Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 80

Find the integral $\displaystyle \int^1_0 xe^{-x^2} dx$

If we let $u = -x^2 $, then $\displaystyle du = -2x dx$, so $\displaystyle xdx = \frac{-du}{2}$. When $x = 0, u =0$ and when $x = 1, u =-1$. Therefore,



$
\begin{equation}
\begin{aligned}

\int^1_0 xe^{-x^2} dx =& \int^1_0 e^{-x^2} x dx
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\int^1_0 xe^{-x^2} dx =& \int^1_0 e^u \cdot \frac{-du}{2}
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\int^1_0 xe^{-x^2} dx =& \frac{-1}{2} \int^1_0 e^u du
\\
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\int^1_0 xe^{-x^2} dx =& \frac{-1}{2} [e^u]^1_0
\\
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\int^1_0 xe^{-x^2} dx =& \frac{-1}{2} [e^{-1} - e^0]
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\int^1_0 xe^{-x^2} dx =& \frac{-1}{2} \left( \frac{1}{e} - 1 \right)
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\int^1_0 xe^{-x^2} dx =& \frac{-1}{2} \left( \frac{1 - e}{e} \right)
\\
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\int^1_0 xe^{-x^2} dx =& \frac{e - 1}{2e}

\end{aligned}
\end{equation}
$