Question

If $f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t\sin (x-t)dt,$ then $f^{\prime \prime }x-f\left(x\right)$ is equal to

• A. $\sin x-\cos x$
• B. $2\sin x$
• C. $\sin x+\cos x$
• D. $2\cos x$

Answer 1

The correct option is C $\sin x+\cos x$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^t\sin (x-t)dt$
Putting $t\to x-t$, we get
$\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^{x-t}\sin tdt\Rightarrow f\left(x\right)=e^x\underset{0}{\overset{x}{\int }}{e}^{-t}\sin tdt$
Using Newton-Leibniz rule, we get
$\Rightarrow f^{\prime }\left(x\right)=e^x\underset{0}{\overset{x}{\int }}{e}^{-t}\sin tdt+e^x\left(e^{-x}\sin x\right)\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)+\sin x\Rightarrow f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)+\cos x\Rightarrow f^{\prime \prime }\left(x\right)=f\left(x\right)+\sin x+\cos x\therefore f^{\prime \prime }\left(x\right)-f\left(x\right)=\sin x+\cos x$