Differentiating both sides with respect to x, we 3.3 Derivative of a Logarithmic Function:
get Sometimes we have to differentiate a
∴ dy = 2x2 5 (3) (3x 7)(4x) function involving complicated expressions
dx (2x2 5)2 f (x)
∴ dy = (6x2 15) (12x2 28x) like f(x).g(x), g(x) and [f(x)]g(x). In this
dx (2x2 5)2 case, we first transform the expression
to a logarithmic form and then find its
∴ dy (6x2 15 12x2 28x) derivative. Hence the method is called
dx = (2x2 5)2 logarithmic differentiation. That is,
∴ dy ( 6x2 28x 15) d (log y) 1 dy
dx = (2x2 5)2 dx = y dx
Examples of Logarithmic Functions.
By derivative of the inverse function
6x 5 5
∴ dx = 1 dy ≠0
dy dy , dx 1) y = 3x2 1 8 2x
dx 2) y = (ex + 1)x × (x + 1)(x+2)
∴ Rate of change of demand with respect to Note : 1) The log function to the base "e" is
called Natural log and the log function to
dx (2x2 5)2 the base 10 is called common log.
price = dy = ( 6x2 28x 15) 2) In (a)bc, a is the base and bc is the index.
Some Basic Laws of logarithms:
EXERCISE 3.2 1) logam n = logam + logan
m
Q.1 Find the rate of change of demand (x) of
a commodity with respect to its price (y) 2) loga n = logam − logan
if
3) loga mn = n loga m
1) y = 12 + 10x + 25x2 4) logn m = log m
a
2) y = 18x + log(x − 4) loga n
3) y = 25x +log(1 + x2) 5) loge = 1 (= logaa)
Q.2 Find the marginal demand of a 6) logaax = x
commodity where demand is x and price
is y SOLVED EXAMPLES
1) y = xe−x + 7 1) Find dy , if y = (3 + x)x
dx
2) y= x+2
x2 +1 Solution: let y = (3 + x)x
3) y= 5x 9 Taking logarithm of both sides, we get
2x 10
∴ logy = log(3 + x)x
∴ logy = x log(3 + x)
92
Differentiating both sides with respect to x, 3) Find dy , if y = (2x 3)5
we get dx (3x 1)3(5x 2)
∴1 dy = x §1 · + log(3 + x) × 1
y dx ©¨ 3 x ¹¸
(2x 3)5
Solution: Let y = (3x 1)3(5x 2)
∴ dy =y ª x § 1 · log(3 x)¼»º
dx ¬« ©¨ ¸¹
3 x 1
∴ dy = (3 + x)x ª x log(3 x)»¼º y = § 3x 2x 3 5 2 ·2
dx ¬« ©¨¨ 1 3 5x ¸¹¸
3 x
2) Find dy , if y = xxx Taking logarithm of both sides, we get
dx
1 °®log § 3x 2x 3 5 2 · °½
Solution: Let y = xxx ∴ logy = 2 ¯° ¨©¨ 1 3 ¸¹¸ ¾
5x °¿
Taking logarithm of both sides, we get 1
∴ logy = log xxx ∴ logy = [5log(2x + 3) − 3log(3x − 1) −
2
log(5x − 2)]
∴ logy = xx log(x) Differentiating both sides with respect to x, we
get
Differentiating both sides with respect to x,
we get
dy = 1
∴1 dy 1 dx x 1 dx 2 ª2 3 5º
y dx = xx x + log(x). dx ∴ y «5 3 »
¬ 3 3x 1 5x 2 ¼
2 x
dy ª x 1 dx x º dy 1 ª 10 9 5º
dx « x dx » dx 2
∴ =y x log( x). ¼ .......... (I) ∴ = y « 2x 3 3x 1 5x 2 »
¬ ¼
¬
Let u = xx
Taking logarithm of both sides, we get ∴ dy = 1 2x 3 5 2 ª 10 3 9 5 2 º
dx 2 1 3 5x « 2x »
2 x ¬ 3x 1 5x ¼
∴ logu = x.log(x)
Differentiating both sides with respect to x, 4) Find dy , if y = xx + (logx)x
we get dx
∴ 1 du = x. 1 + log(x).1 Solution: Let y = xx + (logx)x
u dx x Let u = xx and v = (logx)x
∴ du = u [1 + logx] ∴ y = u + v
dx
Differentiating both sides with respect to x,
∴ du = xx(1 + logx) .......... (II) we get
dx
dy = du + dv
Substituting eqn (II) in eqn (I), we get dx dx dx
∴ dy =y ª x x 1 log( x).x x (1 log x)¼»º Now, u = xx
dx ¬« x
Taking logarithm of both sides, we get
∴ dy = x xx .xx ª 1 log( x).(1 log x)º»¼ ∴ logu = x logx
dx «¬ x
93
Differentiating both sides with respect to x, 3x 1
we get, 3) y = 3 2x 3 5 x 2
∴ 1 du =x 1 + log x.1 Q.3 Find dy if,
u dx x dx
∴ du = u (1 + log x) 1) y = (logx)x + xlogx
dx
2) y = (x)x + (a)x
∴ du = xx (1 + log x) ............. (II) 3) y = 10xx + 10x10 + 1010x
dx
3.4 Derivative of an Implicit Function:
Now, v = (logx)x
If the variable y can be expressed as
Taking logarithm of both sides, we get a function of the variable x. that is,
y = f(x) then the function f(x) is called an
∴ logv = x log(logx) explicit function of x.
Differentiating both sides with respect to x, For Example: f(x) = x2 + x−3, y = logx + e
we get,
If it is not possible to express y as a function
∴ 1 dv =x 1 + log (logx).1 of x or x as a function of y then the function
v dx x.log x is called an implicit function.
∴ dv =v ª1 log(log x)º» For Example: ax2 + 2hxy + by2 = 0 ;
dx « ¼ xm + yn = (x + y)m+n
¬ log x
The general form of an implicit function of
∴ dv = (logx)x ª1 log(log x)»¼º ..... (III) two variables x and y is f(x,y)=0
dx «¬ log x
Solved Examples:
Substituting eqn (II) and eqn (III) in eqn (I), we 1) Find dy if y3− 3y2x = x3 + 3x2y
get
dx
∴ dy = xx (1 + log x) + (logx)x ª1 log(log x)»º Solution: Given y3− 3y2x = x3 + 3x2y
dx « ¼
¬ log x Differentiating both sides with respect
to x, we get
EXERCISE 3.3
∴ 3y2 dy − 3y2 − 3x(2y) dy
Q.1 Find dy if, dx dx
dx
= 3x2 + 3x2 dy + 3y(2x)
1) y = xx2x dx
2) y = xex
3) y = exx ∴ 3y2 dy − 6xy dy − 3x2 dy
dx dx dx
Q.2 Find dy if,
dx = 3x2 + 6xy + 3y2
1) y = §©¨1 1 ·x ∴ (3y2 − 6xy − 3x2) dy = (3x2 + 6xy + 3y2)
x ¸¹ dx
2) y = (2x +5)x ∴ (y2 − 2xy − x2) dy = (x2 +2xy + y2)
dx
94
∴ n dy (m n) dy (m n) m
dy = x2 2xy y2 ∴ y dx dx x
dx y2 2xy x2 x y x y
2) Find dy if xy = yx ∴ § n (m n) · dy § (m n) m ·
dx ¨¨© y ¸¹¸ dx ¨¨© x ¸¸¹
x y x y
Solution: Given xy = yx
∴ ª nx ny my ny º dy ª mx nx mx my º
« » « »
Taking logarithm of both sides, we get ¬ y x y ¼ dx ¬ x x y ¼
∴ y logx = x logy ∴ ª nx my º dy ª nx my º
¬« y ¼» dx ¬« x ¼»
Differentiating both sides with respect to x,
we get ∴ dy = ª nx my º ª y º
dx «¬ x »¼ «¬ nx my ¼»
∴ y 1 + logx dy =x 1 dy +logy.1
x dx y dx dy y
∴ dx = x
∴ logx dy − x dy = logy − y
dx y dx x
EXERCISE 3.4
∴ § log x x · dy § log y y · Q.1 Find dy if,
¨ y ¸ dx ©¨ x ¸¹ dx
© ¹
1) x y
∴ § y.log x x · dy § x.log y y · a
¨ ¸ ¨© x ¸¹
© y ¹ dx 2) x3 + y3 + 4x3y = 0
∴ dy = § x.log y y ·§ y ·
dx ©¨ x ¹¸ ¨ y.log x ¸ 3) x3 + x2y + xy2 + y3 = 81
© x ¹
∴ dy = y § x.log y y · Q.2 Find dy if,
dx ¨ ¸ dx
x © y.log x x ¹
1) y.ex + x.ey = 1
3) If xm.yn = (x + y)(m+n) then show that,
2) xy = e(x−y)
dy = y
dx x 3) xy = log(xy)
Solution: Given xm.yn = (x + y)(m+n) Q.3 Solve the following.
Taking logarithm of both sides, we get 1) If x5.y7 = (x + y)12 then show that,
∴ m.logx + n.logy = (m + n) log(x + y) dy = y
dx x
Differentiating both sides with respect to x, we
get 2) If log(x+y) = log(xy) + a then show
∴ m 1 n 1 dy (m n) x 1 y ©¨§1 dy · that, dy = −y2
x y dx dx ¸¹ dx x2
∴ m n dy (m n) ©§¨1 dy · 3) If ex+ey = e(x+y) then show that,
x y dx dx ¹¸ dy = −ey−x
x y
dx
95
3.5 Derivative of a Parametric Function: ∴ dy = e t d t .......... (I)
dt dt
Now we consider y as a function of x where
both x and y are functions of a variable ‘t’ . Differentiate with respect to t
Here ‘t’ is called a parameter.
Result 3: If x = f(t) and y = g(t) are differentiable ∴ dx = 2 e2t .......... (II)
functions of a parameter ‘t’, then y is a dt
differential function of x and
dy
dy Now, dy = dt
dx dx
dy dt dx
dx dx , dt z 0 dt
dt et 1
SOLVED EXAMPLES ∴ dy = 2t
dx 2e2t
1) Find dy , if x = 2at, y = 2at2 ∴ dy et
dx dx = 4 t e2t
Solution: Given x = 2at, y = 2at2 3) Differentiate log(t) with respect to log(1+t2)
Now, y =2at2 Solution: let y = log(t) and x = log(1+t2)
Differentiate with respect to t Now, y = log(t)
∴ dy = 2a.2t = 4at .......... (I) Differentiate with respect to t
dt
∴ dy = 1 .......... (I)
dt t
x = 2at
Now, x = log(1+ t2)
Differentiate with respect to t
Differentiate with respect to t
∴ dx = 2a
dt .......... (II) ∴ dx 2t
dy dt = 1+ t2 .......... (II)
Now, dy = dt dy
dx dx
Now, dy = dt
dt dx dx
∴ dx = 4at dt
dt 2a
1
∴ dx = 2t ∴ dy = t
dt dx 2t
2) Find dy , if x = e2t, y = e t 1+ t2
dx
∴ dy = 1+ t2
Solution: Given x = e2t, y = e t dx 2t 2
Now, y = e t
Differentiate y with respect to t
96
EXERCISE 3.5 If f ′′(x) is a differential function of x
Q.1 Find dy if, d § d2y · d3y
dx ¨ dx2 ¸ dx3
then © ¹ denoted by or f ′′′(x) is
1) x = at2, y = 2at
2) x = 2at2, y = at4 dx
called the third order derivative of y with respect
to x. It is also denoted by y′′′ or y3.
3) x = e3t, y = e(4t + 5) SOLVED EXAMPLES
Q.2 Find dy if, 1) Find d 2 y , if y = x2
dx dx2
§ 1 · 2 , y = 2 § 1 · Solution: Given y = x2
©¨ u ¹¸ ¨© u ¸¹
1) x= u u
2) x = 1+ u2 , y = log(1 + u2) Differentiate with respect to x
3) Differentiate 5x with respect to log x ∴ dy = 2x
dx
Q.3 Solve the following.
Differentiate with respect to x, again
1) If x =a §¨©1 1 · ,y=a ©¨§1 1 · then, ∴ d 2 y = 2
t ¹¸ t ¸¹
dx2
show that dy = −1 2) Find d 2 y , if y = x6
dx
dx2
2) 4t , y = 3 §1 t2 · then, Solution: Given y = x6
If x = 1+ t2 ¨ t2 ¸
© 1 ¹ Differentiate with respect to x
∴ dy = 6x5
show that dy −9x
dx = 4y dx
Differentiate with respect to x, again
3) If x = t.logt, y = t t then, show that ∴ d 2 y = 6(5x4)
dy − y = 0 dx2
dx
∴ d 2 y = 30x4
3.6 Second Order Derivative: dx2
Consider a differentiable function y = f(x) 3) Find d 2 y , if y = log x
then dy = f '(x)is the first order derivative of dx2
dx Solution: Given y = log x
y with respect to x. It is also denoted by y′ or y1
If f ′(x) is a differentiable function of x Differentiate with respect to x
§ dy · d2y ∴ dy = 1
¨© dx ¸¹ dx2 dx x
then d denoted by or f ′′(x) is called
Differentiate with respect to x, again
dx ∴ d 2 y 1
dx2 x2
the second order derivative of y with respect to
x. It is also denoted by y′′ or y2
97
4) Find d 2 y , if y = e4x Rules of Differentiation:
dx2
If u and v differentiable function of x and if
Solution: Given y = e4x
Differentiate with respect to x 1. y = u + v then dy = du + dv
∴ dy = 4e4x dx dx dx
dx 2. y = u – v then dy = du − dv
∴ d 2 y = 4(4e4x) dx dx dx
dx2 3. y = u.v then dy =u dv +v du
∴ d 2 y = 16e4x dx dx dx
dx2 u dy v du − u dv
v dx dx dx
EXERCISE 3.6 4. y= then = v2 ,v ≠0
Q.1 Find d2y if, Derivative of a Composite Function:
dx2
If y = f(u) is a differentiable function of u
1) y = x and u = g(x) is a differentiable function of x
then
2) y = x5
3) y = x−7 dy dy u du
dx du dx
Q.2 Find d2y if,
dx2
Derivative of an Inverse Function :
1) y = ex
If y = f(x) is a differentiable function of
2) y = e(2x + 1) x such that the inverse function x = f−1(y)
exists, then x is a differentiable function of
3) y = elogx y and
Let's Remember dx = 1 , dy ≠ 0
dy dy dx
Derivative of some standard functions.
dx
y = f(x) dy = f ′(x) Derivative of a Parametric Function:
dx
1 K (constant) If x = f(t) and y = g(t) are differential
2x 0 functions of parameter ‘t’ then y is a
3x 1 differential function of x and
41 1 dy
x
2x dy = dt ,
5 xn dx dx
6 ax −1
7 ex x2 dt
8 logx n.xn−1
ax. loga dx ≠ 0
ex dt
1
x
98
MISCELLANEOUS EXERCISE - 3 7) If ax2+2hxy+by2 = 0 then dy = ?
dx
Q.I] Choose the correct alternative.
1) If y = (5x3 – 4x2 – 8x)9 then dy = ----------- a) ahxx bhyy b ) haxx bhyy
dx ax hy 2ax hy
a) 9(5x3 – 4x2 – 8x)8 (15x2− 8x − 8) c) hx by d) hx 3by
b) 9(5x3 – 4x2 – 8x)9 (15x2 − 8x − 8) 8) If x4.y5 = (x+y)(m+1)
c) 9(5x3 – 4x2 – 8x)8 (5x2− 8x − 8)
d) 9(5x3 – 4x2 – 8x)9 (5x2 − 8x − 8)
2) If y = x + 1 then dy = ? and dy = y then m = ?
x dx dx x
a) x2 1 b ) 1 x2 a) 8 b) 4 c) 5 d) 20
2x2 x2 1 2x2 x2 1 9) If x = et e t , y= et − e−t then dy = ?
2 2 dx
x2 1
c) x2 1 d) 1 x2 a) − y y −x x
2x x 2x x x2 1 x b) x c) y d) y
3) If y = elogx then dy = ? 10) If x = 2at2, y = 4at then dy = ?
dx dx
a) elogx b) 1 c) 0 1 1 1
x x d) a) − 2at2 b) 2at3
2
4) If y = 2x2 + 22 + a2 then dy = ? c) 1 1
dx t d) 4at3
a) x b) 4x c) 2x d) −2x Q.II] Fill in the blanks:
5) If y = 5x.x5 then dy = ? 1 ) If 3x2y + 3xy2 = 0
dx
a) 5x.x4 (5 + log5) b) 5x.x5 (5 + log5) then dy =
dx
c)5x.x4 (5 + xlog5) d)5x.x5 (5 + xlog5) dy .......
2) If xm.yn = (x+y)(m+n) dx x
§ ex · then =
¨ x2 ¸
6) If y = log © ¹ then dy = ? dy −y
dx dx .....
3) If 0 = log(xy) + a then =
a) 2 −x x b) x −x 2
4) If x = t logt and y = tt then dy = ........
e−x d) x−e dx
c) ex ex
5) If y = x.logx then d 2 y = ........
dx2
99
6) If y = ª¬log x ¼º2 then d2y = ....... 3) If y = [log(log(logx))]2 , find dy
dx2 dx
7) 1 then dy = ........ 4) Find the rate of change of demand (x) of a
If x = y + y dx commodity with respect to its price (y)
8) If y = eax, then x. dy = ....... if y = 25+30x – x2.
dx 5) Find the rate of change of demand (x) of a
9) If x = t.logt , y = tt then dy = ....... commodity with respect to its price (y)
dx
if y = 5x 7
m 2x 13
10) If y = x x2 1 6) Find dy , if y = xx
dx
then (x2 −1) dy = .......
dx 7) Find dy , if y = 2xx
dx
Q.III] State whether each of the following is
True or False: 8) Find dy , if y = 3x 4 3
1) If f ′ is the derivative of f, then the derivative dx (x 1)4 (x 2)
of the inverse of f is the inverse of f’
9) Find dy , if y = xx + (7x – 1)x
2) The derivative of logax, where a is constant dx
1
10) If y = x3+3xy2+3x2y Find dy
is . dx
x.log a
11) If x3+y2+xy = 7 Find dy
3) The derivative of f(x) = ax, where a is dx
constant is x.ax−1
12) If x3y3 = x2−y2 Find dy
4) The derivative of polynomial is polynomial. dx
5) d (10x ) x.10x 1 13) If x7.y9 = (x+y)16 then show that
dx
6) If y = log x then dy = 1 Find dy = y
dx x dx x
7) If y = e2 then dy = 2e 14) If xa.yb = (x+y)(a+b) then show that
dx
Find dy = y
8) The derivative of ax is ax.loga dx x
9) The derivative of xm.yn = (x+y)(m+n) is x 15) Find dy if , x = 5t2, y = 10t
y dx
Q.IV] Solve the following: 16) Find dy if , x = e3t , y = e t
1) If y = (6x3 − 3x2 −9x)10 , find dy dx
dx
2) If y = 5 3x2 8x 5 4 , find dy 17) Differentiate log(1+x2) with respective to
dx ax
100
18) Differentiate e(4x+5) with respective to 104x Solution : Let y = 30 + 25x + x2
19) Find d 2 y , if y = log(x) Diff. w.r.to x, we get
dx2
∴ dy = + +
20) Find d 2 y , if y = 2at, x = at2 dx
dx2
∴ dy = 25 + 2x
d2y dx
dx2
21) Find , if y = x2.ex ∴ By derivation of the inverse function
22) If x2+6xy+y2= 10 then show dx = 1 , dy ≠ 0
dy dx
80
that d2y = Rate of change of demand with respect to
dx2 3x y 3
1
price = +
23) If ax2+2hxy+by2 = 0 then show (3): find dy , if y = x(logx)+ 10x
dx
that d2y =0
dx2
Solution:- Let y = x(logx)+ 10x
Activities Let u = xlogx , v = 10x
(1): y = (6x4 − 5x3 + 2x + 3)5 find dy
y=u+v
dx
Solution:- Given Now, u = xlogx
y = (6x4 − 5x3 + 2x + 3)5 Taking log on both sides, we get
Let u = [6x4 − 5x3 + + 3] logu = logxlogx
∴y=u logu = logx.logx.logx
dy
logu = (logx)2
∴ du = 5u4
Diff. w.r.to x, we get
And du = 24x3 − 15( )+2
dx ∴ 1 du 2(log x) u d
u dx ddxx
By chain rule
∴ du ª¬«2 1 º
dy dy × dx u log x u x ¼»
dx dx
=
∴ du xlog x ¬ª«22u 1×»¼º1 .........(II)
dx
∴ dy = 5(6x4 − 5x3 + 2x + 3)
dx Now, v = 10x
× (24x3 −15x2 + ) Diff.w.r.to x, we get
(2): The rate of change of demand (x) of a ∴ dv = 10x.
commodity with respect to its price (y). dt
If y = 30 + 25x + x2 Substitution equation (II) & (III) in
equation (I), we get
101
∴ dy xlog x ª«¬2 log x 1 º 10x.log(10) 5: Find dy if x = et, y = e t
dx x ¼» dx
(4): Find dy , if yx = ex+y Solution:- given, x = et, y = e e t
dx
Now, y = e t
Solution:- Given yx = ex+y Diff. w.r.to t
Taking log on both side, we get, dy d
dt ddt t
∴ log (y)x = log (e)x+y ∴ =e t
∴ x. = .log e ∴ dx = e t . 1 ......... (I)
dt 2 t
∴ x. log y = (x + ). 1
∴ x. log y = x + Now, x = et
Diff. w.r.to x, we get Diff.w.r.to t
∴ x 1 d + log y. 1 = + dy ∴ dx = ....... (II)
y dx dx dt
∴ x 1 dy + log y =1+ dy Now, dy = dy
y dx dx dx dt
∴ x dy − dy = 1 −
y dx dx et
∴ =
∴ dy § x 1¸· = − log y
dx ¨ y ¹ et
©
∴ dy = e t
∴ dy = ( - log y) (y) dx 2 t et
dx x-y
vvv
102
4 Applications of Derivatives
Let's Study Let P (a, f(a)) and Q (a + h, f(a + h)) be two
points on the curve. Join the points P and Q.
• Meaning of Derivatives
• Increasing and Decreasing Functions. The slope of the chord PQ = f (a h) f (a)
• Maxima and Minima h
• Application of derivatives to Economics.
Let the point Q move along the curve such
Introduction that Q→P. Then the secant PQ approaches the
tangent at P as h→0
Derivatives have a wide range of
applications in everyday life. In this chapter, ∴ lim (slope of secant PQ) = lim f (a h) f (a)
we shall discuss geometrical and physical h
significance of derivatives and some of their Q→P ho0
applications such as equation of tangent and
normal at a point on the curve, rate measure in Slope of tangent at P = f ′(a) (if limit exists)
physical field, approximate values of functions
and extreme values of a function. Thus, the derivative of a function y = f(x) at
any point P(a,b) is the slope of the tangent at the
Let's Learn point P(a,b) on the curve.
4.1 Meaning of Derivative: The slope of the tangent at any point P(a,b)
Let y = f(x) be a continuous function of x. is also called gradient of the curve y = f(x) at
It represents a curve in XY-plane. (fig. 4.1).
point P and is denoted by f ′(a) or § dy · .
©¨ dx ¹¸p
Normal is a line perpendicular to tangent,
passing through the point of tangency.
∴ Slope of the normal is the negative
reciprocal of slope of tangent.
−1 1
Thus, slope of normal = f '(a) = § dy ·
¨© dx ¸¹P
Hence,
(i) The equation of tangent to the curve
y = f(x) at the point P(a,b) is given by
(y − b) = f ′(a)(x − a)
(ii) The equation of normal to the curve
y = f(x) at the point P(a,b) is given by
(y − b) = −1 (x − a)
f '(a)
Fig. 4.1
103
SOLVED EXAMPLES ∴ § dy · = −2x1
¨© dx ¸¹x
x1
1) Find the equation of tangent and normal to ∴ The slope of the tangent as P(x1,y1)
the curve y = x2 + 4x + 1 at P(−1, −2). = −2x1
Solution: Given equation of curve is ∴ The slope of the normal at P(x1,y1)
y = x2 + 4x + 1 1
= 2x1
Differentiating with respect to x
1
∴ dy = 2x + 4 Now, slope of x − 4y + 3 = 0 is
dx
4
§ dy · 1
∴ ¨© dx ¸¹p(-1,-2) = 2(−1) + 4 ∴ The slope of the normal = 4 (since
=2 normal is parallel to given line)
∴ The slope of tangent at P(−1, −2) is 2
∴ The equation of tangent is 11
y + 2 = 2(x + 1) ∴ 2x1 = 4
∴ y + 2 = 2x + 2
∴ 2x − y = 0 ∴ x1 = 2
Now, The slope of Normal at P(x1,y1) lies on the curve y = 6 − x2
∴ y1 = 6 − x12
P(−1, −2) is −1 ∴ y1 = 6 − 4
2 ∴ y1 = 2
∴ The point on the curve is (2, 2)
∴ The equation of normal is
∴ The slope of tangent at (2,2) is
y + 2 = −1 (x + 1) −2x1 = −2(2) = −4
2 ∴ The equation of tangent is
2(y + 2) = −1(x + 1) (y − 2) = −4 (x − 2)
2y + 4 = −x − 1 ∴ y − 2 = −4x + 8
x + 2y + 5 = 0 ∴ 4x + y − 10 = 0
2) Find the equation of tangent and normal to ∴ The equation of normal is
the curve y = 6 − x2 where the normal is 1
parallel to the line x − 4y + 3 = 0.
(y − 2) = 4 (x − 2)
Solution: Let P(x1,y1) be the point on the curve y
= 6 − x2 where the normal is parallel to the ∴ 4(y − 2) = 1 (x − 2)
line x − 4y + 3 = 0 ∴ 4y − 8 = x − 2
∴ x − 4y + 6 = 0
Consider, y = 6 − x2
∴ dy = −2x
dx
104
EXERCISE 4.1 Definition: A function y = f(x) is said to
be a decreasing function of x in an interval
Q.1 Find the equation of tangent and normal to (a,b). if f(x2) < f(x1), whenever x2 > x1 for all
the curve at the given points on it. x1, x2 in the interval (a, b).
i) y = 3x2 − x + 1 at (1,3) Fig. 4.3
Geometrically, as we move from left to
ii) 2x2 + 3y2 = 5 at (1,1)
right along the curve y = f(x) in (a,b), then
iii) x2 + y2 + xy = 3 at (1,1) the curve falls. (see fig.4.3)
∴ Slope of tangent f ′(x) < 0
Q.2 Find the equation of tangent and normal to ∴ The slope of tangent is negative.
the curve y = x2 + 5 where the tangent is If f ′(x) < 0 in (a,b) then f(x) is a decreasing
parallel to the line 4x − y + 1 = 0. function in the interval (a,b).
Note: Every function may not be either
Q.3 Find the equation of tangent and normal to increasing or decreasing.
the curve y = 3x2 − 3x − 5 where the tangent
is parallel to the line 3x − y + 1 = 0.
4.2 Increasing and Decreasing Functions:
Definition : The function y = f(x) is said
to be an increasing function of x in the
interval (a,b) if f(x2) > f(x1), whenever
x2 > x1 in the interval (a,b).
SOLVED EXAMPLES
Fig. 4.2 1) Test whether the following function is
increasing or decreasing.
Geometrically, as we move from left to
right along the curve y = f(x) in (a,b), then f(x) = x3 − 3x2 + 3x − 100, x ∈ R
the curve rises. (see fig. 4.2) Solution: Given f(x) = x3 − 3x2 + 3x − 100, x ∈ R
∴ f ′(x) = 3x2 − 6x + 3
∴ Slope of tangent at x: f ′(x) > 0 ∴ f ′(x) = 3(x − 1)2
Since (x − 1)2 is always positive, x ≠ 1
∴ The slope of the tangent is positive.
∴ f ′(x) > 0, ∀x ∈ R − {1}
If f ′(x) > 0 for all x ∈ (a,b) then, y = f(x) is
an increasing function in the interval (a,b) Hence, f(x) is an increasing function,
∀ x ∈ R − {1}
Note: Sign of the Derivative can be used to
find if the function f(x) is increasing.
105
2) Test whether the following function is Now, f ′(x) < 0
increasing or decreasing. ∴ 6(x − 1)(x − 2) < 0
(if ab < 0 either a < 0 and b > 0 or a > 0 and
f(x) = 2 − 3x + 3x2 − x3, ∀x ∈ R
Solution: f(x) = 2 − 3x + 3x2 − x3 b < 0)
∴ f ′(x) = −3 + 6x − 3x2 Case I] (x − 1) < 0 and x − 2 > 0
∴ f ′(x)= −3(x2 − 2x + 1) ∴ x < 1 and x > 2 which is contradiction
∴ f ′(x) = −3(x − 1)2 Case II] x − 1 > 0 and x − 2 < 0
Since (x − 1)2 is always positive, x ≠ 1 ∴ x > 1 and x < 2
∴ 1 < x < 2
∴ f ′(x) < 0, ∀ x ∈ R − {1}
∴ f(x) = 2x3 − 9x2 + 12x + 2 is decreasing
Hence, function f(x) is decreasing function function if x ∈ (1,2).
∀ x ∈ R − {1}
EXERCISE 4.2
3) Find the value of x, for which the function
f(x) = x3 + 12x2 + 36x + 6 is increasing. Q.1 Test whether the following fuctions are
increasing or decreasing
Solution: Given f(x) = x3 + 12x2 + 36x + 6
i) f(x) = x3 − 6x2 + 12x − 16, x ∈ R
∴ f ′(x) = 3x2 + 24x +36 ii) f(x) = x − 1 , x ∈ R, x ≠ 0
∴ f ′(x) = 3(x + 2) (x + 6) x
iii) f(x) = 7 − 3, x ∈ R, x ≠ 0
Now, f ′(x) > 0, as f(x) is increasing.
x
∴ 3(x + 2) (x + 6) > 0 Q.2 Find the values of x, such that f(x) is
(ab > 0 ⇔ a > 0, b > 0 or a < 0, b < 0) increasing function.
i) f(x) = 2x3 − 15x2 + 36x + 1
Case I] x + 2 > 0 and x + 6 > 0 ii) f(x) = x2 + 2x − 5
iii) f(x) = 2x3 − 15x2 − 144x − 7
∴ x > −2 and x > −6 Q.3 Find the values of x such that f(x) is
∴ x > −2 ................. (I) decreasing function.
i) f(x) = 2x3 − 15x2 − 144x − 7
Case II] x + 2 < 0 and x + 6 < 0 ii) f(x) = x4 − 2x3 + 1
iii) f(x) = 2x3 − 15x2 − 84x − 7
∴ x < −2 and x < −6
4.3 Maxima and Minima:
∴ x < −6 ................. (II) a) Maximum value of f(x): A function f(x) is
From case I and II, f(x) is increasing if said to have a maximum value at a point
x < −6 or x > −2 x = c if f(x) < f(c) for all x ≠ c.
The value f(c) is called the maximum value
∴ f(x) = x3 + 12x2 + 36x + 6 is increasing of f(x).
if and only if x < −6 or x > −2
Hence, x ∈ (−∞, −6) or x ∈ (−2, ∞).
4) Find the values of x for which the function
f(x) = 2x3 − 9x2 + 12x + 2 is decreasing.
Soluction: Given f(x) = 2x2 − 9x2 +12x + 2
∴ f ′(x) = 6x2 − 18x + 12
∴ f ′(x) = 6(x − 1)(x − 2)
106
Thus, the function f(x) will have a First Derivative Test : A function
maximum at x = c if f(x) is increasing for y = f(x) is said to have a maximum
x < c and f(x) is decreasing for x > c as value at x = c if the following three
shown in Fig. 4.4 conditions are satisfied.
Fig. 4.4 i) f ′(c) = 0
b) Minimum value of f(x): A function f(x) is
ii) f ′(c − h) > 0
said to have a minimum at a point x = c
if f(x) > f(c) for all x ≠ c. iii) f ′(c + h) < 0 where h a is small
positive number (see fig. 4.4)
A function y = f(x) is said to have a
minimum value at x = c if the following
conditions are satisfied.
i) f ′(c) = 0
ii) f ′(c − h) < 0
iii) f ′(c + h) > 0 where h is a small
positive number (see fig. 4.5)
Remark :
If f′(c) = 0 and f′(c − h) > 0, f′(c + h) > 0
or f′(c − h) < 0, f′(c + h) < 0 then f(c) is neither
maximum nor minimum. In this case x = c is
called a point of inflection (see.fig. 4.6)
Fig. 4.6
Fig. 4.5 A function may have several maxima and
several minima. In such cases, the maxima are
The value of f(c) is called the minimum called local maxima and the minima are called
value of f(x).
local minima. (see. fig. 4.7)
The function will have a minimum at
x = c if f(x) is decreasing for
x < c and f(x) is increasing for x > c as
shown in fig. 4.5
At x = c if the function is neither increasing
nor decreasing, then the function is
stationary at x = c
Note: The maximum and minimum values
of a function are called its extreme values.
To find extreme values of a function, we Fig. 4.7
use the following tests.
107
In this figure the function has a local 2) Divide the number 84 into two parts such
maximum at x = a and a local minimum at x = b that the product of one part and square of
and still f(b) > f(a). the other is maximum.
SOLVED EXAMPLES Solution: Let one part be x then other part will
be 84 - x
1) Find the maximum and minimum value of f(x) = x2 (84 − x)
the function f(x) = 84x2 − x3
f ′(x) = 168(x) − 3x2
f(x) = 3x3 − 9x2 − 27x + 15 f ′′(x) = 168 − 6x
Solution: Given f(x) = 3x3 − 9x2 − 27x + 15
∴ f ′(x) = 9x2 − 18x − 27 For extream value f ′(x) = 0
∴ f ′′(x) = 18x − 18 ∴ 168x − 3x2 = 0
For the extreme values f ′(x) = 0 ∴ 3x (56 − x) = 0
∴ 9x2 − 18x − 27 = 0 x = 0 or x = 56
∴ 9(x2 − 2x − 3) = 0 If x = 0, f ′′(x) = 168 − 6x
∴ (x + 1)(x − 3) = 0 f ′′(0) = 168 − 6(0)
∴ x = −1 or x = 3 = 168 > 0
For x = −1, f ′′(x) = 18x − 18 ∴ f(x) attains minimum at x = 0
f ′′(−1) = 18(−1) − 18 If x = 56, f ′′(x) = 168 − 6x
= −18 − 18 f ′′(x) = 168 − 6(56)
= −36 < 0 = −168 < 0
∴ f(x) attains maximum at x = −1 ∴ f(x) attains maximum at x = 56
Maximum value is ∴ Two parts of 84 are 56 and 28
f(−1) = 3 (−1)3 − 9(−1)2 − 27(−1) + 15 = 30
For x = 3, f ′′(x) = 18x − 18 3) A rod of 108 meter long is bent to form a
rectangle. Find it's dimensions if the area is
maximum.
Solution: Let x be the length and y be the breadth
of the rectangle.
f ′′(3) = 18(3) − 18 ∴ 2x + 2y = 108
= 54 − 18 ∴ 2y = 108 − 2x
= 36 > 0 ∴ 2y = 2(54 − x)
∴ f(x) attains minimum at x = 3 ∴ y = 54 − x ................. (1)
Minimum value is, Now, area of the rectangle = xy
f(3) = 3(3)3 − 9(3)2 − 27(3) + 15 = −66 = x (54 − x)
∴ The function f(x) has maximum value
30 at x = −1 and minimum value − 66 f(x) = 54x − x2
at x = 3
f ′(x) = 54 − 2x
f ′′(x) = −2
108
For extreme value, f ′(x) = 0 3. Total cost function C = f(x), where x is
∴ 54 − 2x = 0 number of items produced,
∴ 2x = 54
∴ x = 27 Marginal cost = Cm = dC
f ′′(27) = −2 < 0 dx
∴ Area is maximum when x = 27, y = 27
Average cost = CA = C
∴ The dimension of rectangle are x
27m × 27m.
4. Total Revenue R = P.D where P is price and
∴ It is a square. D is demand.
R PD
Average Revenue RA = D = D = P
Total profit = R − C
EXERCISE 4.3 With this knowledge, we are now in a
position to discuss price elasticity of demand;
Q.1 Determine the maximum and minimum which is usually referred as 'elasticity of demand'
values of the following functions.
denoted by 'η'.
i) f(x) = 2x3 − 21x2 + 36x − 20
Elasticity of demand K _P . dD
ii) f(x) = x. logx D dP
16 We observe the following situations in the
iii) f(x) = x2 + x formula for elasticity of demand.
Q.2 Divide the number 20 in to two parts such
i) Demand is a decreasing function of price.
that their product is maximum.
dD
Q.3 A metal wire of 36cm long is bent to form ∴ dP < 0
a rectangle. Find it's dimensions when it's
area is maximum. Also, price P and the demand D are always
Q.4 The total cost of producing x units is positive.
Rs. (x2 + 60x + 50) and the price is
Rs. (180 − x) per unit. For what units is the ∴ K _P . dD >0
profit maximum? D dP
ii) If η = 0, it means the demand D is constant
function of price P.
4.4 Applications of derivative in Economics: ∴ dD < 0
dP
We ave discussed the following functions
in XIth standard. In this situation demand is perfectly
inelastic.
1. Demand Function D = f(P).
Marginal demand = Dm = dD iii) If 0 < η < 1, the demand is relatively
dP inelastic.
2. Supply function S = g (P) iv) If η = 1, the demand is exactly proportional
to the price and demand is said to be unitary
dS elastic.
dP
Marginal supply =
109
v) If η > 1, the demand is relatively elastic. Average propensity to save (APS) = S
x
Now let us establish the relation between
marginal revenue (Rm), average revenue Note here that x = Ec + S
(RA) and elasticity of demand (η)
Differentiating both sides w.r.t.x
As, Rm = dR 1 dEc dS
dD dx dx
But R = P.D. ∴ MPC + MPS = 1
∴ Rm = d (P.D) Also as x = EC + S,
dD ES
= P+D dP ∴ 1 c
dD xx
§¨©1 D dP · ∴ 1 = APC + APS
P dD ¹¸
= P ...... (1)
_P dD SOLVED EXAMPLES
But η = dP
. 1) The revenue function is given by
D R = D2 − 40D, where D is demand of the
commodity. For what values of D, the
_ revenue is increasing?
1 D . dP
η . = P dD
Substituting in (1) we get, Solution: Given R = D2 − 40D
Differentiating w.r.t.D
Rm = P ¨§1 1 ·
© K ¸
¹ dR
dD = 2D - 40
Rm = RA ©¨§1 1· (as RA = P) As revenue is increasing
K ¹¸
dR
Marginal propensity to consume: For ∴ dD > 0
any person with income x, his consumption
expenditure (Ec) depends on x. ∴ 2D − 40 > 0
∴ Ec = f(x) ∴ D > 20
Marginal propensity to consume Revenue is increasing for D > 20
(MPC) = dEc 2) The cost C of producing x articles is given
dx as C = x3 − 16x2 + 47x. For what values of
x the average cost is decreasing?
Average propensity to consume
Solution: Given C = x3 − 16x2 + 47x
(APC) = Ec Average cost CA = C
x x
Marginal propensity to save (MPS): If S is a CA = x2 − 16x +47
saving of a person with income x then Differentiating w.r.t.x
MPS = dS
dx
110
ddCxA = 2x − 16 iii) the price p for which elasticity of
demand is equal to one.
Now CA is decreasing if dCA <0 Solution: (i) Elasticity of demand
dx
that is 2x − 16 < 0 − p . dx
η = x dp
∴ x < 8 For x = 200 − 4p,
Average cost is decreasing for x < 8 dx = −4
3) In a factory, for production of Q articles, dp dx
standing charges are 500/-, labour charges η = −xp dp
are 700/- and processing charges are 50Q. ∴ .
The price of an article is 1700 - 3Q. For
what values of Q, the profit is increasing? = (200−−p4 p) (−4) (For p < 50)
Solution: Cost of poduction of Q aricles p
(50 −
C = standing charges + labour charges + ∴ η = p) (For p < 50)
processing charges
(ii) When P = 10
∴ C = 500 + 700 + 50Q 10
η =
∴ C = 1200 + 50Q (50 −10)
Revenue R = P.Q. 10
=
= (1700 − 3Q) Q
40
= 1700Q - 3Q2 = 0.25 < 1
Proit π = R − C ∴ Demand is inelastic for p = 10
= 1700Q − 3Q2 − (1200 + 50Q) When p = 30
∴ π = 1650Q − 3Q2 − 1200 30
η =
Differentiating w.r.t.Q, (50 − 30)
dπ 30
dQ = 1650 − 6Q η =
dπ 20
If profit is increasing, then dQ > 0
∴ 1650 − 6Q > 0 = 1.5 > 1
∴ Demand is elastic when p = 30
That is 1650 > 6Q (iii) To find the price when η = 1
∴ Q < 275 As η = 1,
∴ Profit is increasing for Q < 275 ∴ p p = 1
50 −
4) Demand function x, for a certain commodity ∴ p = 50 − p
is given as x = 200 − 4p, where p is the unit
price. Find ∴ 2p = 50
i) elasticity of demand as a function of p. ∴ p = 25
ii) elasticity of demand when p = 10; ∴ For elasticity equal to 1 then price is
p = 30. Interpret your results. 25/unit.
111
5) If the average revenue RA is 50 and EXERCISE 4.4
elasticity of demand η is 5, find marginal
revenue Rm. 1) The demand function of a commodity at
5P
Solution: Given RA = 50 and η = 5,
price is given as, D = 40 − . Check
Rm = RA ¨§1 1 · 8
© K ¸
¹ whether it is increasing or decreasing
function.
§¨©1 1·
= 50 5 ¹¸ 2) The price P for demand D is given as
P = 183 + 120D − 3D2; find D for which
= 50 § 4· price is increasing.
©¨ 5 ¹¸
Rm = 40 3) The total cost function for production of
articles is given as C = 100 + 600x − 3x2.
6) The consumption expenditure EC of a Find the values of x for which total cost is
person with income x, is given by decreasing.
EC = 0.0006x2 + 0.003x. Find average 4) The manufacturing company produces
propensity to consume, marginal propensity x items at the total cost of Rs. 180 + 4x.
to consume when his income is Rs. 200/- The demand function for this product is
Also find his marginal propensity to save. P = (240 − x). Find x for which (i) revenue
is increasing, (ii) profit is increasing.
Solution: Given EC = 0.006x2 + 0.003x
∴ APC = Ec 5) For manufacturing x units, labour cost is
x 150 − 54x and processing cost is x2. Price
of each unit is p = 10800 − 4x2. Find the
= 0.0006x + 0.003 values of x for which.
At x = 200,
APC = 0.0006 × 200 + 0.003 i) Total cost is decreasing
= 0.12 + 0.003 ii) Revenue is increasing
= 0.123 6) The total cost of manufacturing x articles
C = 47x + 300x2 − x4. Find x, for which
MPC = dEc average cost is (i) increasing (ii) decreasing.
dx
= d (0.0006x2 + 0.003x) 7) i) Find the marginal revenue, if the
dx average revenue is 45 and elasticity of
demand is 5.
= 0.0006 (2x) + 0.003
At x = 200, ii) Find the price, if the marginal revenue
is 28 and elasticity of demand is 3.
MPC = 0.0006 × 400 +0.003
= 0.24 + 0.003 iii) Find the elasticity of demand, if the
= 0.243 marginal revenue is 50 and price is
Rs. 75/-. § p 6 ·
¨ p 3 ¸
As MPC + MPS = 1 8) If the demand function is D= © ¹ ,
∴ MPS = 1 − MPC find the elasticity of demand at p = 4.
= 1 − 0.243
= 0.757
112
9) Find the price for the demand function • A function y = f(x) is said to have local
2p 3 minimum at x = c, if f '(c) = 0 and f ''(c) > 0.
D = 3 p 1 , when elasticity of demand is MISCELLANEOUS EXERCISE - 4
11
I) Choose the correct alternative.
.
14 1) The equation of tangent to the curve
y = x2 + 4x + 1 at (−1, −2) is
10) If the demand function is D = 50 − 3p −p2.
Find the elasticity of demand at (i) p = 5 (a) 2x − y = 0 (b) 2x + y − 5 = 0
(ii) p = 2. Comment on the result.
(c) 2x − y − 1 = 0 (d) x + y − 1 = 0
11) For the demand function D = 100 − p2 .
2 2) The equation of tangent to the curve
x2 + y2 = 5 where the tangent is parallel to
Find the elasticity of demand at (i) p = 10 the line 2x − y + 1 = 0 are
(ii) p = 6 and comment on the results.
(a) 2x − y + 5 = 0; 2x − y − 5 = 0
12) A manufacturing company produces
x items at a total cost of Rs. 40 + 2x. Their (b) 2x + y + 5 = 0; 2x + y − 5 = 0
price is given as p = 120 − x. Find the value
of x for which (i) revenue is increasing. (c) x − 2y + 5 = 0; x − 2y − 5 = 0
(ii) profit is increasing. (iii) Also find
elasticity of demand for price 80. (d) x + 2y + 5; x + 2y − 5 = 0
13) Find MPC, MPS, APC and APS, if the 3) If elasticity of demand η = 1 then demand
expenditure Ec of a person with income I is is
given as
(a) constant (b) in elastic
EC = (0.0003)I2 + (0.075)I
(c) unitary elastic (d) elastic
when I = 1000.
Let's Remember 4) If 0 < η < 1, then the demand is
(a) constant (b) in elastic
• A function f is said to be increasing at a (c) unitary elastic (d) elastic
point c if f '(c) > 0.
5) The function f(x) = x3 − 3x2 + 3x − 100,
• A function f is said to be decreasing at a x ∈ R is
• point c if f '(c) < 0. _P . dD (a) Increasing for all x ∈ R, x ≠ 1
Elasticity of demand K D dP (b) decreasing
(c) Neither, increasing nor decreasing
• Rm = P ©¨§1 1· = RA ©¨§1 1· (d) Decreasing for all x ∈ R, x ≠ 1
K ¸¹ K ¹¸
• For a person with income x, consumption 6) If f(x) = 3x3 − 9x2 − 27x + 15 then
or expenditure Ec and saving S, (a) f has maximum value 66
(b) f has minimum value 30
(i) x = Ec + S (c) f has maxima at x = −1
(ii) MPC + MPS = 1 (d) f has minima at x = −1
(iii) APC + APS = 1
• A function y = f(x) is said to have local
maximum at x = c, if f '(c) = 0 and f''(c) < 0.
113
II) Fill in the blanks: 6x + 3y − 4 = 0 x 2
x 1
1) The slope of tangent at any point (a,b) is 3) Show that the function f(x) = , x ≠ −1
called as .......... is increasing
2) If f(x) = x3 − 3x2 + 3x − 100, x ∈ R then 4) Show that the function f(x) = 3 + 10, x ≠ 0
f ′′(x) is .............. is decreasing x
3) If f(x) = 7 − 3, x ∈ R, x ≠ 0 then f ′′(x) is 5) If x + y = 3 show that the maximum value
................x. of x2y is 4.
4) A rod of 108m length is bent to form 6) Examine the function for maxima and
a rectangle. If area at the rectangle is minima f(x) = x3 − 9x2 + 24x
maximum then its dimension are ...........
Activities
5) If f(x) = x.log.x then its maximum value is
............
III) State whether each of the following is (1) Find the equation of tangent to the curve
True or false: x − y = 1 at P(9,4).
1) The equation of tangent to the curve Solution : Given equation of curve is
y = 4xex at ( −1, −4 e ) is y.e. + 4 = 0.
x− y=1
2) x + 10y + 21 = 0 is the equation of normal
to the curve y = 3x2 + 4x − 5 at (1,2). Diff. w.r.to x
3) An absolute maximum must occur at a ∴ 2 1 − 1 dy =0
critical point or at an end point. x 2 x
4) The function f(x) = x.ex(1−x) is increasing on ∴ 1 dy = 1
( −1 , 1). 2 y dx 2
2
∴ 1 dy = 1
y dx x
IV) Solve the following. ∴ dy = y
dx x
1) Find the equation of tangent and normal to
the following curves § dy · 93
∴ ©¨ dx ¹¸ = =
c p = (9,4) 2
i) xy = c2 at (ct, t ) where t is parameter
ii) y = x2 + 4x at the point whose ordinate 93
is −3 ∴ slope of tangent is= 2
∴ Eqation of the tangent at P(9.4) is
iii) x = 1 , y = t − 1 , at t = 2
tt y − 4 = (x − 9)
iv) y = x3 − x2 − 1 at the point whose ∴ 2(y − 4) = 3(x − 9)
abscissa is −2.
2) Find the equation of normal to the curve ∴ 2y − = + 27
y = x − 3 which is perpendicular to the line
∴ 3x − 2y + 8 + = 0
∴ 3x − 2y + 35 = 0
114
(2): A rod of 108 meters long is bent to form (3): Find the value of x for which the function
rectangle. Find its dimensions if the area of f(x) = 2x3 − 9x2 + 12x + 2 is decreasing.
rectangle is maximum.
Solution: Given f(x) = 2x3 − 9x2 + 12x + 2
Solution: Let x be the length and y be breadth of
the rectangle. ∴ f ′(x) = x2 − +
∴ f ′(x) = 6(x − 1) ( )
∴ 2x + 2y = 108 Now f ′(x) < 0
∴ x + y = ∴ 6(x − 1)(x − 2) < 0
since ab < 0 ⇔ a < 0 & b > 0 or a > 0
∴ y = 54 −
&b<0
Now area of the rectangle = x y Case I] (x − 1) < 0 and x − 2 > 0
∴ x < and x >
= x Which is contradiction
Case II] x − 1 > 0 and x − 2 < 0
∴ f(x) = 54x − ∴ x > and x <
1< <2
∴ f ′(x) = − 2x f(x) is decreasing if and only if x ∈ (1,2).
∴ f ′(x) =
For extream values, f ′(x) = 0
∴ 54 − 2x = 0
∴ −2x =
∴ x = −54
−2
∴ x =
∴ f ′′(27) = −2 < 0
∴ area is maximum when x = 27, y = 27
∴ The dimensions of rectangles are
27m × 27m
vvv
115
5 Integration
Let's Study function is logx. Using the integral sign, we
• Method of Substitution
can write ∫ § 1 · dx = logx, x > 0.
¨© x ¹¸
• Some Special Integrals Let's Learn
• Integration by Parts
• Integration by Partial Fraction 5.1.2 Definition: Integral or primitive or
antiderivative of a function.
d If f(x) and g(x) are two functions such that
dx
Let's Recall [f(x)] = g(x) then f(x) is called an integral of
• Derivatives
g(x) with respect to x. It is denoted by ∫ g(x) dx
= f(x) and read as integral of g(x) w.r.t.x is f(x).
Here, we say that g(x) is the integrand.
5.1.1 Introduction This process of finding the integral of a
function is called integration. Thus, integration
In this chapter, we shall study the operation is the inverse operation of differentiation.
which is an inverse process of differentiation. We
now want to study the problem : the derivative For, example,
of a function is given and we have to determine
the function. The process of determining such a d (x4) = 4x3
function is called integration. dx
∴ ∫ 4x3 dx = x4
Consider the following examples: But, note that
(1) Suppose we want to determine a function d (x4 + 5) = 4x3
whose derivative is 3x2. Since we know that dx
dx3 = 3x2. Therefore, the required function
dx d (x4 − 8) = 4x3
is f(x) = x3. dx
x3 is called integral of 3x2 w.r.t.x and this is What is the observation? Can you
written as ∫ 3x2 dx = x3. generalize from the observation?
In general,
The symbol ∫ , called the integration sign, d (x4 + c) = 4x3
was introduced by Leibnitz. 'dx' indicates dx
that the integration is to be taken with
respect to the variable 'x'. where, c is any real number.
(2) Suppose we want to determine a function Hence, in general, we write
whose derivative is 1 Since we know that ∴ ∫ 4x3 dx = x4 + c
d (log x) = 1. x the required The number 'c' is called constant of
dx x Therefore, integration.
116
Note: (i) From the above discussion, it is clear that integration is an inverse operation of
differentiation. Hence integral is also called antiderivative.
(ii) In ∫ f(x)dx, f(x) is the integrand and x is the variable of integration.
(iii) 'I' is used to denote an integral.
Integrals of some standard functions.
1 d xn = nxn−1 d ª ax b n 1 º
dx « n 1 a »
dx «¬ »¼ ax b n
∴ ∫ xndx = x n +1 + c, n ≠ −1 ∴ ∫ (ax + b)n dx = (ax + b)n+1 + c, where, n ≠ −1
n +1 a(n +1)
2 d logx = 1 d log(ax + b) = 1d a
dx x dx ax + b dx (ax + b) = (ax + b)
∴ ∫ § 1 · dx = log | x | + c ∴∫ § 1 · dx = log ax + b +c
©¨ x ¸¹ ¨© ax ¹¸ a
b
3 d ax = ax log a d apx + q = apx + q(loga) d (px + q) = apx + q.ploga
dx dx dx
ax
∴ ∫ axdx = log a + c, a > 0, a ≠ 1 ∴ ∫ apx + qdx = a px+q + c, a > 0, a ≠ 1
p log a
4 d ex = ex loge = ex d epx + q = epx + q d (px +q) = epx + q.p
dx dx dx
∴ ∫ exdx = ex + c
∴ ∫ epx + qdx = e px+q +c
p
Rules of integration: ³ ³ ³>k1 f1(x) r k2 f2 (x) r ... r kn fn (x)@ dx k1 f1(x)dx r k2
³ ³ ³k1 f1(x)dx r k2 f2 (x)dx r ... r kn fn (x)dx
³5.1.3
integrable
fuTnhcetoiornemo>kf1x1f1:a(nxId)f f is a real valued x) @ dx
rk kis2 fa2 c(xo)nrst.a.n. rt,kthn efnn(
³>k. f (x)@ dx k ³ f (x)dx Result 1:
Theorem 2: If f and g are real valued ³ f (x)dx F (x) c then
integrable functions of x, then
³ f (ax b)dx F (ax b) c
³> f (x) g(x)@ dx ³ f (x)dx ³ g(x)dx a
Theorem 3: If f and g are real valued SOLVED EXAMPLES
integrable functions of x, then
³> f (x) g(x)@ dx ³ f (x)dx ³ g(x)dx (1) Evaluate ∫(7x − 2)2 dx
Generalization of (1), (2) and (3) Solution: I = (7x 2)2 1 c
Corollary 1: If f1, f2, .......... fn are real (2 1) 7
integrable functions of x, and k1, k2, .......... kn are
scalar constants then = (7x 2)3 c
21
117
(2) Evaluate ³ ª««¬§©¨11 t 7 (4t º = x4 3x2 3log x 1 c
3 5)4 » dt 4 2 2x2
·
¹¸ ¼» 1 (8x3 1 12 x 2 6x)
x2 x2
©¨§11 t 7 (6) Evaluate ³ (2x 1)3 dx ³ dx
3
Solution : I = ³ · dt ³(4t 5)4 dt
¸¹
³ x1S2 o(2luxt io1n):3 dIx= ³ (8x3 1 12 x 2 6x) dx
x2
= ©§¨11 t ·7 1
3 ¸¹ u 3 (4t 5)4 1 u 1 c = ³ ¨©§8x 12 6 1 · dx 4x2 12x 6 log x 1 c
x x2 ¸¹ x
7 1 4 1 4
= 3 ©¨§11 t 8 1 (4t 5)5 c (7) Evaluate ³ 5(x6 1) dx ³ 5( x 2 1)(x4 x2 1) dx
8 3 20 x2 1 (x2 1)
·
¹¸
ª º ³ S5(oxxl2u6 t i1o1n) :dxI = ³ 5( x 2 1)(x4 x2 1) dx
«¬ »¼ (x2 1)
(3) Evaluate ³ (6 x 1 5)4 1 dx
(8 3x)9
= ³ 5(x4 x2 1)dx x5 5 x3 5x c
Solution : I = ³(6x 5) 4 dx ³(8 3x) 9 dx 3
(6 x 5) 3 1 ª (8 3x) 8 º 1 (8) Evaluate ³ x3 § 2 3 2 dx
= 3 6 « 8 » 3 ©¨
¬ ¼ ·
x ¹¸
u u c
= §¨© 1 81 · 1 § 1 · 1 Solution: I = ³ x3 § 4 12 9 · dx
¸¹ (6x ©¨ 24 ¹¸ (8 3x)8 ©¨ x x2 ¸¹
5)3 c
= ³(4x3 12x2 9x)dx
(4) Evaluate ³ dx = 4 x4 12 x3 9 x2 c
x x 2 4 32
Solution: I = ³ 1 u x x 2 dx = x4 4x3 9 x2 c
x x 2 x x 2 2
= ³ x x 2 dx 1 x 2)dx (9) Evaluate ³ x3 4x2 6x 5 dx
x (x 2) ³( x x
2
1 ª¬ ³ x1/ 2dx ³ ( x 2)1/ 2 dx º¼ Solution: I = ³ § x2 4x 6 5 · dx
= ©¨ x ¹¸
2
= 1 ª x3/2 (x 2)3/ 2 º c = ³ x2dx 4 ³ x dx 6 ³ dx 5 ³ 1 dx
2 « 32 3 2 » x
¬ ¼
1 ¬ª x3/ 2 (x 2)3/ 2 º¼ c = x3 4 x2 6x 5log x c
= 32
3
§ 1 · 3 § 1 3 ·
©¨ x ¹¸ ¨© x3 x ¸¹
(5) Evaluate: ³ x dx ³ x3 3x dx(10) Evaluate ³(ealog x exloga )dx ³(eloge x a eloge a x )dx
3 1
S³ §©¨oxlu ti1oxn·¹¸: § x3 3 ·
dIx= ³ ¨© x3 3x x ¸¹ dx ³(ealog xS oleuxtloigoan):dxI = ³(eloge x a eloge a x )dx
= x4 1 3x2 3log x c = ³(xa ax )dx xa 1 ax c
4 2x2 2 a 1 log a
118
(11) Evaluate ³ § e(1 5t ) 1 · dt 5.2 Method of Change of Variable or Method
¨© 5t ¸¹ of Substitution
1
In this method, we reduce the given
Solution: I = ³ e(1 5t)dt ³ § 1 1 · dt function to standard form by changing variable
¨© 5t ¸¹ x to t, using some suitable substitution x = φ(t)
e(1 5t ) § log 5t 1 · c Theorem 4 : If x = φ(t) is a differentiable
( 5) ¨ 5 ¸ function of t, then
I= ¹
©
³ f (x)dx ³ f >I(t)@ I '(t)dt
(12) If f ′(x) = 8x3 + 3x2 − 10x − k, f(0) = −3 and
f(−1) = 0, find f(x) 5.2.1 Corollary 1: > f @(x) n 1 c
Solution: By the definition of integral ³> f (x)@n f '(x)dx (n 1)
f(x) = ³ f '(x)dx ³(8x3 3x2 10x k)dx
= 8 ³ x3dx 3 ³ x2dx 10 ³ x dx k ³ dx SOLVED EXAMPLES
= 8 x4 3 x3 10 x2 kx c 1. Evaluate ∫ (log x)7 dx
43 2 x
f(x) = 2x4 x3 5x2 kx c Solution: Put log x = t
Now f(0) = −3 gives c = −3
and f(−1) = 0 gives k = 7 ∴ 1x dx = dt
f(x) = 2x4 x3 5x2 7x 3
∴ I ³ t7dt t7 1 c 1 (log x)8 c
7 1 8
2. Evaluate ³ 1 dx
2x x n
EXERCISE 5.1
Solut ion: I = ³ 1 dx
(i) Evaluate ³ 2 dx § 1 ·
5x 4 5x 2 2x ©¨ xn ¹¸
(ii) Evaluate ³ § x x2 · dx = ³ x( xn dx
¨1 2! ¸ 2 1
¹ n 1)
© Put x(n 1) t
(iii) Evaluate ³ 3x3 2 x dx ∴ (n 1)xndx dt
x
∴ xndx dt
(iv) Evaluate ³(3x2 5)2 dx n 1
(v) Evaluate ³ 1 dx ∴ I = ³ 1 1) u dt 1)
x(x 1) (2t (n
(vi) If f ′(x) = x2 + 5 and f(0) = −1, then find the = 1 ³ dt 1)
value of f(x). (n 1) (2t
(vii) If f ′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and = 1 log 2t +1 + c
f(1) = 4, find f(x) n+1 2
(viii) If f ′(x) = x2 kx 1 , f(0) = 2 and f(3) = 5, I= 1 2xn+1 +1 + c
find f(x) 2 log
2(n +1)
119
3. Evaluate ³ 4x 6 3 dx I ³ 1 dt 11 1
(x2 3x 5)2 = ∫ dt = log t + c
t3 3 t 3
Solution: I = ³ 2(2x 3) 3 dx
(x2 3x 5)2 1 e3x +1 + c
= log
Put (x2 3x 5) t 3
6. Evaluate ³ 1 dx
x(log x 1)
∴ (2x 3)dx dt Solution: Put log x 1 t
³ I 2dt § 3 · 1 dx = dt
x
3 2 ³ t©¨ 2 ¸¹dt
t2 11
ª º§ 1· I ³ log x 1 u x dx
« t ©¨ 2 ¸¹ » 4 c 1
= 2 « » t ³ dt log t c log log x 1 c
« § 1 · » t
¬« ¨© 2 ¹¸ ¼»
d (ex x)
I = 4 c 7. Evaluate ³ ex 1 dx = ³ dx dx
x2 3x 5 ex x
ex x
= log ex + x + c
4. Evaluate ³ (x 1)(x log x)4 dx 8. Evaluate ³ e x 1 xe 1 dx
3x ex xe
Solution: I = § 1 · ³(x log x)4 § x 1 · dx
¨© 3 ¹¸ ¨© x ¹¸
Solution: Put ex + xe = t
= §¨© 1 · §©¨1 1 · ∴ (ex e xe 1)dx dt
3 ¹¸ x ¹¸
³( x log x)4 dx ∴ e(ex 1 xe 1)dx dt
Put x log x t ?¨©§1 1 · dx dt ∴ (ex 1 xe 1)dx dt
x ¸¹ e
= ¨§© 1 · ³(t )4 dt § 1· t5 c I ³ 1 dt 11 1
3 ¸¹ ¨© 3 ¸¹ 5 ³ dt log t c
te e t e
= §©¨ 1 · (x log x)5 c = 1e log ex + xe + c
15 ¹¸
5.2.2 Corollary 2: ³ ª f '(x) º dx log f (x) c 9. Evaluate ∫ 1 dx
«¬ f (x) ¼» x log x.log(log x)
e3x Solution: I = 1 1 dx
e3x ∫. x.log x
5. Evaluate ³ dx log(log x)
1
S olution: Put e3x 1 t Put log (log x) = t
∴ 3e3xdx = dt
11
∴ e3xdx = dt ∴ log x x dx = dt
3
∴ 1
dx = dt
x log x
120
1 ³13. Evaluate : xn 1 dx
I ³ dt log t c 1 xn
t Solution: Put xn = t
= log log (log x) + c
10. Evaluate ³ 10 x9 10x.log10 dx ∴ n xn 1dx dt
10x x10 ∴ xn 1dx dt / n
S olution: Put 10x x10 t ³ ³ I = 1 dt 1 (1 t) 21dt
∴ (10x.log10 10 x9 )dx dt 1 t n n
§1·
1 . (1 t)©¨ 2 ¹¸
I ³ dt log t c 1 2 1 xn c
t = c n
n §1·
©¨ 2 ¸¹
= log 10x + x10 + c
11. Evaluate ³ 1 dx ³ 1 dx 14. Evaluate ³ 3x2 dx
e 1 x3
1 x 1 1
ex
S olution: Put 1 x3 t
ex d (ex 1) ∴ 3x2dx = dt
ex dx
Solution: I = ³ 1 dx ³ ex 1 dx
I ³ 1 3x2dx = dt
dt
log ex 1 c
t
§ e2x 1· 2 t c 2 1 x3 c
¨ ¸
e2x 1dx © e x ¹dx
e2x 1 § e2x ·
³ ³12.
Evaluate I = 1 Integral of Type: ³ (ax b) cx d dx
¨ ex ¸
ex (ex e x ) © ¹
³Solution: I = ex (ex e x ) dx
15. Evaluate ³ (2x 1) x 4 dx
ex e x d (ex e x )
ex e x dx Solution: Put (x 4) t
³ ³ dx ex e x dx ∴ dx = dt
x t 4
I log ex e x c
I = ³>2(t 4) 1@ t dt ³ (2t 9) t dt
5.2.3 Corollary 3: ³ ª f '( x) º dx 2 f (x) c = ³ ¨§ 3 1 · 31
» © 2 2 ¸
¹ 2 t 2 dt 9 t 2 dt
« f (x) »¼ ³ ³ 2t 9t dt
«¬
§5· §3·
5.2.3 Corollary 4: 2 t©¨ 2 ¹¸ 9 t©¨ 2 ¹¸ c
45 3
n n > f ( @x) n 1 = §5· §3· 5 (x 4)2 6(x 4)2 c
³ ª f '(x) º dx c ¨© 2 ¹¸ ©¨ 2 ¹¸
« n f (x) » n 1
«¬ »¼
1
16. Evaluate ³ (5 3x)(2 3x) 2 dx
121
Solution: Put 2 − 3x = t ³1 5t2 30t 45 8t 24 28 dt
∴ −3 dx = dt = §3·
2 4t ©¨ 2 ¹¸
dx = −dt / 3 Also x = (2 − t)/3 ³1 5t2 22t 49
ª 3§¨© 2 t ·º § 1· § dt · = 8 §3· dt
«¬5 3 ¸¹¼» ©¨ 3 ¸¹
³ (t)©¨ 2 ¸¹ t ©¨ 2 ¹¸
I=
³ 1 § 1· ³1 § 5t § 1 · § 1· § 3 · · dt
¨¨© ©¨ 2 ¸¹ ¸¹¸
= (5 2 t)(t)©¨ 2 ¸¹dt = 8 22t¨© 2 ¹¸ 49t©¨ 2 ¸¹
3
³ = 1 3 § 1· ³5 § 1· 22 § 1 · 49 § 3 ·
2 ¸¹
(3 t)(t)¨© 2 ¸¹dt = t ¨© 2 ¹¸dt t ©¨ 2 ¸¹dt t ©¨
³ ³ dt
88 8
1 § ·
= 3 ³ ¨©¨ § 1 · § 1 · ¹¸¸ dt §3· §1· § 1·
2 ¹¸ 2 ¹¸ t ©¨ 2 ¸¹ t ¨© 2 ¸¹ t ¨© 2 ¸¹
3(t )©¨ (t )©¨ §3· §1· § 1·
5 11 49
§ 1 · = 8 4 4 c
t ©¨ 2 ¸¹dt §1·
³ ³ = 3 1 ©¨ 2 ¹¸ ©¨ 2 ¸¹ ©¨ 2 ¸¹
t¨© 2 ¹¸dt
33
1 3 5 (2x 3)3/2 11 (2x 3)1/2
=
= t 2 1 t2 c 12 2
§1· 3§3· 2 2 3x
©¨ 2 ¸¹
¨© 2 ¸¹ 49 2x 1 c
=
2 32
= −2 2 − 3x
23 18. Evaluate ³ x7 dx
(2 3x)2 c (1 x4 )2
9 ³ x7 x4 x3
(1 x4 )2 (1 x4 )2
Solution: Let, I = dx ³ dx
³17. Evaluate 5x2 4x 7 dx
3 Put 1 x4 t
(2x 3)2
Solution: Put 2x + 3 = t ∴ 4x3dx = dt
∴ x3dx = dt
∴ 2 dx = dt 4
Also x4 t 1
∴ dx = dt
2
Also x (t 3) ³ I t 1 dt 1 ³ § 1 1 · dt
2 (t)2 4 4 ¨© t t2 ¹¸
5 ¨§© t 3 2 4 § t 3 · 7
2 ©¨ 2 ¹¸
³ I · dt ³ ³= 1 ³ § 1· dt 1 § 1 · dt 1 ³ § 1 · dt 1 (t 2 )dt
¸¹ 4 ©¨ t ¸¹ 4 ¨© t2 ¸¹ 4 ©¨ t ¹¸ 4
§3· 2
t ©¨ 2 ¸¹
§ t2 6t 9 · 2 t 3 7 = 1 t 1 . t 1 c 1 11
5¨ 4 ¸ log log t c
¹ 4 4 1 4 4 t
³1 ©
dt
= §3·
2 1 11
t ©¨ 2 ¹¸ log 4 1+ x4
= 1+ x4 + +c
4
122
EXERCISE 5.2 SOLVED EXAMPLES
Evaluate the following. ³(1) Evaluate 4ex 25 dx
2ex 5
(i) ³ x 1 x2 dx
³(ii) x3 dx Put Numerator = A (Denominator + B
1 x4 ( d Denominator)
dx
ª d 5)º»¼
³(iii) (ex e x )2 (ex e x ) dx 4 ex − 25 = A(2 ex 5) B ¬« dx (2 ex
³(iv) 1 x dx = A(2 ex 5) B(2 ex )
x e x = (2A 2B)ex 5A
(v) ³ (x 1)(x 2)7 (x 3) dx Comparing the coefficients of ex and
constant term on both sides, we get
(vi) ∫ x 1 dx
logx
x5 2A 2B 4 & 25 5A
x2 1
³(vii) dx ∴ A 5 and B 3
(viii) ³ 2x 6 dx ∴ 4ex 25 5(2ex 5) 3(2ex )
x2 6x 3
∴I = 5(2ex 5) 3(2e x )
2ex 5
³ dx
(ix) ³ 1 dx = ³ ª«5 3(2ex ) º dx
x x ¬ 2ex 5 »
¼
1
= 5 ³ dx 3 ³ 2ex dx
(x) ³ x(x6 1) dx 2ex
5
Activities = 5x 3log 2ex 5 c
For each of these integrals, determine a EXERCISE 5.3
strategy for evaluating. Don't evaluate them,
just figure out which technique of integration Evaluate the following.
will work, including what substitutions you
will use. ³1) 3 e2t 5 dt
4 e2t 5
1) ∫ 1 dx 3 20 12e x
xlogx 3ex 4
2) ³ x2 5x 4 dx ³2) dx
x 5 ex
x2 5x 36 e2x
³ ³3) 7 dx 4) dx 3ex
2ex
³3) 4 dt
8
³5.3 Integrals of the form a ex b dx ³4) 2ex 5 dt
cex d 2ex 1
where a,b,c,d ∈ R
123
5.4.1 Results ³ 1
1 dx
1 log x a c 3. 9x2 25
³1. x2 a2 dx 2a x a
11
Solution: I = 3 ³ dx
³2. 1 1 log a x c x2 25
a2 x2 dx 2a a x 9
³3. 1 dx log x x2 a2 c 1 1 dx
x2 a2 = 3 ³ § 5 · 2
¨© 3 ¹¸
³4. 1 dx log x x2 a2 c x2
x2 a2 1 § 5 · 2
= 3 log x ¨© 3 ¹¸
SOLVED EXAMPLES x2 c
Evaluate the following. 1 11
1 11 4. ³ dx 2³ dx
4x2 9 x2 9
1. ³ 9x2 4 dx 9 ³ x2 4 dx
1 11 4
9 ³ Solution:dIx= 2³ dx
11 4x2 9 x2 9
³Solut ion: I = 9 x2 ¨©§ 32 ·¸¹2 dx 4
2 1 1 dx
= ©§¨ 91 · 1 x = 2 ³ § 3 · 2
¹¸ §2 3 ©¨ 2 ¸¹
©¨ 3 · log 2 c x2
¹¸
2 x
3
1 § 3 · 2
= 2 log x ©¨ 2 ¹¸
1 log 3x 2 c x2 c
=
12 3x 2
1 11 5. ³x 1
dx
2. ³ 16 9x2 dx 9 ³ 16 x2 dx (log x)2 5
1 1 19 Solution: Put log x t ? 1 dx dt
x
³S1o6l ut9ioxn2 :dIx= 9 ³ 16 x2 dx
∴ I ³ 1 dt
9 t2 ( 5)2
11
³ = 9 2 dx
§4·
©¨ 3 ¸¹ x2 = log t t2 ( 5)2 c
4 = log log x (logx)2 ( 5)2 c
= ¨©§ 91 ¹¸· 2¨©§134 ·¸¹ log 334 xx c
1 4 3x c
= log 4 3x
24
124
5.4.2 Integrals of the form ∫ P(x) dx where 5.4.3 Integrals of the type ³ ax2 1
Q( x) dx
bx c
degree (P(x)) ≥ degree (Q(x)). In order to find this type of integrals we
may use the following steps :
P(x)
Method: To evaluate ∫ Q(x) dx Step 1 : Make the coefficient of x2 as one if
11
1. Divide P(x) by Q(x).
it is not, then a ³ x2 b x c dx
After dividing P(x) by Q(x) we get
quotient q(x) and remainder r(x). aa
2. Use Dividend = quotient × divisor Step 2: Add and subtract the square of the
+ remainder
half of coefficient of x that is § b ·2 to complete
¨© 2a ¸¹
P(x) = q(x) × Q(x) + r(x)
11
the square ³a dx =
P(x) r(x) b § b 2 § b 2 c
Q(x) = q(x) + Q(x) x2 a x ¨© 2a ¨© 2a a
· ·
¹¸ ¹¸
P(x) r(x) 11
Q(x) Q(x) ³a dx
∫ dx = ∫q(x) dx + ∫ dx § b 2 § 4ac b2 ·
¨© 2a ¨ 4a2 ¸
x · © ¹
¹¸
3. Using standard integrals, evaluate I.
SOLVED EXAMPLES SOLVED EXAMPLES
x3 x 1 Evaluate the following.
x2 1
1. Evaluate I ³ dx 1
x3 x 1 1. ³ 2x2 x 1 dx
x2 1
Solution: I ³ dx 11
xQ Solution: I = 2 ³ x2 1 x 1 dx
)D x2 1 x3 x 1 22
x3 x 11
= ³ dx
2 1 1 1 1
x2 x
2x 1 R 2 16 16 2
∴ § R · 11
I ³ ¨© Q D ¹¸ dx ³ = 2 2 dx
§ 1 · § 3 ·
2 ¨© x 4 ¸¹ ©¨ 4 ¹¸
I ³ ª x 2x 1º dx ªº
¬« x2 1 »¼
§ x 1 · 3
1 « 1 » ¨© 4 ¹¸ 4
2 x 1 = 2 « §3 » log c
= ³ x dx ³ dx ³ x2 1 dx « ¨© 4 · » § 1 · 3
x 2 1 ¬« ¸¹ »¼ ¨© 4 ¸¹ 4
2 x
³ = x2 log x2 1 1 1
x2 12 dx c x
2
2
x2 log x2 1 1 log x 1 c 1 c
= 2 2 x 1 = 3 log
x 1
125
1 2x 1 4. ³ 1 dx
log
3 2(x 1) c (x 2)(x 3)
Solution: I = ³ 1
dx
2. 1 dx x2 5x 6
³1 x x2 1
1 = ³ x2 5x 25 25 6 dx
Solution: I = ³ 1 1 1 x x2 dx 44
44
= ³ 1 dx
³ 1 dx 22
= ¨©§1 1 · § x 2 x 1· § x 5 · § 1 ·
4 ¸¹ ¨© 4 ¸¹ ©¨ 2 ¹¸ ¨© 2 ¸¹
³ = 1 dx 22
§ 5 2 § x 1 2 = log §©¨ x 52 ¸¹· ©¨§ x 52 ·¹¸ ©¨§ 12 ¸·¹ c
¨ 2 ©¨ 2
© · ·
¸ ¸¹
¹
§ 5·
= 2§¨© 125 ¸¹· log 2255 ¨©§¨©§ xx 1122 ·¸¹¸·¹ c = log ¨© x 2 ¹¸ c
5. ³ 2x 1 dx
x2 2x 3
1 5 1 2x c Solution: I ³ 2x 2 1 dx
= log
5 5 1 2x x2 2x 3
ex = I ³ 2x 2 dx ³ dx
6ex x2 2 x x2 2x 3
³3. dx 3
e2x 5
1
Solution: Put ex = t = 2 x2 2x 3 log
x2 2x 1 2
∴ exdx = dt
I = ³ t 2 dt 5 = 2 x2 2x 3 log (x 1) x2 2x 3 c
6t
= ³ t 2 6t dt 9 5 6. ³ x 1 dx
9 x2 3x 2
³ = (t dt 22 S olution: x 1 A d (x2 3x 2) B
3)2 dx
x 1 A(2x 3) B 2Ax 3A B
1 (t 3) 2 c ∴ 2A 1 and 3A B 1 Solving we get
= 2(2) log (t 3) 2
1 −1
ex +1 A = 2 and B = 2
ex + 5
1 +c 11
= log (2x 3)
2 2
4 x2 3x 2
∴ I ³ dx
126
= 1 2x 3 1 dx 2
x2 3x 2 x2 3x 2
2³ dx 2 ³ = log t t2 5 c
³= 2 x2 3x 2 1 1 = log log x (log x)2 5 c
dx
2 2 22
§ 3 · § 1 ·
©¨ x 2 ¸¹ ©¨ 2 ¸¹ x2dx
2. ³
x6 2x3 3
= x2 3x 2 1 log § x 3 · x2 3x 2 c Solution: Put x3 = t
2 ©¨ 2 ¹¸
=3x2dx d=t : x2dx dt
5.4.4 Integrals reducible to the form 3
³ 1 I = ³ 1 dt
dx t2 2t 3 3
ax2 bx c
11
= ³
To find this type of integrals we use the 3 t2 2t 1 2
following steps: 1 t2 + 2t + 3
Step 1: Make the coefficients of x2 as one = 3 log (t +1) +
if it is not, ie 1 dx = 1 log (t +1) + t2 + 2t + 3 + c
x2 bx c . 3
a³
= 1 log (x3 +1) + x6 + 2x3 + 3 + c
aa 3
Step 2: Find half of the coefficient of x.
1 5.4.5 Integrals of the form ³ px q dx
Step 3: Add and subtract ( coeff.of x)2 ax2 bx c
inside the square root so tha2t the square
root is in the form To find this type of integrals we use the
following steps:
2 2
§ x b 4ac b2 or 4ac b2 § x b Step 1: Write the numerator px + q in the
¨© 2a · 4a2 4a2 ¨© 2a · following form
¸¹ ¸¹
Step 4: Use the suitable standard form for px q A d (ax2 bx c) B
dx
evaluation.
SOLVED EXAMPLES Step 2: Obtain the values of A and B by
equating the coefficients of same power of
1. ³x dx x on both sides.
(logx)2 5
Step 3: Replace px + q by A(2ax + b) + B
Solution : Put log x = t in the given integral to get in the form of
∴ dx = dt ³ px q dx
x ax2 bx c
³I = 1 A 2ax b dx B 1 dx
dt
t2 5 ax2 bx c ax2 bx c
1 dx
= ³ dt
t2 5 = A ax2 bx c B
2
ax2 bx c
127
1 1
(2x 3)
2 2 dx
SOLVED EXAMPLES = ³ x2 3x 2 dx − ³ x2 3x 2
1 1
2 x2 + 3x + 2 − log
= 2
2
1. ³ 2x 8 dx § x 3 · § x 3 2 § 1 2 c
x2 6x 13 ¨© 2 ¸¹ ©¨ 2 ©¨ 2
· ·
¹¸ ¸¹
Let 2x + 8 = A d (x2 + 6x + 13) + B 1
dx
³= 1 (2x 3) dx − ³ 2 dx
2 x2 3x 2 22
§ 3 · § 1 ·
2x + 8 = A (2x + 6) + B ¨© x 2 ¹¸ ©¨ 2 ¸¹
∴ A = 1, B = 2 x2 3x 3 1 log § 3 · x2 3x 2 c
2 ©¨ 2 ¹¸
2x 6 2 = x
2 6x dx
= ³ dx ³
x2 6x 13
x 13
3. ³ 2x 1 dx
x2 2x 1
= x2 6x 13 2 log x 3 x2 6x 13 c Solution : Let 2x + 1 = A d (x2 + 2x +1) + B
f ' xx dx dx
f = A (2x + 2) + B
(using ³ 2 f x c in the 1st integral)
2x + 1 = 2Ax +(2A + B)
2. ³ x 1 dx Comparing the coefficient of x, we get
x 2
2 =2A and 1 = 2A + B
Solut ion: I = ³ xx 12 xx 11 dx A = 1 and B = −1
= ³ x2 x 31x 2 dx I = ³ (2x2x 21)x 11 dx
Let x + 1 = A d (x2 + 3x + 2) + B = ³ x(22 x 2x2) 1 dx − ³ x2 12x 1 dx
dx
³ (2x 2) dx − ³ 1
= A(2x + 3) + B x2 2x 1 dx
(x 1)2
Comparing the coefficient of x, we get = 2 x2 2x 1 log x 1 c
1 = 2A and 1 = 3A + B
1 −1 4. ³ 1 x dx
A = and B = x
22
11
(2x 3)
³x 1 2 dx (1 x)(1 x) dx (1 x) dx
= ³ x2 3x 2 dx = 2 x2 3x 2 Solution: I = ³ x(1 x) ³ x2 x
128
(1 x)(1 x) dx= ³ (1 x) dx ³5) 16 x3 25 dx
x(1 x) x2 x x8
Let x + 1 = A d (x2 + x) + B 1
dx
³6) a2 b2 x2 dx
x + 1 = A(2x + 1) + B = 2Ax +(A + B) 1
Comparing the coefficient of x, we get 7) ³ 7 6x x2 dx
1 = 2A and 1 = A + B ³ 1
dx
11 8) 3x2 8
A = and B =
1
22
11 9) ³ dx
x2 4x 29
x 1 dx (2x 1)
x2 x ³2 2 dx
³ x2 x ³ 1
dx
10) 3x2 5
11
(2x 1)
= ³ 2 x2 x dx ³ 2 dx 11) ³ 1 dx
x2 x x2 8x 20
1 5.5 Integration by Parts.
³ ³= 1 (2x 1) dx 2 dx 5.5.1 Theorem 5: If u and v are two functions of
2 x2 x 22 x then
§ 1 · § 1 ·
©¨ x 2 ¸¹ ©¨ 2 ¸¹
1 1 1 1 22 ³ u.v dx u³ v dx ³ ª«¬³ v dx. du º dx
2 2 § 2 · § 2 · § 1 · dx ¼»
= x2 x log ¨© x ¹¸ ©¨ x ¹¸ ©¨ 2 ¸¹ c
2
The method of integration by parts is used
= x2 x 1 log § x 1 · x2 x c when the integrand is expressed as a product of
2 ¨© 2 ¹¸ two functions, one of which can be differentiated
and the other can be integrated conveniently.
Note:
EXERCISE 5.4 (1) When the integrand is a product of two
functions, out of which the second has to
Evaluate the following. be integrated (whose integral is known),
hence we should make proper choices of
1 first function and second function.
1) ³ 4x2 1 dx (2) We can also choose the first function as
the function which comes first in the word
1 'LAE' where
2) ³ x2 4x 5 dx
1 L - Logarithmic Function
3) ³ 4x2 20x 17 dx
³4) x dx A - The Algebraic Function
20 x 2
4x4 3 E - The Exponential Function
129
SOLVED EXAMPLES = ∫(logt).1 dt
1. ³ x e 2xdx = = t(llooggtt) ³³1.1tdttd t ³ ¬ª«cddt (logt)³1.dtº¼» dt
= tlogt ³ dt c
³ ³Solution: I = x ³ e 2xdx ª d ( x) e 2 x dx º dx
¬« dx »¼ = t(log t 1) c
³ = x e 22x 1. e 2x dx c = (log x).(log (log x) 1) c
2
1 xe 2 x 1 e 2 x c ³5. x.2 3xdx
=
24 ª d º
(2 3x )dx «¬ dx (2 3x »¼
³ ³ ³Solution: I = x x )dx dx
2. ∫logx dx x(2 3x ) (2 3x ) dx c
Solution: I = ∫(logx).1 dx = 3 log2 3 log 2
= (logx) ∫ 1.dx − ³ ª d (logx)³1.dx¼º» dx ³x(2 3x )
«¬ dx
= 3 log 2
1 2 (2 3x ) dx c
1 xdx 3 log
x
= xlogx − ³ c x(2 3x § 2 3x ·
¨¨© ¹¸¸
= xlogx − ∫ dx + c = 3 log ) 3 1 2 3 log 2 c
2 log
= x(log x − 1) + c x 2 3x 1 1
9 (log 2)2
= 3 log 2 2 3x c
3. ∫ x3log x dx
Solution: I = ∫ (log x)x3 dx Integral of the type ³ ex{ f (x) f '(x)}dx
= (logx)³ x3dx ³ ªd (logx)³ x3dx º dx These integrals are evaluated by using
¬« dx ¼»
³ ex{ f (x) f '(x)}dx ex f (x) c
³ = x4 log x 1 x3dx c
1. ³ ex § x log x 1· dx
44 ©¨ x ¸¹
³x4 log x 1 x4 dx c x § 1·
4 44 ¨© 2 ¸¹
= Solution: I = ³ e log x dx
= x4 log x x4 c 1
4 16 =Put log x f=(x) f '(x)
x
log(log x) 1 ³ ex{ f (x) f '(x)}dx ex f (x) c
x x
4. ∫ dx = ∫ log (logx) dx = exlogx + c
Solution: Put log x = t (1 x2 )
(1 x)2
∴ 1 dx = dt ³2. ex dx
x
I = ∫log t dt ³Solution: I = ex (x2 1) 2 dx
(1 x)2
130
³ = ex ªx 1 2 º dx After applying these two steps the integrals
« 1 1)2 » reduces to one of the following two forms
¬ x (x ¼
³∫ a2 x2 dx, x2 a2 dx which can be
Put f (x) x 1
x 1 evaluated easily.
2 SOLVED EXAMPLES
f '(x) x 1 2
³ ex{ f (x) f '(x)}dx ex f (x) c 1. ³ 4x2 5 dx
= ex § x 1 · c Solution: I = ³ 2 x2 5 dx
©¨ x 1 ¸¹ 4
§ 5· 2
x2 ¨©¨ 2 ¹¸¸
3. ³ ex (x 3) dx = 2³ dx
(x 4)2
Solut ion: I = ³ ex ((xx 44 )21) dx = 2 ª x x2 5 5/ 4 x x2 5 º c1
« log 4 »
¬« 2 42 »¼
= ³ ex ª x 1 4 (x 1 º dx = x 4x2 + 5 + 5 log 2x + 4x2 + 5 + c
«¬ 4)2 »¼ 24
Put f (x) 11
and f '(x)
x 4 (x 4)2 2. ³ 9x2 4 dx
³ ex{ f (x) f '(x)}dx ex f (x) c Solut ion: I = ³ 3 x2 94 dx
= ex 1 4 +c
x+
2
§ 2 ·
³ ³Integrals of the type x2 a2 dx, x2 a2 dx = 3³ x2 ¨© 3 ¸¹ dx
³ a2 x2 dx x a2 x2 a2 log x a2 x2 c
3 ª x x2 4º 4/9 x x2 4 c1
22 « 2 » log 9
= ¬ 9 ¼
³ x2 a2 dx x x2 a2 a2 log x x2 a2 c 2
22
= x 9x2 4 2 log 3x 9x2 4 c
23
In order to evaluate integrals of form 3. ³ x2 4x 5 dx
Solution: I = ³ x2 4x 4 9 dx
³ ax2 bx c dx we use the following steps.
= ³ x 2 2 32 dx
Step 1: Make the coefficients of x2 as one
= x − 2 x2 − 4x − 5
by taking a common. 2
Step 2: Add and substract § b ·2 in − 9 log (x 2) x2 4x 5 c
x2 + b x + c to get the perfect square ©¨ 2a ¹¸ 2
aa
∴ § x b 2 4ac b2
¨© 2a 4a2
·
¹¸
131
4. ³ 1 (logx)2 dx 7. ³ x2 x 1 dx
x
13
Solution: I = ³ 1 (logx)2 1 dx Solution: I = ³ x2 x dx
x 44
Put log x = t ³ = § x 1 2 § 2
¨© 2 ©¨¨
∴ 1 dx = dt · 3·
x ¹¸ 2 ¸¹¸ dx
1 § 1 · 2
2= 2 ©¨ 2 ¸¹
§ 3·
¨ ¸
= ³ 1 t2 dt x x2 x 1 © 2 ¹ 1
log x
§ 3· 22 x2
t 1+ t2 + 1 log t + 1+t2 + c ¨¸
= 2 2 © 2 l¹o glog x 1 x2 x 1 c
22
(logx) 1+ (logx)2 Integrals of the form ³ ( px q) ax2 bx c dx
=
2 SOLVED EXAMPLES
+ 1 (logx) + 1+ (logx)2 + c
log
2
³5. ex e2x 1 dx I ³ (3x 2) x2 x 1 dx A d (x2 x 1) B
dx
Let ex = t Solution: We express 3x 2
Solution: Put exdx = dt 3x − 2 = A(2x +1) + B
I = ³ t2 1 dt = 2Ax + (A + B)
= t t2 +1 + 1 log t + t2 +1 + c Comparing coefficients of x and constant
22 term on both sides.
2A = 3 and A + B = −2
= ex e2x +1 + 1 log ex + e2x +1 + c A = 3/2 and B = −7/2
22
∴ I ³ ª 3 (2x 1) 7 º x2 x 1 dx
6. ³ x2 4x 13 dx ¬« 2 2 ¼»
Solution: I = ³ x2 4x 4 9 dx
³ = (x 2)2 32 dx 3 x2 x 1 dx
= 2 ³ (2x 1)
7 x2 x 1 dx
³ 2
Let
= x + 2 (x + 2)2 + 32 ³ I1 (2x 1) x2 x 1 dx,
2
³ 7 x2 x 1 dx
++3322lloogg ((xx++22))++ ((xx++22))22++3322 ++cc
22 I2 2
= x + 2 x2 + 4x +13 + 9 log Put x2 x 1 t in I1
22
³∴ I1 = t dt f t1/2 dt
x + 2 x2 + 4x +13++ 9 log (x + 2) + x2 + 4x +13 + c
22 = t3/2 + c
3/2
132
2 ( x2 + x + 1)3/ 2 + c1 ³7) ex x 1 dx
I1 = 3 (x 1)3
³ 7 x2 x 1 dx ³8) ex ª¬«(logx)2 2logx º dx
x »¼
I2 = 2
7 ª 1 § 1 ·
¬« 2 ©¨ 2 ¸¹
³ = 2
x x2 x 1 31 9x)2 x ¬ª« l1o1»¼ºgx c2 1 º dx
logx (logx)2 ¼»
82
x 1 x2 x 1 c2 10) ³ (1 logx 2 dx
2 logx)
I = I1 + I2 5.6 Integration by method of Partial
EXERCISE 5.5 Fractions:
5.6.1 Types of Partial Fractions.
Evaluate the following. (1) If f(x) and g(x) are two polynomials then
f(x)/g(x) is a rational function where g(x)≠0.
1) ∫ x log x
(2) If degree of f(x) < degree of g(x) then f(x)/
∫2) x2 e4x dx g(x) is a proper rational function.
∫3) x2 e3x dx (3) If degree of f(x) ≥ degree of g(x) then
∫4) x3 ex2 dx f(x)/g(x) is improper rational function.
³5) ex § 1 1 · dx (4) If a function is improper then divide
©¨ x x2 ¹¸ f(x) by g(x) and this rational function
can be written in the following form
³6) ex (x x dx
1)2 f (x) Quotient Remainder and can
g(x) g(x)
be expressed as the sum of partial fractions
using following table.
Type Rational Form Partial Form
1 px r q A B
(x a)(x b) x a x b
2 px2 r qx r r A B C
(x a)(x b)(x c) x a x b x c
px r q A B
3 (x a)2 x a (x a)2
4 px2 r qx r r A B C
(x a)2 (x b) x a (x a)2 x b
5 px2 r qx r r A B C D
(x a)3(x b) x a (x a)2 (x a)3 x b
px2 r qx r r x A Bx C c
6 (x a)(ax2 r bx r c) a ax2 r bx r
where, ax2 ± bx ± c is non factorizable
133
SOLVED EXAMPLES 11 1 11 1
= 4 ³ 1 x dx 2³ x 2 dx 4 ³ x 3 dx
1. ³ x 1 dx 1 11
5x 6 = log x 1 2 log x 2 log x 3 c
x2 44
Solution: I = ³ (x x 1 dx 3. ³ log x dx
2)(x 3) x(1 log x)(2 log
x)
x 1 A B
Consider (x 2)(x 3) x 2 x 3 Solution: Put log x = t
x 1 A(x 3) B(x 2) 1 dx = dt
x
Put x = −2 and we get A = −1
I ³ (1 t)t(2 t) dt
Pur x = −3 and we get B = 2
x 1 1 2 Consider t A B
(x 2)(x 3) x 2 x 3
(1 t)(2 t) 1 t 2 t
I = ³ x2 x 1 6 dx ³ § 1 x 2 3 · dx Put t = −1 A = −1
5x ©¨ x 2 ¹¸
Put t = −2 B = 2
= − ³ dx 2³ 1 I ³ ª 1 2 º dt
x 2 dx «¬1 t 2 »¼
t
x 3
= log x 2 2log x 3 c = ³ 1 1t dt 2³ 1 dt
2 t
x2 2
2. ³ (x 1)(x 2)(x 3) dx = −log|t + 1| + 2log|t + 2| + c
Solution: I = Consider = 2log|log x + 2| − log|logx +1| + c
x2 2 A B C = log|(logx + 2)|2 - log|(logx + 1)| + c
(x 1)(x 2)(x 3) x 1 x 2 x 3
x2 + 2 = A(x + 2)(x + 3) + B (x − 1)(x + 3) + ³4. x3 4x2 3x 11 dx
C(x − 1) (x + 2) x2 5x 6
Put x = 1 A = 1/4 x 1 Q
Put x = −2 B = −2 )[D = x2 5x 6 x3 4x2 3x 11
(x3 5x2 6x)
Put x = −3 C = 11/4 x2 3x 11
x2 2 1/ 4 2 11/ 4 (x2 5x 6)
2x 5 R
(x 1)(x 2)(x 3) x 1 x 2 x 3
³ x2 2 dx Express x3 4x2 3x 11 Q R
(x 1)(x 2)(x 3) x2 5x 6 D
§ 11 · 2x 5
¨ ¸ x2 5x
I = 1/ 4 2 4 ¸ dx ( x 1) 6
³¨ x 1 ¸
x 2 x 3
¨
©¹
134
³ = x3 4x2 3x 11 5 71 5
x2 5x 6 I log x 2 log x 2 c
dx 16 4 (x 2) 16
= ³ (x 1) dx ³ 2x 5 6 dx EXERCISE 5.6
x2 5x
³x2 x 2x 5 dx cv Evaluate:
x2 5x 6
= 2
Express 2x 5 A B 1) ³ ( x 2x 1 2) dx
x2 5x 6 x 2 x 3 1)(x
2x + 5 = A (x − 3) + B (x − 2) 2) ³ 2x 1 dx
1)(x
Put x = 2 we get A = −9 x(x 4)
Put x = 3 we get B = 11 ³3) x2 x 1 dx
x2 x 6
I = 2x 5 9 11 4) ³ x dx
x2 5x 6 x 2 x 3 1)2 (
( x x 2)
x2 x § 9 11 ·
2 ©¨ x 2 x 3 ¸¹
³
∴ I dx cv 3x 2
1)2 ( x
5) ³ ( x 3) dx
∴ I x3 4x2 3x 11 x2 x
³ x2 5x 6 dx x
1
9log x 2 11log x 3 c
6) ³ x(x5 1) dx
3x 1 7) 1
2)2 (x
5. ³ x dx ³ x(xn 1) dx
( 2)
Express ³8) 5x2 20x 6 dx
x3 2x2 x
3x 1 A B C
(x 2)2 (x 2) x 2 (x 2)2 x 2
3x + 1 = A (x − 2) (x + 2) + B (x + 2) +
Activity
C (x − 2)2
Put x = 2 B = 7/4 x 1
(x 3)(x 2)
x = −2, C = −5/16 Evaluate: ³ dx
Comparig Coefficients of x2 on both sides Now, x 1 >@ >@
we get
A+C=0 A = 5/16 (x 3)(x 2) (x 3) (x 2)
5 7 5
3x 1 16 4 16 There is no indicator of what the numerators
(x 2)2 (x 2) should be, so there is work to be done to find
x 2 (x 2)2 x 2 them. If we let the numerator be variables, we
can use algebra to solve. That is we want to find
51 71 51 constants A and B that make equation 2 below
I 16 ³ x 2 dx 4 ³ (x 2)2 dx 16 ³ x 2 dx true for x = 2,3 which are the same constants that
make the following equation true.
135
x 1 > A@ >B@ (1) 2. 1 log ax b c
a
(x 3)(x 2) (x 3) (x 2) ³ (ax b) dx
x − 1 = A (x − 2) + B (x − 3) (2) eax b c
a
[ ] x + [ ] = [ ]x + [ ] (3) ³3. eax b dx
Note: Two polynomials are equal if ³4. abx k dx abx k c
corresponding coefficients are equal. For linear b.loga
functions, this means that ax + b = cx + d for all
x exactly when a = c and b = d ³5. x2 a2 dx
³ = x x2 a2 a2 log x x2 a2 c
Alternately, you can evaluate equation (2)
for various values of x to get equations relating 22
A and B. Some values of x will be more helpful
than others ³6. x2 a2 dx x2 a2 c
³ = x x2 a2 a2 log x
[ ] = [ ]
22
[ ] = [ ]
7. ³ f '(x) dx log f (x) c
continue solving for the constants A and B. f (x)
A= ,B=
∴ x 1 > @ >@ 8. ³ f '(x) dx 2 f (x) c
f (x)
(x 3)(x 2) (x 3) (x 2)
> f @(x) n 1 c, n z 1
∴ ³ x 1 dx ³ >@ dx ³ >@ dx 9. ³> f (x)@n f '(x) dx
3)(x n 1
( x 2) (x 3) (x 2)
I=[ ]+[ ]+c ³10. 1 1 log x a c
x2 a2 dx 2a x a
Let's Remember ³11. 1 1 a x c
a2 x2 dx log a x
Rules of Integration:
1. ∫[f(x) ± g(x)] dx = ∫ f(x) dx ± ∫g(x)dx 2a
2. ∫k f(x) dx = k ∫ f(x) dx ; where k is a constant.
3. If ∫ f(x) dx = g(x) + c then, ³12. dx log x x2 a2 c
∫ f(ax + b) dx = 1 g (ax + b) + c; a ≠ 0
x2 a2
a
Standard Integration Formulae. ³13. dx log x x2 a2 c
x2 a2
³1. (ax b)n dx (ax b)n 1 c;if n z 1
a(n 1)
136
MISCELLANEOUS EXERCISE - 5 c ) − logx + log (1 − x) + c
d) log (x − x2) + c
I. Choose the correct alternative from the 6) ³ dx
following. 8)( x
dx ( x 7)
1 x
1) The value of ³ is x 2
x 1
1 c
a) log
a ) 2 1 x c b) 2 1 x c
15
c) x + c d) x + c b) 115 log xx ++ 87 + c
2) ³ 1 x2 dx c) 1 log x 8 c
15 x 7
a ) x 1+ x2 + 1 log (x + 1+ x2 ) + c
22 d) (x - 8) (x - 7) + c
b) 2 (1+ x2 )3/2 + c 3
3
7) ³ ©§¨ x 1x ¸¹· dx =
1 x 2 + c (x) +c
c) (1 + ) d)
1 2
3 + x
a) 14 ¨©§ 1 · 4
x ¸¹
x c
³3) x2 (3)x3 dx b ) x44 32x2 3log x 21x2 c
(3) x3
a) (3)x3 + c b) 3.log 3 + c c) x44 + 32x2 + 3log x + x12 + c
d) (x - x -1)3 + c
c) log 3(3)x3 + c d) x2(3)x3
4) ³ x 2 dx p ³ 4x 6 dx 1
2x2 6x 5 2x2 6x 5 2
³ dx then P=? ³8) § e2x e 2x · dx
6x ¨ ex ¸
2x2 5 © ¹
a ) 1 b) 1 a ) ex 3e13x c b) ex + 3e13x + c
32
c) 1 d) 2 c) e x 1 c d) e x 1 c
4 3e3x 3e3x
5) ³ ( x dx ) ³9) (1 x) 2 dx =
x2
a) (1 x) 1 c b) (1 x) 1 c
a) logx − log (1 − x) + c c) (1 x) 1 1 c d) (1 x) 1 1 c
b) log (1 − x2) + c
137
³10) ( x3 3x2 3x 1) dx IV. Solve the following:
(x 1)5 1) Evaluate.
a) x 11 c b) §¨© x 11 ¸¹·5 c i) ³ 5x2 6x 3 dx
2x 3
c) log(x +1) + c d) log |x +1|5 + c 4
ii) ³ (5x 1)9 dx
II. Fill in the blanks. 1
1. 5( x 6 1) dx x4 ......x3 5x c iii) ³ (2x 3) dx
x2 1
³ x2 x 6 dx x ...... c iv) ³ x 1 dx
x 2)(x 1) x 4
2.
(
3. If 1 and 5 then v) If f '(x) = x and f(1) = 2 then find the
f '(x) x f (1) = value of f(x).
x 2
f (x) log x x2 ....... vi) ∫ x dx if x < 0
2 x) dx
4. (1 log the 2) Evaluate.
x 1
To find the value of ³
i) Find the primitive of 1+ ex
proper substitution is .........
³5. 1 ª¬log xx 2 dx p(log x)3 c then P =
x3
º¼
...... ³ii) ae x be x dx
(ae x be x
)
III. State whether each of the following is 1
True or False.
iii) ³ 2x 3xlogx dx
1. The proper substitution for
³ x(xx )x (2 log x 1)dx is (xx)x = t iv) ³ 1 dx
x
2. If ∫ x e2x dx is equal to e2xf(x) + c where C is x
constant of integration then f(x) is (2x −1)
2 ³v) 2ex 3 dx
4ex 1
3. If ∫ x f(x) dx = f (x) then f(x) = ex2
2 3) Evaluate.
4. If ³ (x 1) dx A log|x+1| + B log|x−2| ³ dx
(x 1)(x 2)
i) 4x2 5
then A + B = 1 dx
2x
x 1 ii) ³ 3 x 2
(x 1)3
³5. e x dx ex
For = f(x) + c, f(x) =
(x + 1)2. iii) ³ 9 dx 25
x2
138
³iv) ex dx Activities
e2x 4ex 13 1
v) ³ dx 1) ³ (x2 2x dx
4logx 5x 4)
x[(logx)2 1]
dx Solution: > 2x C D
16
vi) ³ 5 x 2 @> @ > @ > x 4@
dx ∴ 2x = C (x − 4) + D (x − 1)
x(logx)2
vii) ³ 25x
∴ C = ,D=
³viii) ex dx ³ (x 1)(x 4) 2x dx ³ ª º dx
4e2x 1 ¬« (x 4) »¼
∴ 1) (x
4) Evaluate. = ³ (x 1) dx ³ (x 4) dx
i) ∫ (logx)2dx
1 x = + + c
(2 x)2
³ii) ex dx
∫iii) xe2xdx ³2) x13/2 (1 x5/2 )1/2 dx
³iv) log(x2 x)dx Solution: ∫xx3/2 (1 + x5/2) dx = ∫(x5/2)2 x3/2
(1 + x5/2) dx
v) ∫ e x dx
let 1 + x5/2 = t
dx = dt
vi) ³ x2 2x 5dx 2
I = ∫ (t − 1)2 t1/2 dt
vii) ³ x2 8x 7dx
5
2
= (t2 − 2t + 1) t1/2 dt
5
5) Evaluate. 2 dt = ∫ dt + dt]
= 5 [
3x 1
i) ³ 2x2 x 1 dx
2 x3 3x2 9x 1 2 − + }+c
2x2 x 10 = {
³ii) dx
5
iii) ³ (1 log x) dx 3) ³ (x dx 1) = .......... (given)
logx)(2 3logx) 2)( x 2
x(3
³ 1 tan 1 x c
x2 1 dx
1 Bx C
Solution: (x 2)(x2 1) (x 2) (x2 1)
139
∴ 1 = A(x2 + 1) + (Bx + C) (x + 2) 1
Put x = −2 1 4) If ³ x5 x dx f (x) c = f(x) + C, then the
we get A =
5
Now comparing the coefficients of x2 and ³ value of x x4 dx is equal to ............
constant term, we get x5
O = A + B ³ ª x4 1 º
»
and 1 = A + 2C I « x x5 ¼ dx
¬
−1 2 11
∴ B = , C =
³ x dx ³ x 5 x dx
55
1 x I c
(x 2)(x2 1) (x 2) (x2 1)
I log x f (x) c1 .......c1 c
I ³ dx ³ x dx ³ dx
(x 2) 1 x2 1
x2
= − + +c
vvv
140
6 Definite Integration
Let's Study SOLVED EXAMPLES
• Definite Integral Ex 1 : Evaluate:
• Properties of Definite Integral
3
i) ∫ x4dx
2
Introduction 11
³ dx
We know that if f(x) is a continuous function ii) (2x 5)
of x, then there exists a function φ(x) such that 0
φ'(x) = f(x). In this case, φ(x) is an integral of
f(x) with respect to x and we denote it by 11
∫ f(x) dx = φ(x) + c. Now, if we restrict the domain of iii) ³ dx
f(x) to (a, b), then the difference φ(b) - φ(a) is called 0 1 x x
definite integral of f(x) w.r.t. x on the interval
Solution:
b i) Here f(x) = x4, φ(x) = x5 + c
[a, b] and is denoted by ³ f x dx . 5
ba
Thus ³ f x dx = φ(b) − φ(a) 3
a
∫ f (x) dx = [φ(x)]32
The numbers a and b are called limits of 2
integration, 'a' is referred to as the lower limit of
integral and b is the upper limit of integral. ∫3 x4dx ª x5 3 35 25
= «
Note that the domain of the variable x is ¬ 5 º
restircted to the interval (a, b) and a, b are finite » 55
numbers. 2 ¼2
243 32 211
=
55 5
11 1
³ ii) 0 (2x 5) dx = 2 [log|2x + 5|]10
Let's Learn
6.1 Fundamental theorem of Intergral 1
Calculus. = 2 [log7 − log5]
Let f be a continuous function defined on 17
(a, b) = 2 log 5
³ f x dx = φ(x) + c. 11
iii) ³ x dx
0 1 x
b
Then ³ f x dx = [φ(x) + c]ba ³1
=
a = [φ(b) + c] − [φ(a) + c] 1 x x dx
= φ(b) − φ(a) 0 1 x x 1 x x
There in no need of taking the constant of ³1 1 x x dx
integration c, because it gets eliminated.
= 0 1 x x
141
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