Find $\int x^{2} e^{-2x}\, dx$ . The curve with equation $y = x e^{-x}$ 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. Apply integration by parts to the given integral.

\text{Let } u = x^2 \Rightarrow \frac{du}{dx} = 2x
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\text{Let } \frac{dv}{dx} = e^{-2x} \Rightarrow v = -\frac{1}{2}e^{-2x}
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\int x^2 e^{-2x} , dx = x^2 \left( -\frac{1}{2}e^{-2x} \right) - \int \left( -\frac{1}{2}e^{-2x} \right) (2x) , dx
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= -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} , dx

2. Apply integration by parts a second time to the remaining integral.

\text{For } \int x e^{-2x} , dx,
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\text{Let } u = x \Rightarrow \frac{du}{dx} = 1
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\text{Let } \frac{dv}{dx} = e^{-2x} \Rightarrow v = -\frac{1}{2}e^{-2x}
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\int x e^{-2x} , dx = x \left( -\frac{1}{2}e^{-2x} \right) - \int \left( -\frac{1}{2}e^{-2x} \right) (1) , dx
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= -\frac{1}{2}x e^{-2x} + \frac{1}{2} \int e^{-2x} , dx

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

\frac{1}{2} \int e^{-2x} , dx = \frac{1}{2} \left( -\frac{1}{2}e^{-2x} \right) = -\frac{1}{4}e^{-2x}
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\text{Substituting this back into the main expression:}
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\int x^2 e^{-2x} , dx = -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C
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= -\frac{1}{4}e^{-2x}(2x^2 + 2x + 1) + C

1. Concisely evaluate the integral from part (i) as it is required for the volume calculation.

$\int x^2 e^{-2x} \, dx$
Using integration by parts twice:
$= -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \, dx$
$= -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C$

2. Set up the integral for the volume of revolution of region $R$.

Volume $= \pi \int_{0}^{1} (y_{\text{curve}}^2 - y_{\text{line}}^2) \, dx$
$= \pi \int_{0}^{1} \left( (x e^{-x})^2 - \left( \frac{1}{e}x \right)^2 \right) \, dx$
$= \pi \int_{0}^{1} x^2 e^{-2x} \, dx - \frac{\pi}{e^2} \int_{0}^{1} x^2 \, dx$

3. Evaluate the definite integrals to find the exact volume.

From part (i):
$\int_{0}^{1} x^2 e^{-2x} \, dx = \left[ -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} \right]_0^1$
$= \left( -\frac{5}{4}e^{-2} \right) - \left( -\frac{1}{4} \right) = \frac{1}{4} - \frac{5}{4e^2}$

For the line (which forms a cone when rotated):
$\frac{\pi}{e^2} \int_{0}^{1} x^2 \, dx = \frac{\pi}{e^2} \left[ \frac{x^3}{3} \right]_0^1 = \frac{\pi}{3e^2}$

Total volume $= \pi \left( \frac{1}{4} - \frac{5}{4e^2} \right) - \frac{\pi}{3e^2}$
$= \pi \left( \frac{1}{4} - \frac{15}{12e^2} - \frac{4}{12e^2} \right)$
$= \pi \left( \frac{1}{4} - \frac{19}{12e^2} \right)$