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(D) To find the area under the curve $y = xe^{-ax}$ from $x = 0$ to $x = \infty$,we evaluate the definite integral:$A = \int_{0}^{\infty} xe^{-ax} dx$

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$\frac{1}{a^2}$

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(D) To find the area under the curve $y = xe^{-ax}$ from $x = 0$ to $x = \infty$,we evaluate the definite integral:$A = \int_{0}^{\infty} xe^{-ax} dx$Using integration by parts,where $u = x$ and $dv = e^{-ax} dx$,we have $du = dx$ and $v = -\frac{1}{a}e^{-ax}$:$A = \left[ -\frac{x}{a}e^{-ax} \right]_{0}^{\infty} - \int_{0}^{\infty} \left( -\frac{1}{a}e^{-ax} \right) dx$$A = (0 - 0) + \frac{1}{a} \int_{0}^{\infty} e^{-ax} dx$$A = \frac{1}{a} \left[ -\frac{1}{a}e^{-ax} \right]_{0}^{\infty}$$A = \frac{1}{a} \left( 0 - (-\frac{1}{a}) \right) = \frac{1}{a^2}$

The area of the region under the graph of $y = xe^{-ax}$ as $x$ varies from $0$ to $\infty$,where $'a'$ is a positive constant,is

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