For $x \in\left(0, \frac{5 \pi}{2}\right)$, define $f(x)=\int_0^x \sqrt{t} \sin t d t$. Then $f$ has
For $x \in\left(0, \frac{5 \pi}{2}\right)$, define $f(x)=\int_0^x \sqrt{t} \sin t d t$. Then $f$ has
local minimum at $\pi$ and $2 \pi$
local minimum at $\pi$ and local maximum at $2 \pi$
local maximum at $\pi$ and local minimum at $2 \pi$
local maximum at $\pi$ and $2 \pi$
Solution
$f^{\prime}(x)=\sqrt{x} \sin x$
Given $x \in\left(0, \frac{5 \pi}{2}\right)$
$f^{\prime}(x)$ changes sign from +ve to -ve at $\pi$
$f^{\prime}(x)$ changes sign from -ve to +ve at $2 \pi$
$f$ has local max at $\pi$, local min at $2 \pi$