NCERT Solutions for Class 12 Maths Chapter 8 Applications of the Integrals

Q. Find the area of the region bounded by the parabola $$ y = x^2 \textbf{ and } y = | x |.$$

$$\textbf{Ans. } \text{Given curves are} \\ y = x^2 \space\space …(i) \\ and \space \space y = | x |\space \space …(ii) \\ โ‡’\space x^2 โ€“ | x | = 0 \\ | x |\space \{ | x | โ€“ 1\} = 0 \\ | x | = 0\space or\space | x | = 1 \\ x = 0\space \space x = โ€“ 1 or + 1 \\ If\space x = 0,\space \space y = 0 \\ If x = ยฑ 1,\space\space y = 1 \\$$

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$$โˆด y = x^2 \text{ is upward parabola with vertex (0, 0)} \\ \text{Required area = }2\int^1_0 (|x|-x^2)\space dx \\ = 2\begin{bmatrix} \frac{x^2}{2}-\frac{x^3}{3} \end{bmatrix}^1_0 \\ =2 \begin{bmatrix} \left(\frac{1}{2}-\frac{1}{3}\right)-(0) \end{bmatrix} \\ = 2\left(\frac{3-2}{6}\right) \\ = \frac{1}{3}\space \text{sq.units.}$$

Q. The area between $$ x = y^2 \textbf{ and x = 4 is divided into two equal parts by the line x = a, find the value of a.}$$

$$\textbf{Ans. } \text{Given } x = y^2 \text{ is a parabola symmetric to positive X-axis with vertex (0, 0).} $$

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$$\text{We have given that} \\ \int^a_0 \sqrt{x}\space dx = \int^4_a \sqrt{x}\space dx \\ \Rightarrow \space \space \begin{bmatrix}\frac{x^{3/2}}{3/2} \end{bmatrix}^a_0 = \begin{bmatrix}\frac{x^{3/2}}{3/2} \end{bmatrix}^4_a \\ \Rightarrow \space \space \frac{2}{3}(a)^{3/2} = \frac{2}{3}(4^{3/2}-a^{3/2}) \\ \Rightarrow \space \space \frac{2}{3}a^{3/2}+ \frac{2}{3}a^{3/2} = \frac{2}{3} ร— 8 \\ \Rightarrow \space \space \frac{4}{3}a^{3/2} = \frac{8ร—2}{3} \\ \Rightarrow \space \space a\sqrt{a} = \frac{16}{3}ร—\frac{3}{4} \\ \Rightarrow \space \space a\sqrt{a} = 4 \\ \Rightarrow \space \space a^{3/2} = 4 \\ \Rightarrow \space \space a = 4^{2/3} = (2^2)^{2/3} \\ \Rightarrow \space \space a = 2^{4/3} = \sqrt[3]{16}$$

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

$$\textbf{Ans. } \text{Given, }\space x^2 + y^2 = a^2 \text{ is a circle with centre (0, 0) and radius a units and x } = \frac{a}{\sqrt{2}} \text{ is a line parallel to Y-axis.} \\ x^2 + y^2 = a^2 \space \space \space …(i) $$

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$$\text{Put} \space \space x= \frac{a}{\sqrt{2}} \text{ in equation (i)} \\ \left(\frac{a}{\sqrt{2}}\right)^2 + y^2 = a^2 \\ \Rightarrow \space \space \frac{a^2}{2} + y^2 = a^2 \\ y^2 = a^2-\frac{a^2}{2}\\ \Rightarrow y^2 = \frac{a^2}{2} \\ \Rightarrow \space \space y = \frac{a}{\sqrt{2}}\\ \text{Required area = }2\int^a_{a/\sqrt{2}} \space \text{Area under circle dx} \\ = 2\int^a_{a/\sqrt{2}}\sqrt{a^2-x^2}\space dx \\ =2 \begin{bmatrix}\frac{x}{2}\sqrt{a^2-x^2}+ \frac{a^2}{2}\space sin^{-1} \frac{x}{a} \end{bmatrix}^a_{a/\sqrt{2}} \\ = 2 \begin{Bmatrix}\frac{a}{2}ร—0 + \frac{a^2}{2}sin^{-1} \left(\frac{a}{a}\right) \end{Bmatrix}-\begin{Bmatrix}\frac{a/\sqrt{2}}{2} \sqrt{a^2-\frac{a^2}{2}}+\frac{a^2}{2} sin^{-1}\left(\frac{a}{\sqrt{2a}}\right) \end{Bmatrix} \\ = 2\begin{bmatrix}0+ \frac{a^2}{2}sin^{-1}(1) – \frac{a}{2\sqrt{2}}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}sin^{-1}\space \frac{1}{\sqrt{2}} \end{bmatrix} \\ = 2\begin{bmatrix}\frac{a^2}{2}ร—\frac{\pi}{2}-\frac{a}{2\sqrt{2}}ร—\frac{a}{\sqrt{2}}-\frac{a^2}{2}ร— \frac{\pi}{2} \end{bmatrix} \\ =2\begin{bmatrix}\frac{a^2\pi}{4}-\frac{a^2}{4}-\frac{a^2\pi}{8} \end{bmatrix} \\ =2 \begin{bmatrix}\frac{2a^2\pi-a^2\pi}{8}-\frac{a^2}{4} \end{bmatrix} \\ =2ร—\frac{a^2\pi}{8}-\frac{a^2}{4}ร—2 \\ = \left(\frac{a^2\pi}{4}-\frac{a^2}{2}\right)\space \text{sq.units.} $$