#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 30 maths textbook solution

4

Hint:

To solve this we will split $1-x^{2}$ in differential form.

Given:

$\int_{-2}^{2}\left|1-x^{2}\right| d x$

Solution:

\begin{aligned} &\int_{-2}^{2}\left|1-x^{2}\right| d x \\\\ &1-x^{2}=0 \\\\ &x^{2}=1 \\\\ &x=\pm 1 \end{aligned}

\begin{aligned} &=\int_{-2}^{-1}\left(1-x^{2}\right) d x+\int_{-1}^{1}\left(1-x^{2}\right) d x+\int_{1}^{2}\left(1-x^{2}\right) d x \\\\ &=\left[\frac{x^{3}}{3}-x\right]_{-2}^{-1}+\left[x-\frac{x^{3}}{3}\right]_{-1}^{1}+\left[\frac{x^{3}}{3}-x\right]_{1}^{2} \end{aligned}

\begin{aligned} &=\left[-\frac{1}{3}+1-\left(-\frac{8}{3}\right)+2\right]+\left[1-\frac{1}{3}-(-1)+\frac{1}{3}\right]+\left[\frac{8}{3}-2-\left(\frac{1}{3}-1\right)\right] \\\\ &=\frac{2}{3}-\left(-\frac{2}{3}\right)+\left[\frac{2}{3}-\left(-\frac{2}{3}\right)\right]+\left[\frac{2}{3}+\frac{2}{3}\right] \end{aligned}

\begin{aligned} &=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{12}{3} \\\\ &=4 \end{aligned}