Q. Let $f(x) = \min{\{ x + 1,\ \sqrt{(1 - x)}}\}$, then area bounded by $f(x)$ 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
Explanation
(c)
$f(x) = \min{\{ x + 1,\sqrt{(1 - x)}\}}$
$= \left\{ \begin{array}{r}
x + 1,\ - 1 \leq x < 0 \\
\sqrt{1 - x},\ 0 < x \leq 1
\end{array} \right.\$
$\therefore$ Required area $= \left| \int_{- 1}^{0}{(x + 1)dx} \right| + \left| \int_{0}^{1}{\sqrt{(1 - x)}dx} \right|$
$= 7/6\ sq\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.