Answer: (a) \(\displaystyle \int_0^1 x e^{-x^2}\,dx = \frac12-\frac{1}{2e}\).

(b) For \(n\ge 1\), \(\displaystyle I_{2n+1}=nI_{2n-1}-\frac{1}{2e}\).

(c) \(\displaystyle I_7=3-\frac{8}{e}\).

First, evaluate

\(I_1=\displaystyle\int_0^1 x e^{-x^2}\,dx\).

Let \(u=-x^2\), so \(du=-2x\,dx\) and therefore \(x\,dx=-\frac12 du\). Then

\(I_1=-\frac12\int_{0}^{-1} e^u\,du\),

or directly by recognising the antiderivative,

\(\displaystyle \int x e^{-x^2}\,dx=-\frac12 e^{-x^2}+C\).

Hence

\(\displaystyle I_1=\left[-\frac12 e^{-x^2}\right]_0^1=-\frac{1}{2e}+\frac12=\frac12-\frac{1}{2e}.\)

Now define

\(\displaystyle I_n=\int_0^1 x^n e^{-x^2}\,dx\).

For \(n\ge 1\),

\(\displaystyle I_{2n+1}=\int_0^1 x^{2n+1}e^{-x^2}\,dx=\int_0^1 x^{2n}\,(x e^{-x^2})\,dx.\)

Integrate by parts with \(u=x^{2n}\) and \(dv=x e^{-x^2}\,dx\). Then \(du=2n x^{2n-1}\,dx\) and \(v=-\frac12 e^{-x^2}\). So

\(\displaystyle I_{2n+1}=\left[-\frac12 x^{2n}e^{-x^2}\right]_0^1+\int_0^1 n x^{2n-1}e^{-x^2}\,dx.\)

The boundary term is \(-\frac{1}{2e}\), and the remaining integral is \(nI_{2n-1}\). Therefore

\(\displaystyle I_{2n+1}=nI_{2n-1}-\frac{1}{2e}.\)

To find \(I_7\), use the recurrence repeatedly:

\(\displaystyle I_3=1\cdot I_1-\frac{1}{2e}=\left(\frac12-\frac{1}{2e}\right)-\frac{1}{2e}=\frac12-\frac{1}{e}.\)

Then

\(\displaystyle I_5=2I_3-\frac{1}{2e}=2\left(\frac12-\frac{1}{e}\right)-\frac{1}{2e}=1-\frac{5}{2e}.\)

Finally,

\(\displaystyle I_7=3I_5-\frac{1}{2e}=3\left(1-\frac{5}{2e}\right)-\frac{1}{2e}=3-\frac{8}{e}.\)