# Evaluate the definite integral. \int_{0}^{1}xe^{-x^{2}}dx

Evaluate the definite integral.
${\int }_{0}^{1}x{e}^{-{x}^{2}}dx$
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Harr1957
Evaluate the definite integral.
${\int }_{0}^{1}x{e}^{-{x}^{2}}dx$
Let ${x}^{2}=t$
2x dx=dt
$xdx=\frac{dt}{2}$
${\int }_{0}^{1}x{e}^{-{x}^{2}}dx={\int }_{0}^{1}{e}^{-t}\frac{dt}{2}$
$=\frac{1}{2}{\int }_{0}^{1}{e}^{-t}dt$
$=\frac{-1}{2}{\left[{e}^{-t}\right]}_{0}^{1}$
$=\frac{-1}{2}\left[{e}^{-1}-1\right]$
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Warajected53
Step 1: Remove parentheses.
${\int }_{0}^{1}x{e}^{-{x}^{2}}dx$
Step 2: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
${\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$
Step 3: In this case, $f\left(x\right)=x{e}^{-{x}^{2}}$. Find its integral.
Step 4: Since $F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(1)−F(0):
$-\frac{{e}^{-{1}^{2}}}{2}-\left(-\frac{{e}^{-{0}^{2}}}{2}\right)$
Step 5: Simplify.
$-\frac{{e}^{-1}}{2}+\frac{{e}^{-0}}{2}$