JEE  >  NCERT Solutions: Application of Integral

# NCERT Solutions: Application of Integral Notes | Study Mathematics (Maths) Class 12 - JEE

## Document Description: NCERT Solutions: Application of Integral for JEE 2022 is part of Mathematics (Maths) Class 12 preparation. The notes and questions for NCERT Solutions: Application of Integral have been prepared according to the JEE exam syllabus. Information about NCERT Solutions: Application of Integral covers topics like and NCERT Solutions: Application of Integral Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Application of Integral.

Introduction of NCERT Solutions: Application of Integral in English is available as part of our Mathematics (Maths) Class 12 for JEE & NCERT Solutions: Application of Integral in Hindi for Mathematics (Maths) Class 12 course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: NCERT Solutions: Application of Integral Notes | Study Mathematics (Maths) Class 12 - JEE
 1 Crore+ students have signed up on EduRev. Have you?

Application of Integral

Question 1: Find the area of the region bounded by the curve y2x & the lines x = 1, x = 4 & the x-axis.

ANSWER : - The area of the region bounded by the curve, y2x, 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 = 9xx = 2, x = 4 & 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.

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, area enclosed by x-axis, the line, and the circle in the first quadrant =

Question 7: Find the area of the smaller part of the circle x2y2a2 cut off by the line

ANSWER : -The area of the smaller part of the circle, x2y2a2, 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, x2y2a2, cut off by the line, , is  units.

Question 8: The area between xy2 and x = 4 is divided into two equal parts by the line xa, find the value of a.

ANSWER : -The line, xa, 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 yx2 and

ANSWER : -The area bounded by the parabola, x2y,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, x2y, and line, yx, is A (1, 1).

Area of OACO = Area ΔOAB – Area OBACO

⇒ Area of OACO = Area of ΔOAB – Area of OBACO

Therefore, required area = units

Question 10: Find the area bounded by the curve x2 = 4y and the line x = 4y – 2

ANSWER : -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 .

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 x = 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.

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.

Question 14: 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

Therefore, the required area OBCDO is units.

Question 15: Find the area bounded by curves (x – 1)2 +  y2 = 1 and x2y 2 = 1

ANSWER : -The area bounded by the curves, (x – 1)2y2 = 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 16: Find the area of the region bounded by the curves yx2 + 2, yxx = 0 and x = 3

ANSWER : -The area bounded by the curves, yx2 +  2, yxx = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

Question 17: 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 18: Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

ANSWER : -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 (ΔACB) = Area (OLBAO) –Area (OLCAO)

Question 19: Smaller area enclosed by the circle x2 +  y2 = 4 and the line xy = 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 20: 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 (ΔOCA) – Area (OCABO)

Thus, the correct answer is B.

Question 21: Find the area under the given curves and given lines:

(i) yx2x = 1, x = 2 and x-axis
(ii) yx4x = 1, x = 5 and x –axis

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

Question 22: Find the area between the curves yx and yx2

ANSWER : -The required area is represented by the shaded area OBAO as

The points of intersection of the curves, yx and yx2, is A (1, 1).

We draw AC perpendicular to x-axis.

∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)

Question 23: Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and y = 4

ANSWER : -The area in the first quadrant bounded by y = 4x2x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as

Question 24: Sketch the graph of and evaluate

ANSWER : -The given equation is

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

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

It is known that,

Question 25: 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 26: Find the area enclosed between the parabola y2 = 4ax and the line ymx

ANSWER : -The area enclosed between the parabola, y2 = 4ax, and the line, ymx, 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 27: 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 28: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 29:Find the area of the smaller region bounded by the ellipse  & 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 30: Find the area of the region enclosed by the parabola x2y, the line yx + 2 and x-axis

ANSWER : -The area of the region enclosed by the parabola, x2y, the line, yx + 2, and x-axis is represented by the shaded region OABCO as

The point of intersection of the parabola, x2y, and the line, yx + 2, is A (–1, 1).

∴ Area OABCO = Area (BCA) Area COAC

Question 31: 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

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

Question 32: 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 33: 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 34: 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

2xy = 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 35: 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 36: Area bounded by the curve yx3, the x-axis and the ordinates x = –2 and x = 1 is

A. – 9
B.
C.
D.

Thus, the correct answer is B.

Question 37:The area bounded by the curvex-axis and the ordinates x = –1 & x = 1 is given by [Hint: yx2 if x > 0 and y = –x2 if x < 0]

A. 0
B.
C.
D.

Thus, the correct answer is C.

Question 38: The area of the circle x2y2 = 16 exterior to the parabola y2 = 6x is

A.
B.
C.
D.

ANSWER : -The given equations are

x2y2 = 16 … (1)

y2 = 6x … (2)

Area bounded by the circle and parabola

Area of circle = π (r)2

= π (4)2

= 16π units

Thus, the correct answer is C.

Question 39: 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.

The document NCERT Solutions: Application of Integral Notes | Study Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
All you need of JEE at this link: JEE

## Mathematics (Maths) Class 12

209 videos|209 docs|139 tests
 Use Code STAYHOME200 and get INR 200 additional OFF

## Mathematics (Maths) Class 12

209 videos|209 docs|139 tests

### Top Courses for JEE

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;