If $f\left(x\right)$ satisfies the conditions for  Rolle's theorem then ${\int }_3^5 f\left(x\right)$  equals 

It is given that $f\left(x\right)$ satisfies all the conditions for

Rolles theorem. Therefore

$f\left(3\right)=f\left(5\right)=0$

$x=3$ and $x=5$ are roots of $f\left(x\right)\text{. }$

$\begin{array}{l}\Rightarrow f\left(x\right)=(x-3)(x-5)=x^2-8x+15\\ \therefore {\int }_3^5 f\left(x\right)dx={\int }_3^5 \left(x^2-8x+15\right)dx={\left[\frac{x^3}{3}-4x^2+15x\right]}_3^5\end{array}$

$\Rightarrow {\int }_3^5 f\left(x\right)dx=\frac{1}{3}(125-27)-4(25-9)+15(5-3)=-\frac{4}{3}$