Question

If $f\left(x\right)$ satisfies the conditions of Rolle's theorem in $[1,2]$ and $f\left(x\right)$ is continuous in $[1,2]$ then ${\int }_1^2{f}^{\prime }\left(x\right)dx$ is equal to

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$0$

Since, $f\left(x\right)$ satisfies the conditions of
Rolle's theorem in $[1,2]$ and $f\left(x\right)$ is continuous in $[1,2]$

$\Rightarrow f\left(2\right)=f\left(1\right)$

Then using Rolle's theorem, we have

${\int }_1^2{f}^{\prime }\left(x\right)dx={\left[f\right(x\left)\right]}_1^2=f\left(2\right)-f\left(1\right)=0$
$\because f\left(2\right)=f\left(1\right)$