Q. Define$f(x) = \int_{0}^{x}{\sin{t\ dt,\ x \geq 0}}$. Then,
A) $f$ is decreasing only in the interval $\left( 0,\frac{\pi}{2} \right)$
B) $f$ is decreasing in the interval $\lbrack 0,\ \pi\rbrack$
C) $f$ attains maximum at $x = \frac{\pi}{2}$
D) $f$ attains minimum at $x = \pi$
Explanation
(a)
$f(x) = \int_{0}^{x}{\sin{t\ dt,\ x \geq 0}}$
${\Rightarrow f^{'}(x) = \sin x
}{Now,\ f^{'}(x) > 0in0 < x < \frac{\pi}{2}}$
$\therefore f(x)$ is increasing in 0$< x < \frac{\pi}{2}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.