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:

img 620f552fe5ea2

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.

img 620f553018ec4

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:

img 620f553065b8e

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

img 620f55308776d

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:

img 620f55310eca8

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

img 620f5531305f4

Question 4:

Find the area of the region bounded by the ellipse img 620f5531ac916

Answer:

The given equation of the ellipse, img 620f5531ac916, can be represented as

img 620f5531cddcb

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

∴ Area bounded by ellipse = 4 × Area of OAB

img 620f5531f0963

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

Question 5:

Find the area of the region bounded by the ellipse img 620f553278c58

Answer:

The given equation of the ellipse can be represented as

img 620f553297b74

img 620f5532ba62c

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

∴ Area bounded by ellipse = 4 × Area OAB

img 620f55330fac1

Therefore, area bounded by the ellipse = img 620f55338a543

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line img 620f5533a4fc3and the circle img 620f5533be7af

Answer:

The area of the region bounded by the circle, img 620f5533d79ba, and the x-axis is the area OAB.

img 620f5533f1d49

The point of intersection of the line and the circle in the first quadrant is img 620f55341a200.

Area OAB = Area ΔOCA + Area ACB

Area of OAC img 620f55343a2a5

Area of ABC img 620f5534aae2b

img 620f5534c3eb5

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 img 620f55354b7ce

Answer:

The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, img 620f55354b7ce, is the area ABCDA.

img 620f553569f70

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

∴ Area ABCD = 2 × Area ABC

img 620f55358b8f9

img 620f553614cc1

Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, img 620f55354b7ce, is img 620f55365e2d8 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

img 620f55367d8c1

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

⇒ Area OED = Area EFCD

img 620f5536a0575

 

From (1) and (2), we obtain

img 620f5536c8b91

Therefore, the value of a is img 620f55371f35a.

Question 9:

Find the area of the region bounded by the parabola x2 and img 620f55373d3e3

Answer:

The area bounded by the parabola, x2 = y, and the line, img 620f55373d3e3, can be represented as

img 620f55376c053

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 img 620f55378dffd  img 620f5537ac49b

Area of OMACO img 620f5537cb295

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

img 620f5537ebf8b

Therefore, required area = img 620f5538172b0units

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.

img 620f553837d9b

Let A and B be the points of intersection of the line and parabola.

Coordinates of point img 620f5538592d6.

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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

img 620f553878132

Similarly, Area OACO = Area OLAC – Area OLAO

img 620f5538c3932

Therefore, required area = img 620f55391debc

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.

img 620f55393cb7e

The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

img 620f55395e357

Therefore, the required area is img 620f5539a9e85units.

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. img 620f5539c871c

C. img 620f5539e6bfe

D. img 620f553a1196d

Answer:

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

img 620f553a3043c

img 620f553a52399

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. img 620f553a9eb9c

C. img 620f553abdf60

D. img 620f553adc4f9

Answer:

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

img 620f553b07555

img 620f553b29129

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.

img 620f553b74595

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection asimg 620f553b95f2a.

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 areimg 620f553bdf675.

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 img 620f553c09b01units

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

img 620f553c80f41

On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as Aimg 620f553ca1ff3and Bimg 620f553cc4053.

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 img 620f553ce3375.

img 620f553d0e3ea

Therefore, required area OBCAO = img 620f553d8e4f7units

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

img 620f553dd7ead

Then, Area OCBAO = Area ODBAO – Area ODCO

img 620f553e0550d

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)

img 620f553e503b9

Equation of line segment AB is

img 620f553e72352

Equation of line segment BC is

img 620f553ee7941

Equation of line segment AC is

img 620f553f6b6ae

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).

img 620f553fe2009

It can be observed that,

Area (ΔACB) = Area (OLBAO) –Area (OLCAO)

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img 620f55400f4e8

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

img 620f554056711

It can be observed that,

Area ACBA = Area OACBO – Area (ΔOAB)

img 620f5540779eb

Thus, the correct answer is B.

Question 7:

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

A. img 620f5540c2a29

B. img 620f5540ddbec

C. img 620f554114758

D. img 620f554130223

Answer:

The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO as

img 620f55414b39c

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)

img 620f55417352bimg 620f55418fd56 img 620f5541abf4bimg 620f5541ca89f

img 620f5541e8981

img 620f5542139fb 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

img 620f554233f26

img 620f55425977f

  1. The required area is represented by the shaded area ADCBA as

img 620f5542cd9e5

img 620f5542f3801

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

img 620f554378186

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)

img 620f554397155

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

img 620f5543e11e0

 

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 img 620f55440bb75and evaluateimg 620f55442757e

Answer:

The given equation is img 620f55440bb75

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

x

– 6

– 5

– 4

– 3

– 2

– 1

y

3

2

1

1

2

3

On plotting these points, we obtain the graph of img 620f55440bb75 as follows.

img 620f554456d9e

It is known that, img 620f554479067

img 620f5544ef524

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

img 620f55459abd5

∴ Required area = Area OABO + Area BCDB

img 620f5545bc582

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

img 620f55461308c

The points of intersection of both the curves are (0, 0) and img 620f55463397f.

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

img 620f55464f4af

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

img 620f5546d7377

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)

img 620f55470aa8e

Question 8:

Find the area of the smaller region bounded by the ellipse img 620f554761ebeand the line img 620f55477e0c0

Answer:

The area of the smaller region bounded by the ellipse, img 620f554761ebe, and the line, img 620f55477e0c0, is represented by the shaded region BCAB as

img 620f55479984a

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

img 620f5547bbb7b

Question 9:

Find the area of the smaller region bounded by the ellipse img 620f554844948 and the line img 620f5548602c8

Answer:

The area of the smaller region bounded by the ellipse, img 620f554844948, and the line, img 620f5548602c8, is represented by the shaded region BCAB as

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img 620f554887205

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

img 620f5548a7535

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

img 620f55492eb1e

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 img 620f554958bcc

[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, img 620f554958bcc, is represented by the shaded region ADCB as

img 620f554975104

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

img 620f5549961dc

Page No 376:

Question 12:

Find the area bounded by curves img 620f5549de389

Answer:

The area bounded by the curves, img 620f5549de389, is represented by the shaded region as

img 620f554a332f5

It can be observed that the required area is symmetrical about y-axis.

img 620f554a5137b

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).

img 620f554acb8d6

Equation of line segment AB is

img 620f554aed752

Equation of line segment BC is

img 620f554b3fd0b

Equation of line segment CA is

img 620f554b89b32

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)

img 620f554bd0fad

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)

img 620f554c5778c

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)

img 620f554c76b98

Question 15:

Find the area of the region img 620f554cf1e4b

Answer:

The area bounded by the curves, img 620f554cf1e4b, is represented as

img 620f554d454f5

The points of intersection of both the curves areimg 620f554d64870.

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

img 620f554dabb8d

img 620f554e0234c

Therefore, the required area is img 620f554e839af 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. img 620f554f034a1

C. img 620f554f1f80d

D. img 620f554f3e7f7

Answer:

img 620f554f5d5e8

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 curveimg 620f554f80be3x-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. img 620f554f9c61d

C. img 620f554fb637a

D. img 620f554fd23f7

Answer:

img 620f554ff1dfc

img 620f555020687

img 620f555067037

Thus, the correct answer is C.

Question 18:

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is

A. img 620f5550aea15

B. img 620f5550c9e67

C. img 620f5550e5887

D. img 620f5550c9e67

Answer:

The given equations are

x2 + y2 = 16       … (1)

y2 = 6x               … (2)

img 620f555110726

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 img 620f55512f2f8

A. img 620f55514b512

B. img 620f55516598b

C. img 620f55517fabf

D. img 620f55519d0c5

Answer:

The given equations are

y = cos x … (1)

And, y = sin x … (2)

img 620f5551bbdef

Required area = Area (ABLA) + area (OBLO)

img 620f5551dd105

Integrating by parts, we obtain

img 620f55522e663

Thus, the correct answer is B.

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