Let f be a twice differentiable function such that f(2)=1.

Question

Let $f$ be a twice differentiable function such that $f(2)=1$. Let $F(x)=xf(x)$, $x\in\mathbb{R}$. If $\displaystyle\int_0^2 xF'(x)\,dx = 6$ and $\displaystyle\int_0^2 x^2F''(x)\,dx = 40$, then $F'(2)+\displaystyle\int_0^2F(x)\,dx$ is equal to:

Accepted Answer

Integrating by parts, $\displaystyle\int_0^2 xF'(x)\,dx = \big[xF(x)\big]_0^2 - \int_0^2 F(x)\,dx = 2F(2) - \int_0^2 F(x)\,dx$.

Since $F(2)=2f(2)=2$: $6 = 4 - \int_0^2F(x)\,dx \implies \int_0^2F(x)\,dx = -2$.

Also, $\displaystyle\int_0^2 x^2F''(x)\,dx = \big[x^2F'(x)\big]_0^2 - 2\int_0^2xF'(x)\,dx = 4F'(2)-12 = 40 \implies F'(2)=13$.

$$F'(2)+\int_0^2F(x)\,dx = 13 + (-2) = 11$$

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