If f(x) satisfies the conditions for Rolle's theorem on [3, 5], then $\int\limits_3^5 f(x) d x$ equals

$-4 / 3$

It is given that $f(x)$ satisfies all the conditions for Rolle's theorem. Therefore,

$f(3)=f(5)=0$

⇒ x = 3 and x = 5 are roots of f(x)

$\Rightarrow f(x)=(x-3)(x-5)=x^2-8 x+15$

∴  $\int\limits_3^5 f(x) d x=\int\limits_3^5\left(x^2-8 x+15\right) d x=\left[\frac{x^3}{3}-4 x^2+15 x\right]_3^5$

$\Rightarrow \int\limits_3^5 f(x) d x=\frac{1}{3}(125-27)-4(25-9)+15(5-3)=-\frac{4}{3}$