#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 1 maths textbook solution

$\frac{\pi }{8}$

Hint:

Using $\int \sqrt{a^{2}-x^{2}} \; d x$

Explanation:

\begin{aligned} &\int_{0}^{1} \sqrt{x(1-x)} d x \\\\ &=\int_{0}^{1} \sqrt{x-x^{2}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1} \sqrt{(x)^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d x \\\\ &=\int_{0}^{1} \sqrt{-\left(x-\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}} d x \\\\ &=\int_{0}^{1} \sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}} d x \end{aligned}

We know

$\int \sqrt{a^{2}-x^{2}}=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$

$I=\left[\frac{x-1 / 2}{2} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}\left(\frac{x-1 / 2}{2}\right)\right]_{0}^{1}$

$I=\left[\frac{2 x-1}{4} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}\left(\frac{x-1 / 2}{1 / 2}\right)\right]_{0}^{1}$

$=\left[\frac{2 x-1}{4} \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(2 x-1)\right]_{0}^{1}$

$=\left(\frac{1}{4} \sqrt{\frac{1}{4}-\left(1-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(1)\right)-\left(\frac{-1}{4} \sqrt{\frac{1}{4}-\left(0-\frac{1}{2}\right)^{2}}+\frac{1}{8} \sin ^{-1}(-1)\right)$

\begin{aligned} &=\left(\frac{1}{4} \sqrt{\frac{1}{4}-\frac{1}{4}}+\frac{1}{8} \times \frac{\pi}{2}\right)-\left(\frac{-1}{4} \sqrt{\frac{1}{4}-\frac{1}{4}}+\frac{1}{8} \times\left(-\frac{\pi}{2}\right)\right) \\\\ &=0+\frac{\pi}{16}-0-\left(-\frac{\pi}{16}\right) \\\\ &=\frac{\pi}{8} \end{aligned}