Question

If $f\left(x\right)=\underset{0}{\overset{x}{\int }}t\sin tdt,$ then the value of $f^{\prime \prime }\left(0\right)$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}t\sin tdt$
Differentiating w.r.t. $x$ we have:
$f^{\prime }\left(x\right)=x\sin x$
Again differentiating w.r.t. $x$ we have:
$f^{\prime \prime }\left(x\right)=\sin x+x\cos x$
$\therefore f^{\prime \prime }\left(0\right)=0$