NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals
NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals deals with the practical applications of integrals in different fields. In this class 12 maths chapter 8, areas under curves and the area between two curves are discussed in such a way that one can learn the best uses of integrals. The solutions are designed to make students understand these things by stepwise problem-solving methods, so that they see the applications of integrals in real life, and further studies in this subject is made easier. These solutions are aimed at affording students a solid comprehension of the class 12 Application of Integrals—an important topic that needs to be given greater attention with respect to competitive and higher studies. Class 12 Maths Chapter 8 pdf explains in detail with examples to establish areas by using integrals. The solutions are done according to CBSE Syllabus in order to give students a glimpse of it for their board examinations. Moreover, this chapter has quite a number of solved examples and exercises that are basically designed for making the students practice and grab the techniques involved in solving the problems related to the project of integrals. This comprehensive approach, therefore, provides a better grasp of the topic to the students.
Download PDF For NCERT Solutions for Maths Application of Integrals
The NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals
Students can access the NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Application of Integrals - Exercise 8.1
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
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.
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.
Find the area of the region bounded by the ellipse
The given equation of the ellipse, 
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
Find the area of the region bounded by the ellipse
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 = 4 x3π/2 = 6π units.
Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x2 + y2 = 4
The area of the region bounded by the circle, x2 + y2 = 4,x = √3y 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 ΔOCA + Area ACB
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, x = a/√2, 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,
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
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
Find the area of the region bounded by the parabola y = x2 and y = |x|
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
The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).
Area of OACO = Area ΔOAM – Area OMACO
Therefore, required area = 2[1/6] = 1/3 units
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
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,
Area OBAO = Area OBCO + Area OACO … (1)
Then, Area OBCO = Area OMBC – Area OMBO
Similarly, Area OACO = Area OLAC – Area OLAO
Therefore, required area =
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
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 8√3 units.
Choose the correct answer:
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. π/2
C. π/3
D. π/4
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as
Therefore, option (A) is correct.
Choose the correct answer:
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
The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as
Application of Integrals - Exercise 8.2
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 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
Then, Area OCBAO = Area ODBAO – Area ODCO
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
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(√2,1/2) and D (-√2,1/2).
It can be observed that the required area is symmetrical about the 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
Therefore, the required area OBCDO is
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
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 A (1/2,√3/2) and B (1/2,√3/2) .
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 (1/2,0).
Therefore, required area OBCAO = 
Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
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)
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units.
Using integration, find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.
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)
Choose the correct answer:
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)
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)
Therefore, option (B) is correct.
Choose the correct answer:
Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3
(B) 1/3
(C) 1/4
(D) 3/4
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)
Therefore, option (B) is correct.
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