How do you find the integral $\int_{0}^{1} x \cdot e^{- x^{2}} d x$ $int_0^1x*e^(-x^2)dx$ ?

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How do you find the integral $\int_{0}^{1} x \cdot e^{- x^{2}} d x$ $int_0^1x*e^(-x^2)dx$ ?

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Answer 1 · 2014-09-22T02:00:36

We begin by making a substitution.

Let $u = - x^{2}$ $u=-x^2$

$\int x e^{u} d x$ $intxe^udx$

$d u = - 2 x d x$ $du=-2xdx$

$\frac{d u}{- 2} = \frac{- 2 x d x}{- 2}$ $(du)/(-2)=(-2xdx)/(-2)$

$\frac{d u}{- 2} = x d x$ $(du)/(-2)=xdx$

$\frac{- 1}{2} \cdot d u = x d x$ $(-1)/2*du=xdx$

$\int e^{u} x d x$ $inte^uxdx$ , Notice that xdx can be replaced by $\frac{- 1}{2} \cdot d u$ $(-1)/2*du$

$\int e^{u} \frac{- 1}{2} d u$ $inte^u(-1)/2du$ , Move the constant to the front of the integral

$\frac{- 1}{2} \int e^{u} d u$ $(-1)/2inte^udu$

After integration look back to the original substitution to find the value for $u$ $u$ . In this case that value is $- x^{2}$ $-x^2$ .

I switched back to $- x^{2}$ $-x^2$ so that I could use the original boundaries.

$\frac{- 1}{2} {[e^{- x^{2}}]}_{0}^{- x^{2}} = \frac{- 1}{2} [e^{- 1} - e^{0}] = \frac{- 1}{2} [\frac{1}{e} - 1] = \frac{- 1}{2 e} + \frac{1}{2} =$ $(-1)/2[e^(-x^2)]_0^1=(-1)/2[e^(-1)-e^0]=(-1)/2[1/e-1]=(-1)/(2e)+1/2=$

$= 0.31606$ $=0.31606$

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How do you find the integral $\int_{0}^{1} x \cdot e^{- x^{2}} d x$ $int_0^1x*e^(-x^2)dx$ ?

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