Let f be twice differentiable on [0,1] with f(0)=1, f(1)=4 and ∫0^1 f(x) dx=13/6. Then ∫0^1 (x^2-x)f''(x) dx equals

Integrate by parts twice, treating $f''$ as the derivative. First, $\int_0^1(x^2-x)f''\,dx=\Big[(x^2-x)f'\Big]_0^1-\int_0^1(2x-1)f'\,dx$. The boundary term vanishes since $x^2-x=0$ at both ends. Next, $\int_0^1(2x-1)f'\,dx=\Big[(2x-1)f\Big]_0^1-2\int_0^1f\,dx=(f(1)+f(0))-2\cdot\dfrac{13}{6}=5-\dfrac{13}{3}=\dfrac23$. Hence the required integral is $-\dfrac23$. Check with $f=1+x+2x^2$: $f''=4$ and $4\int_0^1(x^2-x)\,dx=-\dfrac23$. The idea is that integrating by parts moves derivatives off $f''$ onto the polynomial weight, and choosing the weight $x^2-x$ kills the boundary terms, so only the given data on $f(0)$, $f(1)$ and the integral of $f$ are needed.