$f(0)=6$, $f(1)=5$, and $f'(1)=2 \quad\int_0^1xf''(x)dx=?$

Question

Let $f$ be twice differentiable with $f(0)=6$, $f(1)=5$, and $f'(1)=2$

Evaluate $\int_0^1xf''(x)dx$

Answer 1

Using integration by parts gives $\int_{0}^{1}xf''(x)dx=[xf'(x)]_{0}^{1}-\int_{0}^{1}f'(x)dx=1\times f'(1)-0\times f'(0)-(f(1)-f(0))$

$=1\times2-(5-6)=3$.

See https://en.m.wikipedia.org/wiki/Integration_by_parts for more information on integration by parts, including the formula used here.

Answer 2

Recall integration by parts: $\int u\,dv=uv-\int v\,du$. Taking $u=x$, $dv=f''(x)\,dx$ yields $du=dx$ and $v=f'(x)$, so that $$ \int_0^1 xf''(x)\,dx=xf'(x)\rvert_{x=0}^{x=1}-\int_0^1 f'(x)\,dx. $$ Can you continue on from here?

Answer 3

You know just enough about $f$ to do the problem (well, I think $f$ should actually be twice continuously differentiable but ok). The values given tell you how to evaluate the boundary terms you get with integration by parts.

$$ \int_0^1xf''(x)\mathrm dx=1f'(1)-f(1)+f(0)=2-5+6=3 $$