NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals Miscellaneous Exercise
Page No 375:
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
Answer:
- The required area is represented by the shaded area ADCBA as
- The required area is represented by the shaded area ADCBA as
Question 2:
Find the area between the curves y = x and y = x2
Answer:
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 (ΔOCA) – Area (OCABO) … (1)
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
Answer:
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
Area of ABCDA = ∫14 x dy
=∫14 y2 dy as, y = 4×2
=12∫14y dy
=12×23y3/214
=1343/2 – 13/2
=138 – 1
=13×7
=73 square units
Question 4:
Sketch the graph of 
and evaluate
and evaluateAnswer:
The given equation is
The corresponding values of x and y are given in the following table.
x
|
– 6
|
– 5
|
– 4
|
– 3
|
– 2
|
– 1
|
0
|
y
|
3
|
2
|
1
|
0
|
1
|
2
|
3
|
On plotting these points, we obtain the graph of as follows.
as follows.
It is known that, 
Question 5:
Find the area bounded by the curve y = sin x between x = 0 and x = 2π
Answer:
The graph of y = sin x can be drawn as
∴ Required area = Area OABO + Area BCDB
Question 6:
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Answer:
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and 
.
.
We draw AC perpendicular to x-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Question 7:
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Answer:
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)
Question 8:
Find the area of the smaller region bounded by the ellipse 
and the line 
and the line Answer:
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
, and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 9:
Find the area of the smaller region bounded by the ellipse 
and the line 
and the line Answer:
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
, and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 10:
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
Answer:
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 OACO as
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).
Area of OACO = ∫-12x + 2 dx – ∫-12 x2 dx⇒Area of OACO = x22 + 2x-12 – 13×3-12⇒Area of OACO = 222+22 – -122+2-1 – 1323 – -13⇒Area of OACO = 2 + 4 – 12-2 – 138 + 1⇒Area of OACO = 6 + 32 – 3
⇒Area of OACO = 3 + 32 = 92 square units
Question 11:
Using the method of integration find the area bounded by the curve 
[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]
Answer:
The area bounded by the curve, , is represented by the shaded region ADCB as
, is represented by the shaded region ADCB as
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
Page No 376:
Question 12:
Find the area bounded by curves 
Answer:
The area bounded by the curves, , is represented by the shaded region as
, is represented by the shaded region as
It can be observed that the required area is symmetrical about y-axis.
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)
Answer:
The vertices of ΔABC are A (2, 0), B (4, 5), and C (6, 3).
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment CA is
Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
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
Answer:
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 ΔABC. AL and CM are the perpendiculars on x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Question 15:
Find the area of the region 
Answer:
The area bounded by the curves, , is represented as
, is represented as
The points of intersection of both the curves are
.
.
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
Therefore, the required area is 
units
unitsQuestion 16:
Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B. 
C. 
D. 
Answer:
Required Area =
∫-2 0ydx+∫01ydx
=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. unitsThus, the correct answer is D.
Question 17:
The area bounded by the curve
, x-axis and the ordinates x = –1 and x = 1 is given by
, 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. 
C. 
D. 
Answer:
Thus, the correct answer is C.
Question 18:
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A. 
B. 
C. 
D.
Answer:
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)
Area bounded by the circle and parabola
=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23×3202+2×216-x2+162sin-1×424 =46322-0+20+8sin-11-23+8sin-112=1633+28×π2-23-8×π6=1633+24π-23-4π3=1633+8π-43-8π3=163+24π-43-8π3=16π+1233=434π+3 square units
Area of circle = π (r)2
= π (4)2
= 16π square units
∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units
Thus, the correct answer is C.
Question 19:
The area bounded by the y-axis, y = cos x and y = sin x when 
A. 
B. 
C. 
D. 
Answer:
The given equations are
y = cos x … (1)
And, y = sin x … (2)
Required area = Area (ABLA) + area (OBLO)
Integrating by parts, we obtain
Thus, the correct answer is B.
