Question

Find the interval in which
$f\left(x\right)={\int }_{-1}^x(e^t-1)(2-t)dt$, $(x>1)$ is
increasing

• A. $[3,5]$
• B. $[1,3]$
• C. $[0,3]$
• D. $[0,2]$

Answer 1

The correct option is D $[0,2]$

$f\left(x\right)={\int }_{-1}^x(e^t-1)(2-t)dt$

The function is increasing if

$f^{\prime }\left(x\right)>0$

By Lebinitz's theoram

$f^{\prime }\left(x\right)=(e^x-1)(2-x)$

We see that $f^{\prime }\left(x\right)>0$ for $x\in [0,2]$

So, $f\left(x\right)={\int }_{-1}^x(e^t-1)(2-t)dt$ will be increasing in interval $x\in [0,2]$