Upload
others
View
4
Download
0
Embed Size (px)
Citation preview
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Book Name: NCERT Solutions
EXERCISE- 8.1
Question 1:
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Solution 1:
The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is
the area ABCD.
Area of ABCD = 1
1ydx
1
1xdx
43
2
1
3
2
x
3 3
2 22
4 13
28 1
3
14 units
3
Question 2:
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Solution 2:
The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area
ABCD.
Area of ABCD = 1
2ydx
1
2
43
2
2
3
332
xdx
x
43
2
2
2 x
3 3
2 22 4 2
28 2 2
3
16 4 2 units
Question 3:
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution 3:
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area
ABCD.
Area of ABCD = 1
2xdx
1
2
1
2
2
2
ydx
ydx
43
2
2
322
y
3 3
2 24
4 23
48 2 2
3
32 8 2 units
3
Question 4:
Find the area of the region bounded by the ellipse 2 2x
116 9
y
Solution 4:
The given equation of the ellipse, 2 2x
116 9
y , can be represented as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area of OAB
Area of ABCD = 4
0ydx
24
03 1
16
xdx
42
0
316
4x dx
4
2 1
0
3 1616 sin
4 2 2 4
x xx
1 132 16 16 8sin 1 0 8sin 0
4
3 8
4 2
34
4
3
Therefore, area bounded by the ellipse = 4 × 3π = 12π units
Question5:
Find the area of the region bounded by the ellipse 2 2
14 9
x y
Solution 5:
The given equation of the ellipse can be represented as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
2 2
14 9
x y
2
3 1 ... 14
xy
It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB
2
0
22
0
22
0
Area of OAB= ydx
3 1 Using (1)4
34
2
x
x
2
2 1
0
3 44 sin
2 2 2 2
3 2
2 2
3
2
x xx
Therefore, area bounded by the ellipse = 3
4 6 units2
Question 6:
Find the area of the region in the first quadrant enclosed by x-axis, line 3x y and the circle
2 2 4x y
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Solution 6:
The area of the region bounded by the circle, 2 2 4x y , 3x y , and the x-axis is the area
OAB.
The point of intersection of the line and the circle in the first quadrant is 3,1 .
Area OAB = Area ∆OCA + Area ACB
Area of OAC = 1 1 3
3 12 2 2
OC AC …
Area of ABC = 2
3ydx
22
3
2
2 1
3
1
4
44 sin
2 2 2
3 32 4 3 2sin
2 2 2
x dx
x xx
32
2 3
3 2
2 3
3
3 2
…
Therefore, area enclosed by x-axis, the line 3x y , and the circle 2 2 4x y in the first
quadrant = 3 3
2 3 2 3
units
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Question 7:
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line, 2
ax .
Solution 7:
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, 2
ax , is the area
ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC
Area of ABC = 2
a
a ydx
2 2
2
22 2 1
2
sin2 2
a
a
a
a
a x dx
x a xa x
a
2 2 22 1 1
sin2 2 2 22 2 2
a a a aa
2 2
.4 2 42 2 2
a a a a
2 2 2
2
2
4 4 8
14 2
14 2
a a a
a
a
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
2 2
2 1 14 2 2 2
a aArea ABCD
Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off all the line, 2
ax , is
2
12 2
a
units.
Question 8:
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value
of a.
Solution 8: The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts. ∴ Area OAD = Area ABCD
It can be observed that the given area is symmetrical about x-axis. ⇒ Area OED = Area EFCD
Area OED = 0
a
ydx
0
a
xdx
3
2
0
3
2
a
x
3
22
3a …(1)
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Area of EFCD = 4
0xdx
43
2
0
3
2
x
3
22
8 ... 23
a
From (1) and (2), we obtain
3 3
2 2
3
2
3
2
3
2
2 28
3 3
2. 8
4
4
a a
a
a
a
Therefore, the value of a is 2
34 .
Question 9:
Find the area of the region bounded by the parabola y = x2 and y x
Solution 9:
The area bounded by the parabola, x2 = y, and the line, y x , can be represented as
The given area is symmetrical about y-axis. ∴ Area OACO = Area ODBO
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).
Area of OACO = Area ∆OAB – Area OBACO
13
1 12
0 00
1 1 1Area of OAB= 1 1
2 2 2
1Area of OBACO = ydx x dx
3 3
OB AB
x
⇒ Area of OACO = Area of ∆OAB – Area of OBACO
= 1 1
2 3
= 1
6
Therefore, required area = 1 1
26 3
units.
Question 10:
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Solution 10:
The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area
OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are 1
1,4
.
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Area OBAO = Area OBCO + Area OACO … (1) Then, Area OBCO = Area OMBC – Area OMBO
22 2
0 0
2 22 3
0 0
2
4 4
1 12
4 2 4 3
1 1 82 4
4 4 3
x xdx dx
x xx
3 2
2 3
5
6
Similarly, Area OACO = Area OLAC – Area OLAO 2
0 0
1 1
0 02 3
1 1
2
4 4
1 12
4 2 4 3
x xdx dx
x xx
31 11 1
2 14 2 4 3
1 1 12
4 2 2
1 1 1
2 8 12
7
24
Therefore, required area = 5 7 9
6 24 8
units
Question 11:
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
Solution 11:
The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The area OACO is symmetrical about x-axis. ∴ Area of OACO = 2 (Area of OAB)
Area OACO = 3
02 ydx
3
0
33
2
0
2 2
342
xdx
x
3
28
33
8 3
Therefore, the required area is 8 3 units.
Question 12:
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x =
2 is
A. π
B. 2
C. 3
D. 4
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Solution 12:
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented
as
2
0
22
0
2
2 1
0
Area OAB=
4
44 sin
2 2 2
22
units
ydx
x dx
x xx
Thus, the correct answer is A.
Question 13:
Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
A. 2
B. 9
4
C. 9
3
D. 9
2
Solution 13:
The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
3
0
23
0
33
0
Area OAB =
4
1
4 4
127
12
9 units
4
xdy
ydy
y
Thus, the correct answer is B.
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
EXERCISE- 8.2
Question 1:
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
Solution 1:
The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of
intersection as B 1
2,2
and D 1
2,2
.
It can be observed that the required area is symmetrical about y-axis. ∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are 2,0 .
Therefore, Area OBCO = Area OMBCO – Area OMBO
2 22 2
0 0
2 22 2
0 0
9 4
4 4
1 19 4
2 4
x xdx
x dx x dx
22 32 1
0 0
31
1 9 2 19 4 sin
4 2 3 4 3
1 9 2 2 12 9 8 sin 2
4 2 3 12
x xx x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1
1
1
2 9 2 2 2sin
4 8 3 6
2 9 2 2sin
12 8 3
1 2 9 2 2sin
2 6 4 3
Therefore, the required area OBCDO is
1 11 2 9 2 2 2 9 2 22 sin sin units
2 6 4 3 6 4 3
Question 2:
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1
Solution 2:
The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y2 = 1, is represented by the shaded
area as
On solving the equations, (x – 1)2 + y2 = 1 and x2 + y2 = 1, we obtain the point of intersection
as A 1 3
,2 2
and B 1 3
,2 2
It can be observed that the required area is symmetrical about x-axis. ∴ Area OBCAO = 2 × Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are 1
,02
Area OCAO Area OMAO Area MCAM
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1
12 221
02
1 1 1x dx x dx
1
122 1 2 1
10
2
1 1 11 1 sin 1 1 sin
2 2 2 2
x xx x x x
2 2
1 1 1 11 1 1 1 1 1 1 1 1 11 sin 1 sin 1 sin 1 1 sin
4 2 2 2 2 2 4 2 2 2
3 1 1 1 3 1
8 2 6 2 2 2 2 8 2 6
3
4 12 4 4 12
3
4 6 2
2 3
6 4
Therefore, required area OBCAO = 2 3 2 3
2 units6 4 3 2
Question 3:
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Solution 3:
The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented by the shaded
area OCBAO as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Then, Area OCBAO = Area ODBAO – Area ODCO
3 32
0 0
3 33 2
0 0
2
23 2
x dx xdx
x xx
99 6
2
915
2
21units
2
Question 4:
Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0),
(1, 3) and (3, 2).
Solution 4:
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (∆ACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)
Equation of line segment AB is
3 00 1
1 1
31
2
y x
y x
1
21
11
3 3 3 1 1Area ALBA 1 1 1 3units
2 2 2 2 2 2
xx dx x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Equation of line segment BC is
2 33 1
3 1
17
2
y x
y x
3
23
11
1 1 1 9 1Area BLMCB 7 7 21 7 5units
2 2 2 2 2 2
xx dx x
Equation of line segment AC is
2 00 1
3 1
11
2
y x
y x
3
23
11
1 1 1 9 1Area AMCA 1 3 1 4units
2 2 2 2 2 2
xx dx x
Therefore, from equation (1), we obtain
Area (∆ABC) = (3 + 5 – 4) = 4 units
Question 5: Using integration find the area of the triangular region whose sides have the equations y = 2x
+1, y = 3x + 1 and x = 4.
Solution 5: The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).
It can be observed that,
Area (∆ACB) = Area (OLBAO) –Area (OLCAO)
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
4 4
0 03 1 2 1x dx x dx
4 42 2
0 0
3 2
2 2
x xx x
24 4 16 4
28 20
8units
Question6:
Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
A. 2 (π – 2)
B. π – 2
C. 2π – 1
D. 2 (π + 2) Solution 6: The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the
shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area (∆OAB)
2 2
2
0 04 2x dx x dx
2 2
2 1
0 0
44 sin 2
2 2 2 2
x x xx x
2. 4 22
2 units
Thus, the correct answer is B.
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Question 7: Area lying between the curve y2 = 4x and y = 2x is
A.2
3 C.
1
4
B. 1
3 D.
3
4
Solution 7: The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO
as
The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0). ∴ Area OBAO = Area (∆OCA) – Area (OCABO) 1 1
0 02 2xdx xdx
13
12 2
0
0
2 232
2
x x
41
3
1
3
1 units
3
Thus, the correct answer is B.
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Miscellaneous EXERCISE
Question 1: Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x –axis
Solution 1: i. The required area is represented by the shaded area ADCBA as
Area ADCBA = 2
1ydx
22
1x dx
23
13
x
8 1
3 3
7 units
3
ii. The required area is represented by the shaded area ADCBA as
Area of ADCBA = 5
4
1x dx
55
15
x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
55 1
5 5
4 15
5
1625
5
624.8 units
Question 2: Find the area between the curves y = x and y = x2
Solution 2: The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2, is A (1, 1).
We draw AC perpendicular to x-axis. ∴ Area (OBAO) = Area (∆OCA) – Area (OCABO) … (1) 1 1
2
0 0xdx x dx
1 12 3
0 02 3
x x
1 1
2 3
1 units
6
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Question 3:
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y
= 4
Solution 3:
The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is represented by the
shaded area ABCDA as
4
1
4
1
Area ABCD=
4
2
xdx
dx
43
2
1
1
32
2
y
3
21
4 13
18 1
3
7 units
3
Question 4:
Sketch the graph of 3y x and evaluate 0
63x dx
Solution 4:
The given equation is 3y x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The corresponding values of x and y are given in the following table.
On plotting these points, we obtain the graph 3y x of as follows.
It is known that, 3 0 for -6 x -3 and 3 0 for -3 x 0x x
0 3 0
6 6 33 3 3x dx x dx x dx
3 02 2
6 3
3 32 2
x xx x
2 2 2
3 6 33 3 3 6 0 3 3
2 2 2
9 9
2 2
9
Question 5:
Find the area bounded by the curve y = sin x between x = 0 and x = 2π Solution 5:
The graph of y = sin x can be drawn as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
∴ Required area = Area OABO + Area BCDB
2
0sin sinxdx xdx
2
0cos cosx x
cos cos 0 cos 2 cos
1 1 1 1
2 2
2 2 4 units
Question 6:
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Solution 6:
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented by the
shaded area OABO as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The points of intersection of both the curves are (0, 0) and 2
4 4,
a a
m m
.
We draw AC perpendicular to x-axis. ∴ Area OABO = Area OCABO – Area (∆OCA) 2 2
4 4
0 02
a a
m maxdx mxdx
2
2
4
3 422
0
0
32
2
2 2
23 2
2
4 4 4
3 2
a
m a
mx xa m
a m aa
m m
2 2
3 2
2 2
3 3
32 16
3 2
32 8
3
a m a
m m
a a
m m
2
3
8 units
3
a
m
Question 7:
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Solution 7:
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by
the shaded area OBAO as
The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
2
1 4
2 2
1 33 12
2 4
xx dx dx
4 42 3
2 2
1 3 312
2 2 4 3
x xx
1 124 48 6 24 64 8
2 4
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1 190 72
2 4
45 18
27 units
Question 8:
Find the area of the smaller region bounded by the ellipse 2 2
19 4
x y and the line 1
3 2
x y
Solution 8:
The area of the smaller region bounded by the ellipse, 2 2
19 4
x y , and the line, 1
3 2
x y , is
represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO)
23 3
0 0
3 32
0 0
2 1 2 19 3
2 29 3 1
3 3
x xdx dx
x dx dx
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
33 2
2 1
0 0
2 9 29 sin 3
3 2 2 3 3 2
x x xx x
2 9 2 99
3 2 2 3 2
2 9 9
3 4 2
2 92
3 4
32 units
2
Question 9:
Find the area of the smaller region bounded by the ellipse 2 2
2 21
x y
a b and the line 1
x y
a b
Solution 9:
The area of the smaller region bounded by the ellipse, 2 2
2 21
x y
a b , and the line, 1
x y
a b , is
represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO)
2
20 01 1
a ax xb dx b dx
a a
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
2 2
0 0
a ab ba x dx a x dx
a a
2 22 2 1
0 0
sin2 2 2
a a
b x a x xa x ax
a a
2 22
2 2 2
b a aa
a
2 2
2
4 2
12 2
b a a
a
ba
a
12 2
24
ab
ab
Question 10:
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x axis
Solution 10:
The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis is
represented by the shaded region OABCO as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1). ∴ Area OABCO = Area (BCA) + Area COAC
1 0
2
2 12x dx x dx
1 02 3
2 1
22 3
x xx
2 2 31 2 1
2 1 2 22 2 3
1 12 2 4
2 3
5 units
6
Question 11:
Using the method of integration find the area bounded by the curve 1x y
[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]
Solution 11:
The area bounded by the curve, 1x y , is represented by the shaded region ADCB as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis. ∴ Area ADCB = 4 × Area OBAO
1
04 1 x dx
12
0
42
xx
14 1
2
14
2
2 units
Question 12:
Find the area bounded by curves 2, : and y= xx y y x
Solution 12:
The area bounded by the curves, 2, : and y= xx y y x , is represented by the shaded region
as
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
It can be observed that the required area is symmetrical about y-axis.
Required area = 2 [ Area (OCAO) – Area (OCADO)]
1 12
0 02 xdx x dx
1 12 3
0 0
22 3
x x
1 12
2 3
1 12 units
6 3
Question 13:
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices
are A (2, 0), B (4, 5) and C (6, 3)
Solution 13:
The vertices of ∆ABC are A (2, 0), B (4, 5), and C (6, 3).
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Equation of line segment AB is
5 00 2
4 2
2 5 10
52 ... 1
2
y x
y x
y x
Equation of line segment BC is
3 55 4
6 4y x
2 10 2 8
2 2 18
9 ... 2
y x
y x
y x
Equation of line segment CA is
0 33 6
2 6y x
4 12 3 18
4 3 6
y x
y x
32 ... 3
4y x
Area (∆ABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
4 6 6
2 4 2
5 32 9 2
2 4x dx x dx x dx
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
4 6 62 2 2
2 4 2
5 32 9 2
2 2 2 4 2
x x xx x x
5 38 8 2 4 18 54 8 36 18 12 2 4
2 4
35 8 8
4
13 6
7 units
Question 14:
Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Solution 14:
The given equations of lines are
2x + y = 4 … (1) 3x – 2y = 6 … (2) And, x – 3y + 5 = 0 … (3)
The area of the region bounded by the lines is the area of ∆ABC. AL and CM are the perpendiculars on x-axis.
Area (∆ABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
4 2 4
1 1 2
5 3 64 2
3 2
x xdx x dx dx
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
4 42 2
22
11 2
1 1 35 4 6
3 2 2 2
x xx x x x
1 1 18 20 5 8 4 4 1 24 24 6 12
3 2 2
1 45 11 6
3 2 2
151 3
2
15 15 8 74 units
2 2 2
Question 15:
Find the area of the region 2 2 2, : 4 .4 4 9x y y x x y
Solution 15:
The area bounded by the curves, 2 2 2, : 4 .4 4 9x y y x x y , is represented as
The points of intersection of both the curves are 1 1
, 2 and , 22 2
.
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x-axis. ∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1 3
22 21
02
1 32 2
2 21
02
12 9 4
2
12 3 2
2
xdx x dx
xdx x dx
Question 16:
Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B.15
4
C. 15
4
D. 17
4
Solution 16:
Required area = 1
2ydx
13
2
14
24
x dx
x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
41 ( 2)
4 4
1 154
4 4units
Thus, the correct answer is B.
Question 17:
The area bounded by the curve y x x , x-axis and the ordinates x = –1 and x = 1 is given by
[Hint: y = x2 if x > 0 and y = –x2 if x < 0]
A.0
B. 1
3
C. 2
3
D. 4
3
Solution 17:
Required area = 1
1ydx
1
1
0 12 2
1 0
x x dx
x dx x dx
0 13 3
1 03 3
x x
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1 1
3 3
2 units
3
Thus, the correct answer is C.
Question 18:
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A. 44 3
3
B. 44 3
3
C. 48 3
3
D. 48 3
3
Solution 18:
The given equations are
x2 + y2 = 16 … (1) y2 = 6x … (2)
Area bounded by the circle and parabola
2 4
2
0 2
2 Area OADO Area ADBA
2 16 16xdx x dx
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
23
42
2 1
2
0
162 6 2 16 sin
3 2 2 4
2
x x xx
23
12
0
2 12 6 2 8. 16 4 8sin
3 2 2x
4 62 2 2 4 12 8
3 6
16 3 88 4 3
3 3
44 3 6 3 3 2
3
43 4
3
44 3 units
3
Area of circle = π (r)2
= π (4)2
= 16π units
4Required area = 16 - 4 3
3
44 3 4 3
3
48 3 units
3
Thus, the correct answer is C.
Question 19:
The area bounded by the y-axis, y = cos x and y = sin x when 02
x
A. 2 2 1
B. 2 1
C. 2 1
D. 2
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
Solution 19:
The given equations are
y = cos x … (1) And, y = sin x … (2)
Required area = Area (ABLA) + area (OBLO)
11
21
02
1 11 1
1 1
2 2
cos sin
xdy xdy
ydy xdy
Integrating by parts, we obtain 1
11 2 1 2 2
10
2
cos 1 sin 1y y y x x x
1 1 11 1 1 1 1 1cos 1 cos 1 sin 1 1
2 22 2 2 2
1 11
4 2 2 4 2 2
21
2
2 1units
Thus, the correct answer is B.
Put 2x = t dx = 2
dt
When x = 3
2, t = 3 and when x =
1
2, t = 1
1
3 2 22
0 1
12 3
4xdx t dt
Class XII Chapter 8 – Application of Integrals Maths
______________________________________________________________________________
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a
Top tutor.
1
233
22 1
1
0
1 92 9 sin
3 4 2 2 3
2
x t tt
3
2 2 21 12 1 1 3 9 3 1 9 12 9 3 sin 9 1 sin
3 2 4 2 2 3 2 2 3
1 12 1 9 1 9 10 sin 1 8 sin
4 2 2 2 33 2
1
1
2 1 9 9 12 sin
3 4 4 2 3
2 9 2 9 1sin
3 16 4 8 3
19 9 1 2sin
16 8 3 12
Therefore, the required area is1 19 9 1 2 9 9 1 1
2 sin sin units16 8 3 12 8 4 3 3 2