Math GRE - #4

Assume that $p$ is a polynomial function on the set of real numbers. If $p(0)=p(2)=3$ and $p'(0)=p'(2)=-1$, then $\int_{0}^{2}xp^{\prime\prime}(x)=$

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Solution :

This is an application of integration by parts. Recall that \[\int{u\,dv}=uv-\int{v\,du}.\]
Let $u=x$, $dv=p^{\prime\prime}(x)$. Then we have \[\int_{0}^{2}xp^{\prime\prime}(x)=\left.xp^{\prime}(x)\right|_{0}^{2}-\int_{0}^{2}p^{\prime}(x).\] Using the limit conditions given, we have: \[\left.xp^{\prime}(x)\right|_{0}^{2}-\int_{0}^{2}p^{\prime}(x)=\left(2\cdot p^{\prime}(2)-0\cdot p^{\prime}(0)\right)-\left(p(2)-p(0)\right)=-2.\]

An aside on the usefulness of integration by parts
Let me start off by saying: you don't have to read this part.

However, let me just mention how useful the technique you saw above is in mathematics and physics. In physics, we can use the trick of integration by parts to obtain the first order Euler-Lagrange Equations, which forms the backbone of Lagragian Mechanics. In mathematics, you can use the above method to derive the method of spline interpolation.

So the moral of the story is: Learn how to do integration by parts. Memorize it if you have to, because it comes up a lot more often than you'd think.