Question Statement

Q9. Evaluate $\int x^2{e}^{x}dx$

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Background and Explanation

This problem requires integration by parts, a technique derived from the product rule for differentiation. When integrating a product of a polynomial and an exponential function, we apply integration by parts repeatedly—reducing the power of the polynomial each time until we obtain a simple exponential integral.

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Solution

Let $I=\int x^2{e}^{x}dx$

Applying integration by parts (treating $x^2$ as the first function and $e^x$ as the second function):

$\begin{array}{rl}& =x^2\int e^{x}dx-\int \left(\int e^{x}dx\right)\times \frac{d}{dx}\left(x^2\right)dx\\ & =x^2\cdot e^x-\int e^x\cdot 2xdx\\ & =x^2\cdot e^x-2\int x{e}^{x}dx\end{array}$

Again applying integration by parts to evaluate $\int x{e}^{x}dx$ (treating $x$ as the first function and $e^x$ as the second):

$\begin{array}{rl}& =x^2{e}^x-2\left[x\int e^{x}dx-\int \left(\int e^{x}dx\right)\times \frac{d}{dx}(x)dx\right]\\ & =x^2{e}^x-2\left[x\cdot e^x-\int e^x\cdot 1dx\right]\\ & =x^2{e}^x-2x{e}^x+2\int e^{x}dx\\ & =x^2{e}^x-2x\cdot e^x+2e^x+c\\ & =e^x\left(x^2-2x+2\right)+c\end{array}$

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Key Formulas or Methods Used

• Integration by parts: $\int uvdx=u\int vdx-\int \left(\int vdx\right)\frac{du}{dx}dx$ (or equivalently $\int udv=uv-\int vdu$)
• Power rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Exponential integral: $\int e^{x}dx=e^x+C$

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Summary of Steps

1. Set up the integral $I=\int x^2{e}^{x}dx$ and apply integration by parts with $u=x^2$ and $dv=e^{x}dx$.
2. Compute to obtain $x^2{e}^x-2\int x{e}^{x}dx$.
3. Apply integration by parts again to the remaining integral with $u=x$ and $dv=e^{x}dx$.
4. Evaluate to get $x{e}^x-e^x$ and substitute back into the expression from step 2.
5. Simplify to $x^2{e}^x-2x{e}^x+2e^x+c$.
6. Factor out $e^x$ to obtain the final answer $e^x(x^2-2x+2)+c$.