Find $\displaystyle\int x^{2}e^{-2x}\,\mathrm{d}x$. The curve with equation $y = xe^{-x}$ and the line with equation $y=\tfrac{1}{e}x$ meet at the origin $O$ and the point $P$ with x-coordinate 1. The region $R$ is bounded by the curve and the line (see diagram). Find the exact volume of the solid formed when $R$ is rotated through $360^\circ$ about the x-axis.

1. Set up the first integration by parts

Let $u = x^2 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2x$
Let $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-2x} \implies v = -\frac{1}{2}e^{-2x}$

2. Apply the integration by parts formula

$\int x^2 e^{-2x} \,\mathrm{d}x = x^2 \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right)(2x) \,\mathrm{d}x$
$= -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \,\mathrm{d}x$

3. Apply integration by parts a second time to the new integral

For $\int x e^{-2x} \,\mathrm{d}x$, let $u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1$
and $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-2x} \implies v = -\frac{1}{2}e^{-2x}$.
$\int x e^{-2x} \,\mathrm{d}x = x \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right)(1) \,\mathrm{d}x$
$= -\frac{1}{2}x e^{-2x} + \frac{1}{2}\int e^{-2x} \,\mathrm{d}x$

4. Evaluate the final integral and state the complete result with the constant of integration

$\frac{1}{2}\int e^{-2x} \,\mathrm{d}x = \frac{1}{2} \left(-\frac{1}{2}e^{-2x}\right) = -\frac{1}{4}e^{-2x}$
Substituting this back into the original expression gives:
$\int x^2 e^{-2x} \,\mathrm{d}x = -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C$

1. Calculate the indefinite integral from part (i), which is required for this part, using integration by parts.

Let $I = \displaystyle\int x^{2}e^{-2x}\,\mathrm{d}x$. Using integration by parts with $u = x^2$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-2x}$:
$I = -\frac{1}{2}x^2e^{-2x} - \int \left(-\frac{1}{2}e^{-2x}\right)(2x)\,\mathrm{d}x = -\frac{1}{2}x^2e^{-2x} + \int xe^{-2x}\,\mathrm{d}x$
Using integration by parts again for $\displaystyle\int xe^{-2x}\,\mathrm{d}x$ with $u = x$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-2x}$:
$\int xe^{-2x}\,\mathrm{d}x = -\frac{1}{2}xe^{-2x} - \int \left(-\frac{1}{2}e^{-2x}\right)(1)\,\mathrm{d}x = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}$
$\therefore \int x^{2}e^{-2x}\,\mathrm{d}x = -\frac{1}{2}x^2e^{-2x} - \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C$

2. Set up the integral for the volume of revolution for the region bounded between the curve and the line.

The volume of revolution $V$ about the x-axis is given by:
$V = \pi \int_{0}^{1} \left( y_{\text{curve}}^2 - y_{\text{line}}^2 \right) \mathrm{d}x$
$V = \pi \int_{0}^{1} \left( (xe^{-x})^2 - \left(\frac{1}{e}x\right)^2 \right) \mathrm{d}x$
$V = \pi \int_{0}^{1} x^2e^{-2x} \,\mathrm{d}x - \pi \int_{0}^{1} \frac{x^2}{e^2} \,\mathrm{d}x$

3. Evaluate the exact volume generated by rotating the region under the curve.

Using the integral found in part (i):
$V_{\text{curve}} = \pi \left[ -\frac{1}{2}x^2e^{-2x} - \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \right]_0^1$
$= \pi \left( \left(-\frac{1}{2}e^{-2} - \frac{1}{2}e^{-2} - \frac{1}{4}e^{-2}\right) - \left(0 - 0 - \frac{1}{4}\right) \right)$
$= \pi \left( \frac{1}{4} - \frac{5}{4}e^{-2} \right)$

4. Evaluate the exact volume generated by rotating the region under the line.

$V_{\text{line}} = \pi \int_{0}^{1} \frac{x^2}{e^2} \,\mathrm{d}x$
$= \frac{\pi}{e^2} \left[ \frac{x^3}{3} \right]_0^1 = \frac{\pi}{3e^2}$

5. Subtract the two volumes to find the final exact volume.

$V = V_{\text{curve}} - V_{\text{line}} = \pi \left( \frac{1}{4} - \frac{5}{4e^2} \right) - \frac{\pi}{3e^2}$
$V = \pi \left( \frac{1}{4} - \frac{15}{12e^2} - \frac{4}{12e^2} \right)$
$V = \pi \left( \frac{1}{4} - \frac{19}{12e^2} \right)$