = lim [−e − = + ⇒
0 k→∞
∫Thus, lim k e−xdx exists and is finite. Hence the given integral is convergent.
k→∞ 0
∫The integral 0 1 dx, a ≠ 0 is
−∞ a2 + x2
(a) Convergent and equal to π (b) Convergent and equal to π
a 2a
(c) Divergent and equal to π (d) Divergent and equal to π
a 2a
0 dx 0 dx
-∞ a2 + x2 k a2 + x2
∫ ∫I =
= lim
k→−-∞
⇒ I = lim 1 tan−1 x 0 = lim 1 tan−1 0 − 1 tan−1 k ⇒ I = 0 − 1 tan −1(−∞) = − 1 −π = π
a a k a a a a a 2 2a
k →−∞ k →−∞
Hence integral is convergent.
∫The integral ∞1 dx is
−∞ e−x + ex
(a) Convergent and equal to π/6 (b) Convergent and equal to π/4
(d) Convergent and equal to π/2
(c) Convergent and equal to π/3
∞ 1 dx ∞ ex dx
−∞ e−x + ex −∞ 1 + e2x
∫ ∫I = =
Put ex = t ⇒ exdx = dt
∫∴ I = ∞ 1 1 dt ⇒ I = [tan−1 t]∞0 = [tan−1 ∞ − tan−1 0] ⇒ I = π /2 , which is finite so convergent.
0 + t2
Example: 35 ∫2 x + 1 dx is
Solution: (a) 1 x −1
Example: 36
(a) Convergent and equal to 14 (b) Divergent and equal to 3
Solution: (c) 3 14
(c) Convergent and equal to ∞ (d) Divergent and equal to ∞
∫ ∫2 2 2 dx = 2 (x − 1)3 / 2 2 + [4 x − 1]12 = 14/3 which is finite so convergent.
3 1
I = x − 1dx +
1 1 x −1
∫2 x2 dx dx is
1 − 5x + 4
(a) Convergent and equal to 1 log 2 (b) Convergent and equal to 3/log2
3
(c) Divergent (d) None of these
∫ ∫I =2 (x − dx − 4) = 1 2 x 1 4 − x 1 1 dx = 1 [log 2 − ∞] = −∞
1 1)(x 3 1 − − 3
So the given integral is not convergent.
Some Important results of Definite Integral.
In =∫(1) If π/4 tan n xdx then I n + I n−2 = 1
n−1
0
∫(2) If In = π / 4 cot n xdx then In + I n−2 = 1
1−n
0
In =∫(3) If π/4 sec n xdx then In = ( 2)n−2 + n − 2 I n−2
n−1 n − 1
0
In =∫(4) If π / 4 cosec n xdx then In = ( 2 )n − 2 + n−2 In−2
n−1 n−1
0
∫(5) If In = π / 2 x n sin xdx then In + n(n − 1)In−2 = n(π / 2)n−1
0
In =∫(6) If π / 2 x n cos xdx then In + n(n − 1)In−2 = (π /2)n
0
∫(7) If a > b > 0, then π / 2 dx 2 tan−1 a+b
0 a + b cos x = a2 − b2 a−b
∫(8) If 0 < a < b then π / 2 dx 1 log b+a − b−a
0 a + b cos x = b2 − a2 b+a + b−a
∫(9) If a > b > 0 then π / 2 dx 2 tan −1 a−b
0 a + b sin x = a2 − b2 a+b
∫(10) If 0 < a < b , then π / 2 dx 1 log b+a + b−a
0 a + b sin x = b2 − a2 b+a − b−a
∫(11) If a > b, a 2 > b 2 + c 2 , then π / 2 dx 2 tan−1 a−b+c
0 a + b cos x + c sin x = a2 − b2 − c2 a2 − b2 − c2
∫(12) If a > b, a 2 < b 2 + c 2 , then π / 2 dx 1 log a − b + c − b2 + c2 − a2
0 a + b cos x + c sin x = b2 + c2 − a2 a − b + c + b2 + c2 − a2
a 2 < b 2 + c 2 then π / 2 dx − 1 log b − a − c − b2 + c2 − a2
∫(13) If a < b, 0 a + b cos x + c sin x = b2 + c2 − a2 b − a − c + b2 + c2 − a2
Important Tips
x
∫ f(x)dx
• lim 0 = f(0)
x→0 x
b 1 f[(b − a) t + a]dt
f(x)dx = (b − a)
∫ ∫•
a0
Integration of Piecewise Continuous Functions.
Any function f(x) which is discontinuous at finite number of points in an interval [a, b] can be made
continuous in sub-intervals by breaking the intervals into these subintervals. If f(x) is discontinuous at points
x1, x2, x3..........xn in (a, b), then we can define subintervals (a, x1), (x1, x2).............(xn−1, xn), (xn, b) such that f(x)
is continuous in each of these subintervals. Such functions are called piecewise continuous functions. For integration
of Piecewise continuous function. We integrate f(x) in these sub-intervals and finally add all the values.
Example: 37 ∫20 [cot−1 x] dx , where [.] denotes greatest integer function
Solution: (b) −10
Example: 38 (a) 30 + cot 1 + cot 3 (b) 30 + cot 1 + cot 2 + cot 3
Solution: (a)
(c) 8 30 + cot 1 + cot 2 (d) None of these
∫Let I = 20 [cot−1 x] dx,
−10
we know cot−1 x ∈(0,π ) ∀ x ∈ R
3, x ∈ (−∞, cot 3)
2, x ∈ (cot 3, cot 2)
thus, [cot −1 x] = 1 x ∈ (cot 2, cot 1)
0 x ∈ (cot 1, ∞)
cot 3 cot 2 cot 1 20
∫ ∫ ∫ ∫Hence, I = 3 dx + 2 dx + 1 dx + 0 dx = 30 + cot 1 + cot 2 + cot 3
−10 cot 3 cot 2 cot 1
∫2 [x2 − x + 1] dx , where [.] denotes greatest integer function
0
(a) 7− 5 (b) 7+ 5 (c) 5 −3 (d) None of these
2 2 2
2 [x2 − x + 1] dx = 1+ 5 [x2 − x + 1] dx + 2 [x2 − x + 1] dx = 1+ 5 2 7− 5
2 2 2
0 1+ 5 1+ 5
∫ ∫ ∫ ∫ ∫Let I = 0 2 0 2 2dx =
1 dx +
Definite integral and area under curves
Introduction .
We know the methods of evaluating definite integrals. These integrals are used in evaluating certain types of
bounded regions. For evaluation of bounded regions defined by given functions, we shall also require to draw
rough sketch of the given function. The process of drawing rough sketch of a given function is called curve
sketching.
Procedure of Curve Sketching .
(1) Symmetry:
(i) Symmetry about x-axis: If all powers of y in equation of the given curve are even, then it is symmetric about
x-axis i.e., the shape of the curve above x-axis is exactly identical to its shape below x-axis.
For example, y 2 = 4ax is symmetric about x-axis.
(ii) Symmetry about y-axis: If all power of x in the equation of the given curve are even, then it is symmetric
about y-axis
For example, x 2 = 4ay is symmetric about y-axis.
(iii) Symmetry in opposite quadrants or symmetry about origin: If by putting –x for x and –y for y, the equation
of a curve remains same, then it is symmetric in opposite quadrants.
For example, x 2 + y 2 = a 2 and xy = a 2 are symmetric in opposite quadrants.
(iv) Symmetry about the line y = x : If the equation of a given curve remains unaltered by interchanging x and y
then it is symmetric about the line y = x which passes through the origin and makes an angle of 450 with the
positive direction of x-axis.
(2) Origin: If the equation of curve contains no constant terms then it passes through the origin.
Find whether the curve passes through the origin or not.
For examples, x2 + y2 + 4ax = 0 passes through origin.
(3) Points of intersection with the axes: If we get real values of x on putting y = 0 in the equation of
the curve, then real values of x and y = 0 give those points where the curve cuts the x-axis. Similarly by putting
x = 0, we can get the points of intersection of the curve and y-axis.
For example, the curve x2 y2 =1 intersect the axes at points (± a, 0) and (0, ± b) .
a2 + b2
(4) Special points: Find the points at which dy = 0, at these points the tangent to the curve is parallel to
dx
x-axis. Find the points at which dx = 0. At these points the tangent to the curve is parallel to y-axis.
dy
(5) Region: Write the given equation as y = f(x), and find minimum and maximum values of x which
determine the region of the curve.
For example for the curve xy2 = a2(a − x) ⇒y=a a−x
x
Now y is real, if 0 ≤ x ≤ a , So its region lies between the lines x = 0 and x = a
(6) Regions where the curve does not exist: Determine the regions in which the curve does not exists. For
this, find the value of y in terms of x from the equation of the curve and find the value of x for which y is imaginary.
Similarly find the value of x in terms of y and determine the values of y for which x is imaginary. The curve does not
exist for these values of x and y.
For example, the values of y obtained from y2 = 4ax are imaginary for negative value of x, so the curve does
not exist on the left side of y-axis. Similarly the curve a2y2 = x2(a − x) does not exist for x > a as the values of y are
imaginary for x > a.
Sketching of Some Common Curves .
(1) Straight line: The general equation of a straight line is ax + by + c = 0 . To draw a straight line, find the
points where it meets with the coordinate axes by putting y = 0 and x = 0 respectively in its equation. By joining
these two points, we get the sketch of the line.
(2) Region represented by a linear inequality: To find the region represented by linear inequalities
ax + by ≤ c and ax + by ≥ c, we proceed as follows.
(i) Convert the inequality into equality to obtain a linear equation in x, y.
(ii) Draw the straight line represented by it.
(iii) The straight line obtained in (ii) divides the xy-plane in two parts. To determine the region represented by
the inequality choose some convenient points, e.g. origin or some point on the coordinate axes. If the coordinates
of a point satisfy the inequality, then region containing the point is the required region, otherwise the region not
containing the point is the required region.
(3) Circle: The equation of a circle having centre at (0,0) and radius r is given by 2
x 2 + y 2 = r 2. The equation of a circle having centre at (h, k) and radius r is given by 1
(x − h)2 + (y − k)2 = r 2. The general equation of a circle is x 2 + y 2 + 2gx + 2 fy + c = 0. – 2 – 1 – 11 2
–2
This represents the circle whose centre is at (-g,-f) and radius equal to g 2 + f 2 − c.
The figure of the circle x2 + y2 = (2)2 is given. Here centre is (0,0) and radius is 2.
(4) Parabola: There are four standard forms of parabola with vertex at origin and the axis along either of
coordinate axis. Y
(i) y 2 = ±4ax : For this parabola Directrix
(a) Vertex: (0,0) (b) Focus: (± a, 0) X' Z A S (a,0) X
(c) Directrix: x ± a = 0 (d) Latus rectum: 4a
(e) Axis y = 0 (f) Symmetry : It is symmetric about x-axis. Y´
(ii) x 2 = ±4ay : For this parabola Y
(a) Vertex: (0,0) (b) Focus: (0, ± a) (0, a)
(c) Directrix: y ± a = 0 (d) Latus rectum: 4a A
Z Directrix
(e) Axis x = 0 (f) Symmetry: It is symmetric about y-axis X' X
Y'
(5) Ellipse: The standard equation of the ellipse having its centre at the origin Y
and major and minor axes along the coordinate axes is x2 + y2 =1 Here a > b. (- a, 0) B (0,b)
a2 b2 X′ A′ (a, 0)
AX
The figure of the ellipse is given.
B (0, – b)
Example: 1 Sketch the region bounded by 3x + 4y ≤ 12 Y´
Solution:
Converting the inequality into equation we get 3x + 4y = 12 . This line meets x- X´ Y
Example: 2 (0, 3)
Solution: axis at (4,0) and y-axis at (0,3). Joining these two points we obtain the straight
line represented by 3x + 4y = 12. This straight line divides the plane in two 3x + 4y = 12
Example: 3 parts. One part contains the origin the other does not contain the origin. Clearly, O (4, 0) X
Solution: (0,0) satisfies the inequality 3x + 4y ≤ 12 . So, the region represented by
Y'
3x + 4y ≤ 12 is the region containing the origin as shown in figure.
Sketch the parabola y2 = 8x. Y
y2 = 8x X' (0, 0) (2, 0) X
y2 = 4(2)x
Here vertex is (0,0) and focus is (2,0) Y'
Sketch the graph for y = x 2 − x. Y- axis
We note the following points about the curve. (0,0) X-axis
(1,0)
(i) The curve does not have any kind of symmetry.
(ii) The curve passes through the origin and the tangent at the origin is obtained by (½, – ¼)
equating the lowest degree term to zero.
The lowest degree term is x + y . Equation it to zero, we get x + y = 0 as the equation of tangent at the origin.
(iii) Putting y=0 in the equation of curve, we get x2 − x = 0 ⇒ x = 0, 1 . So, the curve crosses x-axis at (0,0) and (1,0).
Putting x=0 in the equation of the curve, we obtain y=0. So, the curve meets y-axis at (0,0) only.
(iv) y = x2 − x ⇒ dy = 2x −1 and d2y =2
dx dx 2
Now, dy = 0⇒ x = 1 ,
dx 2
At x = 1 , d2y > 0 , So x= 1 is point of local minima.
2 dx 2 2
(v) dy >0 ⇒ 2x −1 > 0 ⇒ x> 1 , So the curve increases for all x > 1 and decreases for all x< 1
dx 2 2 2
Area of Bounded Regions.
(1) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by
∫ ∫Area = by dx = bf(x)dx Y
aa
y = f(x)
y dx X
O x=a x=b
(2) If the curve y = f(x) lies below x-axis, then the area bounded by the curve y = f(x), the x-axis and the
ordinates x =a and x =b is negative. So, area is given by by dx
∫a
(3) The area bounded by a cartesian curve x =f(y), y-axis and abscissa y = c and y = d is given by
∫ ∫Area = d x dy = df(y)dy Y
cc
y=d x x = f(y)
y=c dy
OX
(4) If the equation of a curve is in parametric form, say x = f(t), y = g(t) then the area = by dx = t2 g(t) f ' (t) dt
∫ ∫a t1
where t1 and t2 are the values of t respectively corresponding to the values of a and b of x.
Sign convention for finding the Areas using Integration .
While applying the discussed sign convention, we will discuss the three cases.
∫Case I: In the expression bf(x) dx if b > a and f(x) > 0 for all a ≤ x ≤ b, then this integration will give the
a
area enclosed between the curve f(x), x-axis and the line x = a and x = b which is positive. No need of any
modification.
∫Case II: If in the expression bf(x)dx if b > a and f(x) < 0 for all a ≤ x ≤ b, then this integration will calculate
a
to be negative. But the numerical or the absolute value is to be taken to mean the area enclosed between the curve
y = f(x), x-axis and the lines x = a and x = b.
∫Case III. If in the expression bf(x)dx where b > a but f(x) changes its sign a Y
a
a cd e f b
numbers of times in the interval a ≤ x ≤ b, then we must divide the region [a, b] in X
such a way that we clearly get the points lying between [a, b] where f(x) changes its
sign. For the region where f(x)>0 we just integrate to get the area in that region
and then add the absolute value of the integration calculated in the region where
f(x)<0 to get the desired area between the curve y = f(x), x-axis and the line x = a and x = b.
Hence, if f(x) is as in above figure, the area enclosed by y = f(x), x-axis and the lines x = a and x = b is given
by
∫ ∫ ∫ ∫ ∫A = c f(x)dx + df(x)dx + e f(x)dx + f f(x)dx + bf(x)dxa f
cd e
Example: 4 The area (in square units) enclosed by the curve x 2y = 36, the x-axis and the lines x = 6 and x = 9 is
Solution: (a)
Example: 5 (a) 2 (b) 1 (c) 4 [Kerala (Engg.) 2000]
Solution: (d)
(d) 3
Example: 6
Solution: (a) 9 9 36 x 2y = 36 ⇒ y = 36
Example: 7 6 x2 x2
Solution: (a) y dx =
Example: 8 6
∫ ∫Required area = dx [Given ]
= − 36 9 = − 36 − 36 = −[4 − 6] = 2.
x 6 9 6
The area bounded by the x-axis, the curve y = f(x) and the lines x = 1, x = b is equal to b2 + 1 − 2 for all b > 1, then
f(x) is [MP PET 2000]
(a) x − 1 (b) x + 1 (c) x2 − 1 (d) x
1 + x2
∫b b2 + 1 − 2 = x2 + 1 b
f(x) dx = 1
1
⇒ f(x) = 1 . 2x
2 x2 +1
Hence f(x) = x .
1 + x2
The area of the region bounded by the curve y = x − x 2 between x = 0 and x = 1 is [Pb. CET 1994, 89]
(a) 1 (b) 1 (c) 1 (d) 5
6 3 2 6
∫Required Area = 1 − x 2 )dx = x2 − x3 1 = 1 − 1 = 1 .
2 3 6
(x 2 3 0
0
Find the area bounded between the curve y2 = 2y − x and y-axis .
(a) 4 (b) 2 (c) 1 (d) 5
3 3 3
The area between the given curve x = 2y − y2 and y- axis will be as shown in diagram
∫∴ Required Area = 2 − y 2 )dy Y
(0, 2)
(2y
0
= y 2 − y3 2 = 4
3 3
0
O (0, 0) X
Find the area bounded by the curves x = a cost, y = b sin t in the first quadrant
(a) πab (b) πa 2b (c) πab2 (d) None of these
4 4 4
Solution: (a) Clearly the given equation are the parametric equation of ellipse x2 + y2 =1. Curve meet the x-axis in the first quadrant
a2 b2
at (a,0)
a 0 = ab π / 2 sin2 t dt πab
0 4
y dx π (b sin t)(−a cos t)dt
o 2
∫ ∫ ∫∴ Required area = = = ( At x = 0, t = π / 2 and x = a, t = 0)
Symmetrical Area .
If the curve is symmetrical about a coordinate axis (or a line or origin), then we find the area of one symmetrical
portion and multiply it by the number of symmetrical portions to get the required area.
Example: 9 Find the whole area of circle x2 + y2 = a2
Solution: (b)
(a) π (b) πa 2 (c) πa 3 (d) a2
Example: 10
Solution: (a) The required area is symmetric about both the axis as shown in figure Y
∫a a2 − x2 dx = 4 x a2 − x2 + a2 sin−1 x a
2 2
∴ Required area = 4 a 0 (a, 0)
X
0 O (0, 0)
= 4 π × a2 = π a2
2
2
Find the area bounded by the parabola y2 = 4 x and its latus rectum [Rajasthan PET 1997, 94, 92, 84]
(a) 8 (b) 4 (c) 16 (d) None of these
3 3 3
Y
Since the curve is symmetrical about x-axis, therefore the required area S
11 O (1, 0) X
∫ ∫= 2 y dx = 2 4x dx
00
= 4. 2 3 1 = 8
3 x 2 3
0
Area between Two curves .
(1) When both curves intersect at two points and their common area lies between these points:
If the curves y1 = f1(x) and y2 = f2(x), where f1(x)> f2(x) intersect in two Y
points A(x = a) and B(x = b), then common area between the curves is y = f1(x) B
b A y = f2(x)
∫= (y1 − y2 )dx
a
b O x = a dx x=b X
∫= [ f1(x) − f2 (x)]dx
a
(2) When two curves intersect at a point and the area between them is bounded by x-axis:
Area bounded by the curves y = f1(x), y2 = f2(x) and x − axis is Y
αb y1 = f1(x) P(α,β)
y2 = f2(x)
∫ ∫= f1(x)dx + f2 (x)dx
aα
Where P(α, β ) is the point of intersection of the two curves. O X
(3) Positive and negative area : Area is always taken as positive. If some part of the area lies above the x-
axis and some part lies below x-axis, then the area of two parts should be calculated separately and then add their
numerical values to get the desired area.
Important Tips
• The area of the region bounded by y2 = 4ax and x2 = 4 by is 16ab square units.
3
• The area of the region bounded by y2 = 4ax and y = mx is 8a2 square units
3m3
• The area of the region bounded by y2 = 4ax and its latus rectum is 8a2 square units
3
• The area of the region bounded by one arch of sin (ax) or cos (ax) and x-axis is 2 sq. units
a
• Area of the ellipse x2 y2 = 1 is π ab sq. units.
a2 + b2
• Area of region bounded by the curve y = sin x , x-axis and the line x =0 and x= 2π is 4 unit.
Example: 11 The area of the region bounded by the curves y = x 2 and y =|x| is [Roorkee (Qualifying) 2000]
Solution: (b)
(a) 1 (b) 1 (c) 5 (d) 5
Example: 12 6 3 6 3
Solution: (a)
Example: 13 1 1 2 x2 1 x 3 1 Y
Solution: (d) 2 0 x 2dx 2 0 3 0
∫ ∫Required area = xdx − = − B(–1, 1) y=–x y=x A(1, 1)
0 x2=y
X
= 2 1 − 0 − 1 − 0 = 2 1 − 1 = 2 1 = 1 .
2 3 2 3 6 3 O
The area (in square units) bounded by the curve y = x3 , y = x 2 and the ordinates x = 1, x = 2 is [EAMCET 2000]
(a) 17 (b) 12 (c) 2 (d) 7
12 17 7 2
∫Required area = 2 (x3 − x 2 )dx = x4 − x3 2 = 4 − 8 − 1 − 1 = 4 + 1 = 16 + 1 = 17 .
1 4 3 3 4 3 3 12 12 12
1
The area of the region bounded by the curve y=2x– x 2 and line y =x is [Pb. CET 2000; Roorkee 1992]
(a) 1 (b) 1 (c) 1 (d) 1
2 3 4 6
The given curve is y = 2x − x 2
⇒ y = −(x 2 − 2x + 1) + 1
⇒ y − 1 = −(x − 1)2 , it represents a downward parabola with vertex (1,1) Y
Its points of intersection with the line y = x are (0,0) and (1,1).
Required area = shaded region (1, 1)
∫ ∫ ∫1 1 = 1 (x − x 2 )dx = x2 − x3 10 = 1 − 1 = 1 . O X
0 2 3 2 3 6 (0, 0) [MP PET 1996]
= (2x − x2)dx − x dx
Y
00
Example: 14 Area bounded by the lines y = 2 + x, y = 2 – x and x = 2 is
Solution: (b)
Example: 15 (a) 3 (b) 4
(c) 8 (d) 16 B(0,2)
Given lines are y = x + 2, y = –x + 2, x = 2
Hence required area = Area of ∆CAB = 1 (2)(4) = 4 sq. units. C(–2,0) A(2,0) X
2 x =2
The area bounded by the curve y 2 = 4 x and x 2 = 4y is
[Karnataka CET 1999,2003; MP PET 1997; SCRA 1986; Rajasthan PET 1988, 99,97]
(a) 16 sq. units (b) 3 sq. units (c) 14 sq. units (d) 3 sq. units
3 16 3 14
Solution: (a) 4 4 x2 dx = 16 square unit.
0 4 3
Example: 16 (OABC − ODBC) Region
Solution: (b)
0
Example: 17 ∫ ∫Required area = = 4x − Y
Trick : From Important Tips’ the area of the region bounded by y 2 = 4ax and (4, 4) D
x2 = 4 y A
x2 = 4by is 16ab square unit. B
3 CX
X´ (0, 0) O y2 = 4 x
Here y2 = 4 x and x2 = 4y, so a = 1 and b = 1 Y´
Required area = 16 (1) (1) = 16 square unit.
3 3
The area of the bounded region by the curve y = sinx, the x-axis and the line x = 0 and x = π is
[Rajasthan PET 1989, 92]
(a) 4 (b) 2 (c) 0 (d) None of these
∫Required area = πsin x dx Y
0 1
X´ π/2 π 3π/2 X
π /2 –1
sin x dx Y´
0
∫ ]= 2 2[− ]π / 2 [
= cos x =2 (− cos π / 2)– (− cos 0) = 2(1)
0
= 2 square unit.
Trick : For the curve y = sin x or cos x, the area of
π / 2 π 3π / 2 2π
sin x dx = 3, sin x dx = 4 and so on.
∫ ∫ ∫ ∫sin x dx = 1, sin x dx = 2,
0 00 0
The area enclosed by the parabola y2 = 8x and the line y = 2x is [EAMCET 1996; Rajasthan PET 1990, 98]
(a) 4 (b) 3 (c) 1 (d) 1
3 4 4 2
Solution: (a) Solve the equation y2 = 8x and the line y = 2x, we get the point of intersection. Then find the required area bounded by
this region. It is 4 .
3
Trick : Required area = 8(2)2 = 32 = 4 [ Area bounded by y2 = 4ax and y = mx is 8a2 . Here a = 2, m = 2 ]
3(2)3 24 3 3m3
Example: 18 If the area bounded by y = ax2 and x = ay2 , a > 0, is 1, then a = [IIT Screening 2004]
Solution: (b)
(a) 1 (b) 1 (c) 1 (d) 13
3 3 −
The x coordinate of A is 1
a
Y
According to the given condition A
1 1 OX
a 3a
∫1 = x − ax 2 dx = 1 . 2 x 2 − a [ x 3 ]10 / a
a a 3 0 3
0
⇒a= 1
3
Volumes and Surfaces of Solids of Revolution .
If a plane curve is revolved about some axis in the plane of the curve, then the body so generated is known as
solid of revolution. The surface generated by the perimeter of the curve is known as surface of revolution and the
volume generated by the area is called volume of revolution.
For example, a right angled triangle when revolved about one of its sides (forming the right angle) generates a
right circular cones. Y
(1) Volumes of solids of revolution: Q (x+δ x, y+δy)
S x=b
(i) The volume of the solid generated by the revolution, about the x-axis, of (x,y) P
the area bounded by the curve y = f(x), the ordinates at x = a, x = b and the AR
x-axis is equal to π b . x=a
∫ay 2 dx
O K NM L X
(ii) The revolution of the area lying between the curve x = f(y), the y-axis and
the lines y = a and y = b is given by (interchanging x and y in the above formulae) ∫a b x2 dy .
π
(iii) If the equation of the generating curve be given by x = f1(t) and y = f2 (t) and it is revolved about x-axis,
b y2 becomes t2π 2 d {f1(t)} , where and are the values of t
∫ ∫then the formula corresponding to dx { f 2 (t)} f1 f2
a π t1
corresponding to x = a and x = b
(iv) If the curve is given by an equation in polar co-ordinates, say r = f(θ ) , and the curve revolves about the
initial line, the volume generated
b y2 dx = π β y 2 dx .dθ , where and are the values of corresponding to x = a and x = b
α dθ
a
∫ ∫= π α β θ
∫Now x = r cosθ , β r 2 sin2 θ d
y = r sinθ . Hence the volume =π dθ (r cosθ )dθ
α
(v) If the generating curve revolves about any line AB (which is different from either of the axes), then the
∫volume of revolution is π (PN)2 d (ON), D
Q
P
C
A B
O NM
Note : The volume of the solid generated by revolving the area bounded by the curve r = f(θ ) and the radii
and θ = β about the initial line is 2 β r 3 sinθ dθ .
∫vectors 3 π
θ =α α
The volume in the case when the above area is revolved about the line π is 2 β r 3 cosθ dθ .
∫ θ = 2 3 π
α
(2) Area of surfaces of revolution: Y C
A SQ
(i) The curved surface of the solid generated by the revolution, about the x-
axis, of the area bounded by the curve y = f(x) , the ordinates at x = a, x = b BP R
and the x-axis is equal to 2π x=b y ds . x=a x=b
∫x =a O D NM E X
(ii) If the arc of the curve y = f(x) revolves about y-axis, then the area of
∫the surface of revolution (between proper limits) = 2π x ds, where ds = 1 + dy 2 dx
dx
(iii) If the equation of the curve is given in the parametric form x = f1(t) and y = f2(t), and the curve revolves
t=t2 yds = 2π t=t2
t =t1
t =t1
∫ ∫about x-axis, then we get the area of the surface of revolution f2 (t)ds
= 2π
∫= 2πt2 f2 (t) dx 2 + dy 2 dt , where t1 and t2 are the values of the parameter t corresponding to x = a and x = b.
t1 dt dt
(iv) If the equation of the curve is given in polar form then the area of the surface of revolution about x-axis
y ds = 2π sinθ ) dS .dθ r 2 dr 2 between proper limits.
∫ ∫ ∫= 2π (r dθ r sinθ . dθ dθ
= 2π +
Example: 19 The part of circle x2 + y2 = 9 in between y = 0 and y = 2 is revolved about y-axis. The volume of generating solid will be
Solution: (a)
[UPSEAT 1999]
(a) 46 π (b) 12π (c) 16π (d) 28π
3
The part of circle x 2 + y 2 = 9 in between y = 0 and y = 2 is revolved about y- axis. Then a frustum of sphere will be
formed.
2 2dy 2 2
∫ ∫The volume of this frustum=π =π − y )dy
x (9
00
= π 9y − 1 y3 2 = π 9 × 2 − 1 (2)3 − (9.0 − 1 .0) = 46 π cubic unit.
3 0 3 3 3
Example: 20 The part of straight line y = x + 1 between x = 2 and x = 3 is revolved about x-axis, then the curved surface of the solid
Solution: (d)
thus generated is [UPSEAT 2000]
(a) 37π (b) 7π (c) 37π (d) 7π 2
3 2
∫Curved surface = b + dy 2 . Given that a = 2, b=3 and y = x + 1 on differentiating with respect to x
1 dx dx
2πy
a
∫dy dy =1. 3 [1 + (1)2] dx
dx
dx 2π (x + 1)
=1+ 0 or Therefore, curved surface =
2
3 =2 2π 3 2π (x + 1)2 3 22 π [(3 + 1)2 − (2 + 1)2]
2 2 2
2π (x + 1) (x + 1)dx = 2
2 2
∫ ∫= 2 dx =
= 2π (16 − 9) = 7 2π = 7π 2
Indefinite integral
Introduction .
A function φ(x) is called a primitive or an antiderivative of a function f(x) if φ' (x) = f(x).
For example, x5 is a primitive of x 4 because d x5 = x4
5 dx 5
Let φ(x) be a primitive of a function f(x) and let c be any constant.
Then d [φ (x) + c] = φ ' (x) = f(x) [ φ' (x) = f(x)]
dx
⇒ φ(x) + c is also a primitive of f(x).
Thus, if a function f(x) possesses a primitive, then it possesses infinitely many primitives which are contained in
the expression φ(x) + c, where c is a constant.
For example x5 , x5 + 2, x5 − 1 etc. are primitives of x4.
5 5 5
Note : If F1(x) and F2(x) are two antiderivatives of a function f (x) on an interval [a, b], then the difference
between them is a constant.
Definition .
Let f(x) be a function. Then the collection of all its primitives is called the indefinite integral of f(x) and is
∫denoted by f(x)dx.
∫Thus,d f(x)dx = φ(x) + c
dx (φ(x) + c) = f(x) ⇒
where φ(x) is primitive of f(x) and c is an arbitrary constant known as the constant of integration.
∫Here is the integral sign, f(x) is the integrand, x is the variable of integration and dx is the element of
integration.
The process of finding an indefinite integral of a given function is called integration of the function.
It follows from the above discussion that integrating a function f(x) means finding a function φ(x) such that
d (φ(x)) = f (x).
dx
Comparison between Differentiation and Integration .
(1) Differentiation and integration both are operations on functions and each gives rise to a function.
(2) Each function is not differentiable or integrable.
(3) The derivative of a function, if it exists, is unique. The integral of a function, if it exists, is not unique.
(4) The derivative of a polynomial function decreases its degree by 1, but the integral of a polynomial function
increases its degree by 1.
(5) The derivative has a geometrical meaning, namely, the slope of the tangent to a curve at a point on it. The
integral has also a geometrical meaning, namely, the area of some region.
(6) The derivative is used in obtaining some physical quantities like velocity, acceleration etc. of a particle. The
integral is used in obtaining some physical quantities like centre of mass, momentum etc.
(7) Differentiation and integration are inverse of each other.
Properties of Integrals .
(1) The differentiation of an integral is the integrand itself or the process of differentiation and integration
[∫ ]neutralize each other, i.e., d
dx f(x)dx = f(x).
(2) The integral of the product of a constant and a function is equal to the product of the constant and the
∫ ∫integral of the function, i.e., cf(x)dx = c f(x)dx.
(3) Integral of the sum or difference of two functions is equal to the sum or difference of their integrals, i.e.,
∫ ∫ ∫{f1(x) ± f2(x)}dx = f1(x)dx ± f2(x)dx
∫ ∫ ∫ ∫In the general form, {k1 . f1(x) ± k2 . f2 (x) ± k3 . f3 (x) ± .....}dx = k1 f1(x)dx ± k2 f2 (x)dx ± k3 f3 (x)dx ± .......
Fundamental Integration Formulae .
∫(1) (i) x ndx = x n+1 + c, n ≠ −1 d x n+1 = xn ∫(ii) dx = x + c
n+1 dx n
+1
∫(iii) 1 dx =2 x + c, ∫(iv) (ax + b)n dx = 1 . (ax + b)n+1 +c
x a n+1
∫(2) (i) 1 dx = log | x | +c d (log | x |) = 1 ∫(ii) 1 b dx = 1 (log |ax + b | +c
x dx x ax + a
∫(3) e xdx = e x + c d (e x ) = e x
dx
∫(4) a x dx = ax a + c d ax a = ax
log e dx log e
∫(5) sin x dx = − cos x + c d (− cos x) = sin x
∫(6) cos x dx = sin x + c dx
∫(7) sec 2 x dx = tan x + c
d (sin x) = cos x
dx
d (tan x) = sec 2 x
dx
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