Question

The interval in which $f\left(x\right)={\int }_0^x(e^t-1)(t^2+2t-8)(t+1)dt$ increases is

• A. $(-\infty ,-4)$
• B. $(-1,0)$
• C. $(2,\infty )$
• D. All of these

Answer 1

The correct option is D All of these

So to find interval in which $f\left(x\right)$ increases

If $f\left(x\right)$increases in an interval then the slope of $f\left(x\right)$ in that interval will always be positive that is $f^{\prime }\left(x\right)>0$ for that interval

Using Newton-Leibuitz Rule

$f^{\prime }\left(x\right)=(e^x-1)(x^2+2x-8)(x+1)=(e^x-1)(x+4)(x-2)(x+1)f^{\prime }\left(x\right)>0$

For $x\in (-\infty ,-4)\cup (-1,0)\cup (2,\infty )$

Hence correct answer is $x\in (-\infty ,-4)\cup (-1,0)\cup (2,\infty )$