Solution
= Solution
Apply <integration by parts> twice. Taking an <antiderivative> of $e^{x/2}$ to be $2e^{x/2}$ gives
$$
\int x^2e^{x/2}\,dx=2x^2e^{x/2}-4\int xe^{x/2}\,dx,
$$
and the second application gives $\int xe^{x/2}\,dx=2xe^{x/2}-4e^{x/2}+C$. Therefore the required <definite integral> is
$$
\left[2e^{x/2}(x^2-4x+8)\right]_0^1=\boxed{10\sqrt e-16}.
$$