Question

Define $f\left(x\right)=$integral from $0\mathrm{to}x\mathit{sin}\left(t\right)dt\mathit{,}\mathit{}x\mathit{\ge }\mathit{0}$. Then

• A. $f$is increasing only in the interval $\left(0,\frac{\pi }{2}\right)$
• B. $f$is decreasing in the interval $\left(0,\pi \right)$
• C. $f$attains maximum at $x=\frac{\pi }{2}$
• D. $f$attains minimum at $x=\pi$
• E. $f$ attains maximum at $x=\pi$

Answer 1

The correct option is A

$f$is increasing only in the interval $\left(0,\frac{\pi }{2}\right)$

Explanation for correct option:

Step-1: Integrate the function.

Given, $f\left(x\right)={\int }_0^x\sin \left(t\right)dt$

$$\begin{gathered} ={-\left[\cos \left(t\right)\right]}_0^x \\ =-\left[\cos \left(x\right)-\cos \left(0\right)\right] \end{gathered}$$

$\Rightarrow f\left(x\right)=\left[1-\cos \left(x\right)\right]$

Step-2: Differentiate the function.

$f\text{'}\left(x\right)=0-\left(-\sin \left(x\right)\right)$

$\Rightarrow f\text{'}\left(x\right)=\sin \left(x\right)$

We know that $\sin \left(x\right)$is greater than zero.

Therefore, $f$is increasing in interval $\left(0,\frac{\pi }{2}\right)$

Hence, correct answer is option $\left(A\right)$.