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 1/(sqrt(e))-1/(e^{5/2)}

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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

\frac{1}{e} - \frac{1}{e ^{\frac{5}{2}}}

Decimal Notation

0.52444 \dots

Solution steps

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

Apply u-substitution

: \int_{- 1}^{- 5} - \frac{e ^{\frac{u}{2}}}{2} d u

= \int_{- 1}^{- 5} - \frac{e ^{\frac{u}{2}}}{2} d u

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

= - \int_{- 5}^{- 1} - \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_{- 5}^{- 1} e^{\frac{u}{2}} d u)

Apply u-substitution

: \int_{- \frac{5}{2}}^{- \frac{1}{2}} e^{v} \cdot 2 d v

= - (- \frac{1}{2} \cdot \int_{- \frac{5}{2}}^{- \frac{1}{2}} e^{v} \cdot 2 d v)

Take the constant out:

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

= - (- \frac{1}{2} \cdot 2 \cdot \int_{- \frac{5}{2}}^{- \frac{1}{2}} e^{v} d v)

Use the common integral:

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

= - (- \frac{1}{2} \cdot 2 [e^{v}]_{- \frac{5}{2}}^{- \frac{1}{2}})

Simplify

- (- \frac{1}{2} \cdot 2 [e^{v}]_{- \frac{5}{2}}^{- \frac{1}{2}}) : [e^{v}]_{- \frac{5}{2}}^{- \frac{1}{2}}

= [e^{v}]_{- \frac{5}{2}}^{- \frac{1}{2}}

Compute the boundaries:

\frac{1}{e} - \frac{1}{e ^{\frac{5}{2}}}

= \frac{1}{e} - \frac{1}{e ^{\frac{5}{2}}}

\int_{1}^{4} - x - (2 - x) d x

\int_{0}^{h} \frac{b}{h} (h - y) d y

\int_{1}^{4} (2 x) d x

\int_{0}^{π} cos (\frac{1}{2}) x d x

x \to \frac{π}{2} lim (\frac{cos ( x )}{1 - sin ( x )})

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 1/(sqrt(e))-1/(e^{5/2)}