# NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals)Exercise 8.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Class 12 Mathematics

Chapter-8 (Application Of Integrals)

Questions and answers given in practice

Chapter-8 (Application Of Integrals)

### Exercise 8.1

Q1. Find the area of the region bounded by the curve ${y}^{2}=x$ and the lines x = 1, x = 4 and the x-axis in the first quadrant.

Answer. The area of the region bounded by the curve, ${y}^{2}$ = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

Q2. Find the area of the region bounded by ${y}^{2}=9x$, x = 2, x = 4 and the x-axis in the first quadrant.

Answer. The area of the region bounded by the curve, ${y}^{2}$ = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.

Q3. Find the area of the region bounded by ${x}^{2}=4y$, y = 2, y = 4 and the y-axis in the first quadrant.

Answer. the area of the region bounded by ${x}^{2}$ = 4y, y = 2, y = 4 and the y-axis is the area ABCD.

Q4. Find the area of the region bounded by the ellipse $\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$

Answer. The given equation of the ellipse, $\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$, 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  $\begin{array}{rl}& =\frac{3}{4}{\left[\frac{x}{2}\sqrt{16-{x}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\frac{x}{4}\right]}_{0}^{4}\\ & =\frac{3}{4}\left[2\sqrt{16-16}+8{\mathrm{sin}}^{-1}\left(1\right)-0-8{\mathrm{sin}}^{-1}\left(0\right)\right]\\ & =\frac{3}{4}\left[\frac{8\pi }{2}\right]\\ & =\frac{3}{4}\left[4\pi \right]\\ & =3\pi \end{array}$ Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Q5. Find the area of the region bounded by the ellipse $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$

Answer. The given equation of the ellipse can be represented as $\begin{array}{l}\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\\ ⇒y=3\sqrt{1-\frac{{x}^{2}}{4}}\end{array}$ It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB

Q6. Find the area of the region in the first quadrant enclosed by x-axis, line

Answer. The area of the region bounded by the circle,${x}^{2}+{y}^{2}=4,x=\sqrt{3}y$ , and the x-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is $\left(\sqrt{3},1\right)$ . Area OAB = Area ΔOCA + Area ACB  $\begin{array}{rl}& ={\int }_{\sqrt{3}}^{2}\sqrt{4-{x}^{2}}dx\\ & ={\left[\frac{x}{2}\sqrt{4-{x}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\frac{x}{2}\right]}_{\sqrt{5}}^{2}\\ & =\left[2×\frac{\pi }{2}-\frac{\sqrt{3}}{2}\sqrt{4-3}-2{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]\\ & =\left[\pi -\frac{\sqrt{3}}{2}-2\left(\frac{1}{3}\right)\right]\\ & =\left[\pi -\frac{\sqrt{3}}{2}-\frac{2\pi }{3}\right]\\ & =\left[\frac{\pi }{3}-\frac{\sqrt{3}}{2}\right]...\left(ii\right)\end{array}$

Q7. Find the area of the smaller part of the circle ${x}^{2}+{y}^{2}={a}^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$

Answer. It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC  $\begin{array}{rl}& =\frac{{a}^{2}\pi }{4}-\frac{a}{2\sqrt{2}}\cdot \frac{a}{\sqrt{2}}-\frac{{a}^{2}}{2}\left(\frac{\pi }{4}\right)\\ & =\frac{{a}^{2}\pi }{4}-\frac{{a}^{2}}{4}-\frac{{a}^{2}\pi }{8}\\ & =\frac{{a}^{2}}{4}\left[\pi -1-\frac{\pi }{2}\right]\\ & =\frac{{a}^{2}}{4}\left[\frac{\pi }{2}-1\right]\\ ⇒\text{Area ABCD}& =2\left[\frac{{a}^{2}}{2}\left(\frac{\pi }{2}-1\right)\right]=\frac{{a}^{2}}{2}\left(\frac{\pi }{2}-1\right)\end{array}$

Q8. The area between x = ${y}^{2}$ and x = 4 is divided into two equal parts by the line x = a, find the value of a.  It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Area OED=Area EFCD Therefore,required area
Q10. Find the area bounded by the curve ${x}^{2}$ = 4y and the line x = 4y – 2 