Question

If $y=f\left(x\right)$ satisfies the conditions of Rolle's theorem in $\left[2,6\right],$ then ${\int }_2^6{\left(\text{f}\right)}^{{}^{\prime }}\left(\text{x}\right)dx$ is equal to

• A. $2$
• B. $0$
• C. $4$
• D. $6$

Answer 1

Answer: (B)

Since $f\left(x\right)$ satisfies the conditions of Rolle's Theorem so $f\left(2\right)=f\left(6\right)$
Now, given integral $={\left(f\left(x\right)\right)}_2^6=f\left(6\right)-f\left(2\right)=0$