Question

If the function, $F\left(x\right)={\int }_{-1}^x(t^2-t)dt,x\in R$. Then find, $F^{\prime }\left(x\right)$ and $F^{\prime \prime }\left(x\right)$

• A. $F^{\prime }\left(x\right)=(x^2-x),F"\left(x\right)=2x-1,$
• B. $F^{\prime }\left(x\right)=(x^2-x+1),F"\left(x\right)=2x-1,$
• C. $F^{\prime }\left(x\right)=(x^2+x-1),F"\left(x\right)=2x+1,$
• D. $F^{\prime }\left(x\right)=(x^2+x+1),F"\left(x\right)=2x+1,$

Answer 1

The correct option is A $F^{\prime }\left(x\right)=(x^2-x),F"\left(x\right)=2x-1,$

Consider, $F\left(x\right)={\int }_{-1}^x(t^2-t)dt$

$F\left(x\right)={[\frac{t^3}{3}-\frac{t^2}{2}]}_{-1}^x$

$F\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}+\frac{1}{3}+\frac{1}{2}$

Differentiating on both sides, we get,

$\therefore F^{\prime }\left(x\right)=x^2-x$

Differentiating on both sides, we get,

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