Question

If $y=f\left(x\right)$ satisfies the conditions of Rolle's theorem in $\left[2,6\right],$ then $\displaystyle \int _{2}^{6} \left(\text{f}\right)^{'} \left(\text{x}\right) dx$ 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 $=\left(f \left(x\right)\right)_{2}^{6}=f\left(6\right)-f\left(2\right)=0$