Mathematics & Statics Commerce Part 1

1 ª¬ x3 x2 1 0

= ³ 1 x x dx ax¼º0
0

ª2 3 2 3º 1 ii) (1 1 a) 0 0

= « 3 (1 x)2 3 x2 » 0 2+a−0=0
¬ ¼

ª2 3 2 3º ª2 3 2 3º a = −2

= « (1 1) 2 (1) 2 » « (1 0) 2 (0) 2 » a
¬ ¼ ¬ ¼
3 3 3 3 ³ 3x2 dx 8

0

2ª 3 º 2 ª x3 º a
= «22 1» [1 0] « 3 » 0
3 ¬ ¼ 3 ¬ ¼ 8
3

= 23 ¬ª«232 2»¼º (a3 03 ) 8
a3 = 8

2 ª¬2 2 2º¼ a=2
=
b b 2
3
x3 dx 0 and x2 dx
4 ³ ³iii) 3
3 ¬ª 2 1º¼ aa
=
ª x4 ºb ª x3 ºb
∴ « » 0 and « » 2
¬ ¼a ¬ ¼a 3
Ex 2 : Evaluate: 4 3

1 ∴ 1 b4 a4 0 and 1 (b3 a3) 2
4 33
i) If ³ (3x2 2x a) dx 0 ; find a.
0 ∴ b4 a4 0 and b3 a3 2
∴ b4 a4 ? b ra
a But b = a does not satisfy b3 − a3 = 2
∴ b ≠ a
ii) If ³ 3x2dx 8 ; find the value of a. ∴ b = −a
0 Substituting b = −a in b3 a3 2
We get (−a)3 − a3 = 2, −2a3 = 2
bb 2 We get a = −1
. Find the ∴ b = −a = 1
iii) ³ x3dx 0 and ³ x2dx ∴ a = −1 , b = 1
aa 3
values of a and b.

a

iv) If ³ 4x3dx 16 , find α
0

v) If f(x) = a + bx + c x2, show that

³ 1 1 ª f (0) 4 f § 1 · f (1)¼»º
6 «¬ ¨© 2 ¸¹
f (x) dx

0

Solution: a

iv) ³ 4x3dx 16
0

1 ∴ 4 ª x4 º a 16
« 4 » 0
i) ³ (3x2 2x a) dx 0 ¬ ¼
0

Then ª x3 2 x2 1 0 4 ¬ªa4 0¼º 16
«3 3 2 4
¬ º
ax»

¼0

142

a4 = 16 Solution:
∴ a = 2
21
1
³i) 0 4 x x2 dx

v) ∫ f (x)dx 21
0
³= 0 x2 x 4 dx
1
³2 1 dx
³= (a bx cx2 )dx 1
0
= 1
11 1 0 x 2 x 4

= a³1dx b³ x dx c³ x2dx 44

00 0 ³ = 02 §¨© x 12 ·¸¹2 1¨§© 127 ¹·¸2 dx

= ª«ax bx2 cx3 1
¬ 2 3
º
»
¼0

= a + b + c ............(1) 2 1
2 3
= ³0 § dx
Now f(0) = a + b(0) + c(0)2 = a 2 2
¨ 17 § x 1
© 2 · ©¨ 2 ·
f(1/2) = a + b(1/2) + c(1/2)2 = a + b/2 + c/4 ¸ ¸¹
and f(1) = a + b + c ¹

ª º 2
»
ª º 1 « 17 § x 1 · »
«¬ (1)»¼ = ««log 2 ¨© 2 ¹¸
∴ 1 § 1 ·
6 f (0) 4 f ©¨ 2 ¹¸ f 17 17 § 1 · »
2 ©¨ 2 ¹¸ »
« x ¼0

1 ¬ª«a b c c)¼»º ¬
= 6 2 4
4(a ) (a b

ª º 2
«log »
1 17 2x 1
=
= 1 >a 4a 2b c a b c@ 17 ¬« 17 2x 1 »¼0

6

= 1 >6a 3b 2c@ = 1 ­®°log § 17 4 1 · log § 17 1 ·°½
17 °¯ ¨¨© 17 4 1 ¸¹¸ ¨©¨ 17 1¸¹¸¾¿°
6

= a + b + c .............(2)
2 3
1 °®­log § 17 3 · § 17 1 ·°½
From (1) and (2) = 17 °¯ ¨©¨ 17 3 ¹¸¸ log ¨¨© 17 1¸¸¹°¿¾

³1 1 ª f (0) 4 f § 1 · f º 1§ 17 3 u 17 1·
6 ¬« ¨© 2 ¹¸ (1)»¼ 17 3 17 1 ¸¹¸
= f (x)dx = log ¨¨©
17
0

Ex 3 : Evaluate: 1 log § 20 4 17 ·
21 = ¨¨© 20 4 17 ¸¸¹

³ i) 0 4 x x2 dx 17

4 1 log § 5 17 ·
0 = ¨©¨ 5 17 ¸¹¸
dx
2x 17

³ ii) x2 3 dx

143

³4 dx ∫ii) 2 log x dx
1 x2
ii) 0 x2 2x 3

41 ∫2 1
³ = dx x2
0 x2 2x 1 1 3 = log x. .dx

1

41 1 ª § 2 2 1 § x 1 ·
log x. x2 .dx «¬(log ©¨ 1 x ¨ 1 ¸
1 ·º © ¹
x ¹¸»¼1
= ³ dx ³ ³= x dx

0 (x 1)2 ( 2 )2

= ª¬«log 1 2
§ x · 1 º
= «¬ªlog (x 1) 4 x ©¨ ¸¹ x »¼1
º
(x 1)2 ( 2)2 »¼0

4 = §¨© 21 log 2 .log1·¸¹ § 1 1 ·
= ª¬«log (x 1) º ©¨ 2 1 ¹¸
x2 2x 3 ¼»0

= log(5 16 8 3) log (1 3) 1 log 2 1
= 2 2

= log(5 3 3) log (1 3) 1
= ( log 2 1)
ª5 3 3º
» 2

= log « (1 3) ¼ 1
¬ = ( log 2 log e)

2

Ex. 4: Evaluate: 1 log e
=
22

2 Ex. 5: Evaluate:

i) ∫ log x dx
1

2 log x 21
1 x2 ³ dx
ii) ∫ dx i) ( x 1)( x 3)
1

Solution: 31

³ ii) 1 x(1 x2 ) dx

2 Solution:

i) I ³ log x dx
1

21
i) ³ dx
2 ( x 1)( x 3)
1
I ³ log x.1.dx
1 A B
1 Let (x 1)(x 3) (x 1) (x 3)

³I >log x. x@2 2 1 x dx 1 = A(x + 3) + B(x + 1) ............. (1)

1 1x

= > x log x x @2 Putting x + 1 = 0

1

= >(2log2 1log1)@ [2 1] i.e. x = −1 in equation (i) we get 1
A=
2

= (log4 − 0) − 1 Putting x + 3 = 0
= log 4 −1
i.e. x = −3 in equation (i) we get B 1
2

144

11 1
= (log 3 − 2 log 5)
1 2 2 = log 3 log 51/2 log 3 log 5

(x 1)(x 3) (x 1) (x 3)

21 § 3·
³ dx ©¨ 5 ¸¹
(x 1)(x 3) = log
1

³ ³1 2 dx 1 2 dx EXERCISE 6.1
1 x 1 2 1 x 3
= 2

12 Evaluate the following definite intergrals:
= ¬ªlog x 1 log x 3 º¼1
2

11 91
= 2 (log 3 − log 2) − 2 (log 5 − log 4) ∫ x dx
1.
4

1 «¬ªlog 3 5 º 31
= 2 4 ¼» ³2. dx
log 2 x 5
2

1 log ª 6 º ³3. 3 x dx
= «¬ 5 »¼ 2 2
x 1
2

³ii) 31 ³4. 1 x2 3x 2 dx
1 x(1 x2 ) dx
0x
Let x(1 1 x2 ) Ax B1 x x2c
³5. 3 x dx
1 A(1 x2 ) (Bx c) x .........(1) 2 (x 2)(x 3)

Putting x = 0 in equation (i) we get A = 1 ³6. 2 x2 dx 5 dx
Comparing the coefficient of x2 and x, we 1 6x
get A + B = 0, B = −1 & C = 0
a
1 1 x
7. If ³ (2x 1) dx 2 , find the real value of a.
x(1 x2 ) x 1 x2 0

³3 dx a

1 x(1 x2 ) 8. If ³ (3x2 2x 1) dx 11, find a.
1

11
³ dx
³ ³3 1 dx 3 x dx 9. 1 x x
1 1 x2 0
=
1x

= ¬ªlog x 3 1 ¬ªlog 1 x2 3 ³10. 2 3x 1) dx
2 1 (9x2
¼º1 ¼º1

1 3
= (log 3 − log1) − (log10 − log 2)
11. ∫ log x dx
2 1

= log § 3 · 1 log § 10 ·
©¨ 1 ¸¹ 2 ©¨ 2 ¸¹

145

6.2 Properties of definite integrals ³b f (x) dx

In this section we will study some properties 4. a f (x) f (a b x)
of definite integrals which are very useful in
evaluating integrals. ³5. 7 (11 x2 ) dx
4 x2 (11 x2 )
a
Solution:
Property 1 : ³ f (x) dx 0
a 1 01

ba 1. ³ f (x) dx = ³ f (x) dx ³ f (x) dx
1 1 0
Property 2 : ³ f (x) dx ³ f (x) dx
ab

bb 01

Property 3 : ³ f (x) dx ³ f (t) dt = ³ (1 2x) dx ³ (1 2x) dx
aa 1 0

bc = ¬ªx x2 0 ª¬ x x2 1

³Property 4 : f (x) dx ³ f (x) dx ¼º 1 ¼º0
aa

b = >0 ( 1 1)@ >(1 1) 0@

³ f (x) dx, where a c b = 2 2 4
c

b b 1

Property 5 : ³ f (x) dx ³ f (a b x) dx 2. ³ x(1 x)n dx
a 0
a aa
a
a By property ³ f (x) dx ³ f (a x) dx
Property 6 : ³ f (x) dx 00
0 ³ f (a x) dx
1n
0
I = ³ (1 x)>1 (1 x)@
2a a a

Property 7 : ³ f (x) dx ³ f (x) dx ³ f (2a x) dx 0

000 1

[If f(−x) = f(x), f(x) is an even function. = ³ (1 x) xn dx
If f(−x) = −f(x), f(x) is an odd function.] 0

aa 1

Property 8 : ³ f (x) dx 2³ f (x) dx if f is an ³ = (xn xn 1) dx
a 0 0

even function = ª« x n 1 1 ª xn 2 1
¬ « n 2
º ¬ º
» »
= 0 if f is an odd function n 1 ¼ 0 ¼0

SOLVED EXAMPLES 11 (n 2) (n 1)
= n 1 n 2 (n 1) (n 2)
Ex. Evaluate the following integrals: =

­1 1 2 x;xd0 1
³1. f (x) dx where f(x) = ®¯ 1 2 x; xt0
= n 1 n 2
1

a ³3 3 (x 4) dx

2. ³ x(1 x)n dx 3. Let I = 0 3 (x 4) 3 (7 x) ............ (1)
0

³3 3 x 4 dx aa

3. 0 3 x 4 3 7 x By property ³ f (x) dx ³ f (a x) dx
00

146

³3 3 (3 x) 4 dx ³ b f (x) f (a b x) dx
a f (x) f (a b x)
I 0 3 (3 x) 4 3 7 (3. x)

3 3 7 x dx b
0 3 7 x 3 x 4
³ ³ 1dx
................. (2)
a

On adding equations (1) and (2) > @xb
a

³ 2I 3 ª 3 x 4 3 7 x º 2I b a
0 « 7 x 3 x x 3 x » dx I b a
¬ 3 4 3 7 4 ¼

³ 2I 3 3 x 4 3 (7 x) dx 2

0 3 x 4 3 (7 x) ³∴ b f (x) f (x) dx b a
a f (a b x) 2

3

= ∫1 dx ³5. 7 (11 x2 ) dx ......(1)
0 4 x2 (11 x2 )

= > x@3 ³7 x2
0

= (11 x2 ) x2 dx ......(2)
4
2I = 3

3 By Property
I =
bb
2
³ f (x) dx ³ f (a b x) dx
³3 3 (x 4) 3 aa
dx
∴ 2 Adding equations (1) and (2)
0 3 (x 4) 3 (7 x)

³b f (x) dx ....... (1) 7 (11 x2 ) 7 x2
4 x2 (11 x2 ) 4 x2)
4. ³ ³ 2I dx (11 x2 dx

a f (x) f (a b x)

bb ³ 7 (x2 ) (11 x2 ) dx
4 (x2 ) (11 x2 )
By property ³ f (x) dx ³ f (a b x) dx
aa

7

³b f (a b x) dx ³1dx

I a f (a b x) f [(a b (a b x)] 4

= > x@7
4

³b f (a b x) dx .............. (2) 2I = 3
a f (a b x) f (x) 3
I
I =
2

Adding equations (1) and (2) we get, ³ 7 (11 x2 ) dx 3
4 x2 (11 x2 ) 2
³b f (x) dx ∴

2I a f (x) f (a b x)

³ b f (a b x) dx
a f (a b x) f (x)

147

EXERCISE 6.2 b cb

Evaluate the following integrals: 3) ³ f (x) dx ³ f (x) dx ³ f (x) dx
a ac
³1) 9 x3 dx
9 4 x2 bb

a 4) ³ f (x) dx ³ f (t) dt
aa
³2) x2 (a x)3/2 dx
aa

5) f (x) dx f (a x) dx

00

0 bb

³3 3 x 5 dx 6) f (x) dx f (a b x) dx

3) aa

1 3 x 5 3 9 x 2a a a

³5 x dx 7) ³ f (x) dx ³ f (x) dx ³ f (2a x) dx
0 00
4)
2 x 7 x aa

³2 x dx 8) ³ f (x) dx 2³ f (x) dx,, if f is even function,
a 0
5)
1 3 x x =0 , if f is odd function

³6) 7 x dx MISCELLANEOUS EXERCISE - 6
2 x 9 x
I) Choose the correct alternative.

³7) 1 log § 1 1¸¹· dx ³1) 9 x3 dx =
0 ¨© x 9 4 x2

1 a) 0 b) 3 c) 9 d) −9

³8) x(1 x)5 dx 3 dx
0 2 x 5

³2) =

Let's Remember a) log § 8 · b) log § 8 ·
¨© 3 ¸¹ ©¨ 3 ¸¹
 Rules for evaluating definite integrals.
c) log § 3 · d) log § 3 ·
b bb ©¨ 8 ¸¹ ¨© 8 ¹¸

1) ³[ f (x) r g(x)]dx ³[ f (x)dx r ³ g(x)dx ³3) 3 x x 1 dx =
a aa 2 2

bb a) log § 8 · b) log § 8 ·
¨© 3 ¹¸ ¨© 3 ¸¹
2) ³ k f (x)dx k ³ f (x) dx
aa c) 1 log § 8 · d) −1 log 8
2 ¨© 3 ¸¹ 23
 Properties of definite integrals

a

1) ³ f (x) dx 0
a

ba

2) ³ f (x) dx ³ f (x) dx
ab

148

∫4) 9 dx = b) 4 c) 2 d) 0 II) Fill in the blanks.

4x 2
a) 9
1) ∫ exdx = .................
a 0

5) If ³ 3x2dx 8 then a = ? 3
0
2) ∫ x4dx = .................
a) 2 8 d) a 2
b) 0 c)
³1 dx
3
3) 0 2x 5 = .................
3
a
6) ∫ x4 dx =
2 4) If ³ 3x2 dx 8 then a = ..............
0
a) 1 5 5 211
b) c) d)
22
211 5

2 ³5) 91 .................
dx
7) ∫ ex dx = 4x
0
b) 1 − e ³6) 3 x x 1 dx .................
a) e − 1 d) e2 − 1 2 2

c) 1 − e2

b 7) ³3 dx .................

8) ∫ f (x) dx = 2 x 5
a
a b ³8) 9 x3 dx .................
9 4 x2
a) ∫ f (x) dx b) ³ f (x) dx
b a III) State whether each of the following is
True or False
a a

c) ³ f (x) dx d) ∫ f (x) dx
b 0

³9) 7 x3 7 dx = b a
7 x2
1) ³ f (x) dx ³ f (x) dx
a) 7 b) 49 c) 0 7 a b
d)
bb
2
2) ³ f (x) dx ³ f (t) dt
³7 x dx = aa
2 x 9 x
10) a0

3) ³ f (x) dx ³ f (a x) dx
0a

75 d) 2 bb
a) b) c) 7
4) ³ f (x) dx ³ f (x a b) dx
22 aa

³5) 5 x3 dx 0
5 x2
7

149

³2 x 1 2
dx
6) 1 3 x x 2 12) ∫ x2 dx
1

³7 x 9 ³13) 1 1
dx dx
7) 2 x 9 x 2 4 x

7 (11 x)2 3 ³1 1 dx
4 (11 x)2 x2 x
³8) dx 14) 0 1 x
2

³4 1 dx

IV Solve the following. 15) 0 x2 2x 3

³3 x dx 4 x
2 2
1) 2 (x 2)(x 3) ³16) x 1 dx

³2) 2 x 3 dx 11
17) ³ dx
1 x(x 2) 2 x 3
0

3 ³18) 2 x2 5x2 3 dx
1 4x
3) ∫ x2 log x dx
1

1 ³19) 2 dx
1 x(1 log x)2
∫4) ex 2 x3 dx
0

2 e2x § 1 1 · 91
³5) 1 ©¨ x 2x2 ¸¹ dx 20) ³ dx
1 x
0

91
6) ∫ dx
x Activities
4

31 1) Complete the following activity.

³7) dx b

2 x 5 If ³ x3 dx 0 then

³8) 3 x 1 dx a
2 x2
§ x4 ·b
³9) 1 x2 3x 2 dx ¨¸ 0

0x © ¹a

³5 dx ∴ 1 0

10) 4

3 x 4 x 2 ∴ b4 0

³11) 3 x dx ∴ (b2 a2 )( ) 0
2 x2
1 ∴ b2 0 as a2 b2 z 0

∴ b r

150

³2) 02 4 dxx x2 ³1 § · dx ............(2)
¨© ¹¸
= log 1 x

0

³2 dx Adding (i) and (ii), we get

= 0 x2 ³ 2I 1 log ª1 x u x º dx
0 « x »
¬ ¼
³2 dx
x2 x 1 4 11
=
0 4 dx 2I = ³ log dx ³ o dx
00
2
2 § 1 · dx
¨© 2 ¹¸
= ³ 2

0 x 8 x5
8 1 x2
³4) dx

1 § 20 4 17 · f (x) x5
17 log ©¨¨ 20 4 17 ¸¹¸ 1 x2

1 f ( x) ( x)5 1 x2
0 1 x2
³3) log § 1 1¸·¹ dx
¨© x
Hence f is function

³1 § 1 x · dx ........(1) 8 x5
©¨ ¸¹ 8 1 x2
= log ³∴ dx

0

1 § 1 1 x · dx
¨ ¸
= ³ log
0© ¹

vvv

151

7 Applications of Definite Integration

Introduction 4. x2 = − 4by

The theory of integration has a large variety Fig. 7.1
of applications in Science and Engineering. In
this chapter we shall use integration for finding 7.2 Standard forms of ellipse
the area of a bounded region. For this, we first
draw the sketch (if possible) of the curve which 1) x2 y2 1 (a ! b)
encloses the region. For evaluation of area a2 b2
bounded by the certain curves, we need to know
the nature of the curves and their graphs.

The shapes of different types of curves are
discussed below.

7.1 Standard forms of parabola & their shapes

1. y2 = 4ax

Fig. 7.2

2. y2 = − 4ax x2 y2
a2 b2
2) 1 (a b)

3. x2 = 4by

Fig.7.3
152

7.3 Area under the curve (3) The area of the shaded region bounded by
two curves y = f(x), y = g (x) as shown in
To find the area under the curve, we state fig. 7.6 is obtained by
only formulae without proof.

(1) The area "A" bounded by the curve
y = f(x), X-axis and bounded between the
lines x = a and x = b (fig 7.4) is given by

A = Area of the region PRSQ

b xb

= ³ ydx ³ f (x) dx
a xa

Fig. : 7.6

xb xb

A ³ f (x) dx ³ g (x) dx
xa xa

where the curve y = f(x) and y = g (x)
intersect at points (a, f(a)) and (b, f(b)).

Fig. : 7.4 Remarks:

(2) The area A bounded by the curve x = g(y), (i) If the curve under consideration is below
Y-axis and bounded between the lines y = c the X-axis, then the area bounded by the
and y = d (Fig. 7.5) is given by curve, X-axis and lines x = a, x = b is
negative (fig. 7.7).
d yd
We consider the absolute value in this case.
A ³ x.dy ³ g ( y) dy
c yc xb

³ Thus, required area = f (x) dx
xa

Fig. : 7.5 Fig. : 7.7

153

(ii) The area of the portion lying above the 00
X-axis is positive.
A ³ y dx ³ (2x) dx
(iii) If the curve under consideraion lies above 1
as well as below the X-axis, say A1 lies
below X-axis and A2 lies above X-axis (as x 2 2
in Fig. 7.8), then A, the area of the region is
given by, 0

A = A1 + A2 = 2.³ x dx
2

Fig. : 7.8 = ¬ª«2. x22 º»¼0 2

tb = 0 4 4 sq. units

³ A1 = f (x) dx and A2 ³ f (x) dx ³A2 4 2 ª x2 º 4 =(42-02) = 16-0 = 16
at ¬« 2 ¼» 0
Area A bounded by the curve y = 2x, X-axis 2x dx

and lines x = −2 and x = 4 is A1 +A2. 0

A = A1 + A2 = 4 + 16 = 20 sq. units

SOLVED EXAMPLES

1. Find the area of the regions bounded by the
following curves, the X-axis and the given
lines.

(a) y = x2, x = 1, x = 3

(b) y2 = 4x, x = 1, x = 4

(c) y = −2x, x = −1, x = 2

Solution: Let A denote the required area in
each case.

Fig. : 7.9 Fig. : 7.10

154

3 Required area A = A1 + |A2|

(a) A = ∫ y dx 02
1
3 A2 ³ ( 2x) dx ³ 2x dx
1 0
= ∫ x2 dx
1 = «ª 2 x 2 º 0 ª 2x 2 º 2
¬ 2 » 1 « 2 » 0
1 1 33 13 1 27 1 ¼ ¬ ¼
= 3
[ x3 ]3 = 3 3

1

26 = ¬ª x 2 ¼º 0 ¬ª x2 º¼ 2
= 3 sq. units 1 0

4 = (0 1) (4 0)
A = 5 sq. units
(b) A = ∫ y dx
1

4 2. Find the area of the region bounded by the
parabola y2 = 16x and the line x = 4.
= ∫ 2 x dx
1

2 [ x3/ 2 ]4 = 4 Solution: y2 = 16x
= 2. (43/2 - 13/2) ∴ y = ± 4 x
1 ∴ A = Area POCP + Area QOCQ
3 3 = 2(Area POCP) (why?)

4 28 sq. units
= (8 - 1) =
3 3

4

= 2∫ y dx
0

4

= 2∫ 4. x dx
0

Fig. : 7.11 Fig. : 7.13
(c) A = (Area below X-axis) +
 y lies above X-axis
(Area above X-axis)
= 8 . 2 . [ x3/ 2 ]4
3 0
Fig. : 7.12
16 128
= 3 . [8] = 3 sq.units

155

3. Find the area of the region bounded by the
curve x2 = 16y, y = 1, y = 4, and the Y - axis
lying in the first quadrant. 4

Solution: Required area = ∫ x.dy
1

44
1

∴ A = ∫ 16 y dy = 4∫ y 2.dy
11

2 4 8
= ¬ª«4. 3 º u7
3 »¼1
3
y2

56 Fig. : 7.15
= sq. units.
3 y = b a2 − x2
a


 In first quadrant, y > 0

a

∴ A = 4.∫ y dx
0

³a b a2 x2 dx
a
= 4.

0

Fig. : 7.14 = 4b ª x a2 x2 a2 . sin 1 § x ·ºa
a « 2 2 ©¨ a ¹¸¼»0
¬

4. Find the area of the ellipse x2 y2 1. = 4b ­a2 sin 1 (1) a2 sin 1 (0) ½
a ® 2 ¾
a2 b2 ¯2 ¿
Given

x a2 § x · = 4ab . a2 .π - 0
2 2 ©¨ a ¸¹ 2 2
³ a2 x2 dx a2 x2 sin 1

π = π ab sq.units.
sin−1(1) = 2 , sin−1(0) = 0
5. Find the area of the region bounded by the
Solution: From the equation of ellipse curve y = x2 and the line y = 4.
Solution:
x2 y2 1
a2 b2 Equation of curve is y = x2 .......... (i)
and equation of line is y = 4 .... (ii)
∴ y2 = 1− x2 Because of symmentry,
b2 a2 Required area = 2 [Area in first quadrant]

y2 = ba22 (a2 − x2 ) 4

y = r b a2 x2 A = 2.∫ x.dy
a 0

156

4 MISCELLANEOUS EXERCISE - 7

= 2.∫ y dy
0

× 2 y3/ 2 ]4 4 I) Choose the correct alternative.
3 0 (43/2 - 03/2)
= 2 [ = 1) Area of the region bounded by the curve
3 x2 = y, the X-axis and the lines x = 1 and
x = 3 is ___________
4 32 sq.units.
= (8 - 0) =
3 3

26 3
a) sq. units b) sq.units
3 26

c) 26 sq. units d) 3 sq. units

2) The area of the region bounded by y2 = 4x,
the X-axis and the lines x = 1 & x = 4 is
___________

a) 28 sq. units b) 3 sq. units

28 3
c) sq. units d) sq. units

3 28

Fig. : 7.16 3) Area of the region bounded by x2 = 16y,
y = 1 & y = 4 and the Y=axis. lying in the
EXERCISE 7.1 first quadrant is ___________

1. Find the area of the region bounded by the a) 63 sq. units 3
following curves, the X-axis and the given b) sq. units
lines:
56
i) y = x4, x = 1, x = 5
ii) y = 6x + 4 , x = 0, x = 2 56 63
c) sq.units d) sq.units
iii) y = 16 − x2 , x = 0 , x = 4
iv) 2y = 5x + 7, x = 2, x = 8 3 7
v) 2y + x = 8, x = 2, x = 4
vi) y = x2 + 1, x = 0, x = 3 4) Area of the region bounded by y = x4, x = 1,
vii) y = 2 − x2, x = −1, x = 1 x = 5 and the X-axis is ___________
2. Find the area of the region bounded by the
3142 3124
parabola y2 = 4x and the line x = 3. a) sq.units b) sq. units
3. Find the area of circle x2 + y2 = 25 55
4. Find the area of ellipse x2 y2 1
3142 3124
4 25 c) sq. units d) sq. units
33

5) Using definite integration area of circle
x2 + y2 = 25 is ___________

a) 5π sq. units b) 4π sq. units

c) 25πsq. units d) 25 sq. units

157

II. Fill in the blanks. 3) Find the area of the region bounded by the
curve y = x2 and the line y = 10.
1) Area of the region bounded by y = x4, x = 1,
x = 5 and the X-axis is ___________ 4) Find the area the ellipse x2 y2 1.
16 9
2) Using definite integration area of the circle
x2 + y2 = 49 is _________ 5) Find the area of the region bounded by
y = x2, the X-axis and x = 1, x = 4.
3) Area of the region bounded by x2 = 16y,
y = 1, y = 4 and the Y-axis lying in the first 6) Find the area of the region bounded by the
quadrant is _____________ curve x2 = 25y, y = 1, y = 4 and the Y-axis.

4) The area of the region bounded by the 7) Find the area of the region bounded by the
curve x2 = y, the X-axis and the lines x = 3 parabola y2 = 25x and the line x = 5.
and x = 9 is ___________ Activities

5) The area of the region bounded by y2 = 4x, From the following information find the area
the X-axis and the lines x = 1 & x = 4 is of the shaded regions.
___________ 1)

III) State whether each of the following is 2)
True or False.

1) The area bounded by the curve x = g(y),
Y-axis and bounded between the lines y = c

d yd

and y = d is given by ³ x dy ³ g( y) dy
c yc

2) The area bounded by two curves y = f(x),

ba

y = g(x) and X-axis is ³ f (x) dx ³ g(x) dx
ab

3) The area bounded by the curve
y = f(x), X-axis and lines x = a and x = b is

b

∫ f (x) dx

a

4) If the curve, under consideration, is below
the X-axis, then the area bounded by curve,
X-axis and lines x = a, x = b is positive.

5) The area of the portion lying above the
X-axis is positive.

IV) Solve the following.

1) Find the area of the region bounded by
the curve xy = c2, the X-axis, and the lines
x = c, x = 2c.

2) Find the area between the parabolas
y2 = 7x and x2 =7y.

158

3) 5)
4)

vvv

159

8 Differential Equations and Applications

Let's Study 4) r dr eT 8
dT
• Differential Equation
• Ordinary differential equation 5) 1 dy d 2 y
• Order and degree of a differential equation dx dx2
• Solution of a differential equation
• Formation of a differential equation 6) x dx y dy 0
• Applications of differential equations
8.1.1 Ordinary differential equation

A differential equation in which the
dependent variable, say y, depends only on one
independent variable, say x, is called an ordinary
differential equation.

Let's Recall 8.1.2 Order of a differential equation

• Independent variable It is the order of the highest order derivative
• Dependent variable occurring in the differential equation.
• Equation
• Derivatives dy y x is of order 1
• Integration dx

x2 d2y x dy y 0 is of order 2
dx2 dx

Let's Learn § d2y ·2 x dy 2 y is of order 2
¨ dx2 ¸ dx
© ¹

8.1 Differential Equations: 2dy = ex is of order 1
dx
Definition: An equation involving
dependent variable(s), independent variable 8.1.3 Degree of a differential equation
and derivative(s)of dependent variable(s) with
respect to the independent variable is called a It is the power of the highest order
differential equation. derivative when all the derivatives are made free
from fractional indices and negative sign,if any.
For example :

1) dy y x For example -
dx
1) x2 § d2y 1 x dy y 0
¨ dx2 dx
d2y dy © ·
2) x2 dx2 x dx y ¸
¹

0

In this equation, the highest order derivative

3) d2y = 2t is d2y and its power is one. Therefore this
dt 2 dx2

160

equation has degree one. 2) Particular Solution:

2) d2y 3 1 § dy 2 A solution of the differential equation
dx2 ©¨ dx which can be obtained from the general solution
· by giving particular values to the arbitrary
¸¹ constants is called a particular solution.

In this equation, the highest order

derivative is d2y but to determine the degree of SOLVED EXAMPLES
dx2

this equation, first we have to remove the cube

root by raising both sides to the power 3. 1) Verify that the function y = aex + be−2x, a,b

∈ R is a solution of the differential equation

§ d 2 y 3 § dy 2 d2y dy 2y .
©¨ dx dx2 dx
¨ · 1 · ∴ the degree of this
© ¸ ¸¹
dx2 ¹ Solution: Consider the function

equation is 3. y = aex + be−2x ........... (I)

3) dy 2x 7 Differentiating both sides of equation I
dx dy with respect to x, we get

dx dy
dx
The equation can be written as = aex − 2be−2x .................... II and

§ dy 2 2x 7 . Now highest order derivative is Differentiating both sides of equation II
with respect to x, we get
·
©¨ dx ¹¸

dy and its power is two.Hence the equation has d2y = aex + 4be−2x .................... III
dx dx2

degree two. Now, L.H.S = d2y + dy
dx2 dx

We have learnt: = (aex + 4 be−2x) + (aex − 2be−2x)
(from II and III)
To find the degree of the differential = 2 aex + 2be−2x
equation, make all the derivatives free from = 2(aex + be−2x) = 2 y (from I)
fractional indices and negative sign, if any. = R.H.S.

8.1.4 Solution of a Differential Equation: Therefore, the given function is a general
solution of the given differential equation.
A function of the form y = f(x) + c which
satisfies the given differential equation is called 2) Verify that the function y = e−x + ax
the solution of the differential equation. + b, where a, b ∈ R is a solution of the

Every differential equation has two types differential equation ex § d2y · 1
of solutions: 1) General and 2) Particular ¨ dx2 ¸
© ¹
1) General Solution:
Solution: y = e−x + ax + b ..................... I
A solution of the differential equation
in which the number of arbitrary constants is Differentiating both sides of equation I
equal to order of differential equation is called a with respect to x, we get
general solution.

161

∴ dy = − e−x + a ........................ II iii) y = ex, dy =y
dx dx

Differentiating both sides of equation II iv) y = 1 − logx d2y =dd1yx 2y
with respect to x, we get dx2
x2

d2y =ddeyx−x 2y v) y = aex + be−x d2y =ddyyx
dx2 dx2
d2y =ddeyxx 2y
Consider L.H.S. = ex dx2 (e 2−xy)

= e0 = 1 = R.H.S. vi) ax2 + by2 = 5 d2y +ddxyx¨§© ddyx2¹¸·y2 dy
dx2 dx
Therefore, the given function is a general xy = y.
solution of the given differential equation.

EXERCISE 8.1 8.1.5 Formation of a differential equation:
By eliminating arbitary constants

1. Determine the order and degree of the If the order of a differential equation is n,
following differential equations. differentiate the equation n times to eliminate
arbitrary constants.
i) d2x § dx 2 8 0
dt 2 ©¨ dt SOLVED EXAMPLES
·
¹¸ 1. Form the differential equation of the line
having x-intercept 'a' and y-intercept 'b'.
ii) § d2y ·2 § dy ·2 ax
¨ dx2 ¸ ©¨ dx ¹¸ Solution: The equation of a line is given by,
© ¹ x y 1 ............... I

iii) d4y ª § dy 2 3 ab
dx4 «1 ¨© dx
¬« · º Differeentiating equation I with r. t. x we
¸¹ » get,
¼»

iv) y ''' 2 2 y '' 2 6 y ' 7 y 0

1 § dy 3/ 2

v) 1 ·
¨© dx ¹¸
§ dy 2 1 1 dy 0, ? 1 dy 1

·
©¨ dx ¸¹
a b dx b dx a

vi) dy = 7 d 2 y ? dy b ................. II
dx dx2 dx a

§ d 3 y 1/ 6 Differentiating equation II with r. t. x we

vii) · 9 d2y =dd0yx
¨ ¸ dx2
© dx3 ¹ get, is2tyhe required differential

2. In each of the following examples, verify equation.
that the given function is a solution of the
corresponding differential equation. 2. Obtain the differential equation from the
relation Ax2 + By2 = 1, where A and B are
Solution D.E. constant.

i) xy = log y + y' (1 − xy) = y2 Solution: The given equation is
k
Ax2 + By2 = 1 .................... I
ii) y = xn d2y
dx2 −ddnyx× x2dyy Differentiating equation I twice with
x2 dx +ny=0 respect to x, we get,

162

2Ax + 2 By dy =0 that is 2x3 - y3 + 3xy2 dy = 0.
dx dx

Ax + By dy = 0 .................... II and is the required differential equation.
dx
We have learnt:
§ d2y § dy · 2 ·
©¨¨ ax2 ©¨ dx ¸¹ ¸¸¹ To form a differential equation by
A+B y = 0 ............... III eliminating arbitrary constants, if 'n' arbitrary
constants are present in the given equation
since the equations I, II & III are consistent then differentiate the given equation 'n' times.

in A and B,

x2 y2 1

?x y dy 0 0 EXERCISE 8.2
dx
1. Obtain the differential equation by
1 y d2y § dy 2 0 eliminating arbitrary constants from the
dx2 ¨© dx following equations.
·
¸¹

°­ ª d2y § dy 2 º dy °½ i) y = Ae3x + B.e−3x
® « dx2 ©¨ dx » dx ¾ c1
? ¯° x «¬ y · ¼» 1u y. ¿° 0 ii) y = c2 + x
¹¸

xy d2y x dy 2 y dy 0 iii) y = (c1 + c2x) ex
dx2 dx dx
iv) y = c1e3x + c2e2x

is the required differential equation. v) y2 = (x + c)3

3. Form the differential equation whose 2. Find the differential equation by eliminating
general solution is x3 +y3 = 4ax arbitrary constants from the relation
x2 + y2 = 2ax
Solution: Given equation is
3. Form the differential equation by
x3 + y3 = 4ax .............I eliminating arbitrary constants from the
relation bx + ay = ab.
Since the given equation contains only
one arbitrary constant, the required differential 4. Find the differential equation whose
equation will be of order one. general solution is x3 + y3 = 35ax.

Differentiating equation I with respect to x, 5. Form the differential equation from the
relation x2 + 4y2 = 4b2.
we get, dy
dx
3x2 + 3y2 = 4a .................. II

To eliminate a from the equations I & II, 8.2.1 Solution of a Differential Equation:

substitute the value of 4a from equation II Variable Separable Method.

in I dy Sometimes, a differential equation of first
dx order and first degree can be written in the form
x3 + y3 = x (3x2 + 3y2 ) f(x) dx + g(y) dy = 0 ........................ I

x3 + y3 = 3x3 + 3xy2 dy where f(x) and g(y) are functions of x and y
dx respectively.

163