Question

Let $f\left(x\right)=$min$\{x+1,\sqrt{1-x}\}$ then area bounded by $f\left(x\right)$ and $x-$axis is:

• A. $\frac{1}{6}$ sq.unit
• B. $\frac{5}{6}$ sq.unit
• C. $\frac{7}{6}$ sq.unit
• D. $\frac{11}{6}$ sq.unit

Answer 1

The correct option is C $\frac{7}{6}$ sq.unit

$f\left(x\right)=$min$\{x+1,\sqrt{(1-x)}\}$
$=\{\begin{array}{c}x+1;-1\le x<0\\ \sqrt{1-x};0<x\le 1\end{array}$
$\therefore$ Required Area$=\left|{\int }_{-1}^0(x+1)dx\right|+\left|{\int }_0^1\sqrt{(1-x)}dx\right|$
$={[\frac{x^2}{2}+x]}_{-1}^0+{\left[\frac{-{(1-x)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$
$=0-[\frac{1}{2}-1]-[0-\frac{2}{3}]$
$=\frac{1}{2}+\frac{2}{3}$
$=\frac{7}{6}$.sq.unit