Class 12 Maths NCERT Solutions

NCERT Class 12 Maths Solutions Chapter 8 – Application of Integrals

NCERT Class 12 Maths Solutions Chapter 8 – Application of Integrals: I welcome all the students on this page. You will get the full free solution of Application of Integrals. So read the full article till the end.

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NCERT Class 12 Maths Solutions Chapter 8 – Application of Integrals

The topics and sub-topics included in the Integrals chapter are the following:

Section NameTopic Name
8Application of Integrals
8.1Introduction
8.2Area under Simple Curves
8.3Area between Two Curves

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Page No 365:

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.

Answer:

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.

Question 2:

Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, y2 = 9xx = 2, and x = 4, and the x-axis is the area ABCD.

Page No 366:

Question 3:

Find the area of the region bounded by x2 = 4yy = 2, y = 4 and the y-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, x2 = 4yy = 2, and y = 4, and the y-axis is the area ABCD.

Question 4:

Find the area of the region bounded by the ellipse 

Answer:

The given equation of the ellipse, , can be represented as

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Question 5:

Find the area of the region bounded by the ellipse 

Answer:

The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse = 

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line and the circle 

Answer:

The area of the region bounded by the circle, , and the x-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of OAC 

Area of ABC 

Therefore, required area enclosed =

32 + π3 – 32 = π3 square units

Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line 

Answer:

The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the area ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis.

∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, , is  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.

Answer:

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

From (1) and (2), we obtain

Therefore, the value of a is .

Question 9:

Find the area of the region bounded by the parabola x2 and 

Answer:

The area bounded by the parabola, x2 = y,and the line,, can be represented as

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x2 = y, and line, x, is A (1, 1).

Area of OACO = Area ΔOAM – Area OMACO

Area of ΔOAM   

Area of OMACO 

⇒ Area of OACO = Area of ΔOAM – Area of OMACO

Therefore, required area = units

Question 10:

Find the area bounded by the curve x2 = 4y and the line x = 4– 2

Answer:

The area bounded by the curve, x2 = 4y, and line, x = 4– 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 .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area = 

Question 11:

Find the area of the region bounded by the curve y2 = 4x and the line x = 3

Answer:

The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

Question 12:

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and = 2 is

A. π

B. 

C. 

D. 

Answer:

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

Thus, the correct answer is A.

Question 13:

Area of the region bounded by the curve y2 = 4xy-axis and the line y = 3 is

A. 2

B. 

C. 

D. 

Answer:

The area bounded by the curve, y2 = 4xy-axis, and y = 3 is represented as

Thus, the correct answer is B.

Page No 371:

Question 1:

Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y

Answer:

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.

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.

Therefore, Area OBCO = Area OMBCO – Area OMBO

=∫02(9-4×2)4dx-∫02x24dx

=∫02322-x2dx-14∫02x2dx

=x2322-x2+98sin-12×302-14×3302

=24+98sin-1223-11223

=122+98sin-1223-132

=162+98sin-1223

=1226+94sin-1223Therefore, the required area OBCDO is units

Question 2:

Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1

Answer:

The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by the shaded area as

On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as Aand B.

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 .

Therefore, required area OBCAO = units

Question 3:

Find the area of the region bounded by the curves y = x+ 2, xx = 0 and x = 3

Answer:

The area bounded by the curves, y = x+ 2, xx = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

Question 4:

Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

Answer:

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain

Area (ΔABC) = (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 = 4.

Answer:

The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and = 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 (ΔACB) = Area (OLBAO) –Area (OLCAO)

Page No 372:

Question 6:

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)

Answer:

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 (ΔOAB)

Thus, the correct answer is B.

Question 7:

Area lying between the curve y2 = 4x and y = 2x is

A. 

B. 

C. 

D. 

Answer:

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 (OCABO) – Area (ΔOCA)

 square units

Thus, the correct answer is B.

Page No 375:

Question 1:

Find the area under the given curves and given lines:

(i) y = x2x = 1, x = 2 and x-axis

(ii) y = x4x = 1, x = 5 and x –axis

Answer:

  1. The required area is represented by the shaded area ADCBA as
  1. 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 = 4x2x = 0, y = 1 and = 4

Answer:

The area in the first quadrant bounded by y = 4x2x = 0, y = 1, and = 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

Answer:

The given equation is 

The corresponding values of and y are given in the following table.

x– 6– 5– 4– 3– 2– 10
y3210123

On plotting these points, we obtain the graph of  as follows.

It is known that, 

Question 5:

Find the area bounded by the curve y = sin 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 

Answer:

The area of the smaller region bounded by the ellipse, , 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 

Answer:

The area of the smaller region bounded by the ellipse, , 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 – = 11]

Answer:

The area bounded by the curve, , 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

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 – 3+ 5 = 0

Answer:

The given equations of lines are

2x + y = 4 … (1)

3x – 2y = 6 … (2)

And, x – 3+ 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

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

Question 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 curvex-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.

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