Question

Consider the function, $F\left(x\right)={\int }_{-1}^x(t^2-t)dt,x\in R$. Find the interval on which $F\left(x\right)$ is increasing and the intervals on which $F\left(x\right)$ is decreasing.

• A. decreasing $(-\infty ,0)\cup (2,\infty )$, increasing $(0,1)$
• B. increasing $(-\infty ,0)\cup (1,\infty )$, decreasing $(0,1)$
• C. decreasing $(-\infty ,0)\cup (1,\infty )$, increasing $(0,2)$
• D. increasing $(-\infty ,0)\cup (2,\infty )$, decreasing $(0,2)$

Answer 1

The correct option is B increasing $(-\infty ,0)\cup (1,\infty )$, decreasing $(0,1)$

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

$={|\frac{t^3}{3}-\frac{t^2}{2}|}_{-1}^x=\frac{x^3}{3}-\frac{x^2}{2}+\frac{1}{3}+\frac{1}{2}=\frac{({2x}^3-{3x}^2+5)}{6}$

Applying Leibnitz rule

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

$f^{\prime }\left(x\right)=0\Rightarrow x=0,1$

$f"\left(0\right)=-1maxima$

$f"\left(1\right)=1minima$

Hence the correct answer is

$f\left(x\right)$ increasing $(-\infty ,0)\cup (1,\infty )$

$f\left(x\right)$ decreasing $(0,1)$