What is the integral from 0 to 2 of xe^{-(x^2-1)/2} ?

The integral from 0 to 2 of xe^{-(x^2-1)/2} is sqrt(e)-1/(e^{3/2)}

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integral from 0 to 2 of xe^{-(x^2-1)/2}

\int_{0}^{2} x e^{- \frac{x ^{2} - 1}{2}} d x

e - \frac{1}{e ^{\frac{3}{2}}}

Decimal Notation

1.42559 \dots

Solution steps

\int_{0}^{2} x e^{- \frac{x ^{2} - 1}{2}} d x

Apply u-substitution

: \int_{- 1}^{3} \frac{e ^{- \frac{u}{2}}}{2} d u

= \int_{- 1}^{3} \frac{e ^{- \frac{u}{2}}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- 1}^{3} e^{- \frac{u}{2}} d u

Apply u-substitution

: \int_{\frac{1}{2}}^{- \frac{3}{2}} - 2 e^{v} d v

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{- \frac{3}{2}} - 2 e^{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{2} (- \int_{- \frac{3}{2}}^{\frac{1}{2}} - 2 e^{v} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- (- 2 \cdot \int_{- \frac{3}{2}}^{\frac{1}{2}} e^{v} d v))

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} (- (- 2 [e^{v}]_{- \frac{3}{2}}^{\frac{1}{2}}))

Simplify

\frac{1}{2} (- (- 2 [e^{v}]_{- \frac{3}{2}}^{\frac{1}{2}})) : [e^{v}]_{- \frac{3}{2}}^{\frac{1}{2}}

= [e^{v}]_{- \frac{3}{2}}^{\frac{1}{2}}

Compute the boundaries:

e - \frac{1}{e ^{\frac{3}{2}}}

= e - \frac{1}{e ^{\frac{3}{2}}}

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What is the integral from 0 to 2 of xe^{-(x^2-1)/2} ?
The integral from 0 to 2 of xe^{-(x^2-1)/2} is sqrt(e)-1/(e^{3/2)}