Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 76

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

If we let $u = -x^2$, then $du = -2x dx$, 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
\\
\\
\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} (e^{-1} - 1)
\\
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\int^1_0 xe^{-x^2} dx =& \frac{- (e^{-1} - 1)}{2}
\\
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\int^1_0 xe^{-x^2} dx =& \frac{\displaystyle 1 - \frac{1}{e}}{2}
\\
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\int^1_0 xe^{-x^2} dx =& \frac{e - 1}{2e}

\end{aligned}
\end{equation}
$