Question

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

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

Answer 1

Answer: (A)

$\frac{7}{6}$

$f\left(x\right)=min\{x+1,\sqrt{1-x}\}$
$y=x+1$ and $y=\sqrt{1-x}$ curves intersect at $(0,1)$ and
$f\left(x\right)=x+1$ for $x\in (-1,0)$
$f\left(x\right)=\sqrt{1-x}$ for $x\in (0,1)$
$\therefore$ Area under the curve and x-axis is
$A={\int }_{-1}^0(x+1)dx+{\int }_0^1\sqrt{1-x}dx$
=${[\frac{x^2}{2}+x]}_{-1}^0-{\left[\frac{{(1-x)}^{3/2}}{3/2}\right]}_0^1=\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$
$\therefore$ required area is $\frac{7}{6}$