What is the integral from 0 to 2pi of xe^{-x^2} ?

The integral from 0 to 2pi of xe^{-x^2} is (e^{4pi^2}-1)/(2e^{4pi^2)}

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integral from 0 to 2pi of xe^{-x^2}

\int_{0}^{2 π} x e^{- x^{2}} d x

\frac{e ^{4 π^{2}} - 1}{2 e ^{4 π^{2}}}

Decimal Notation

0.5

Solution steps

\int_{0}^{2 π} x e^{- x^{2}} d x

Apply u-substitution

: \int_{0}^{- 4 π^{2}} - \frac{e ^{u}}{2} d u

= \int_{0}^{- 4 π^{2}} - \frac{e ^{u}}{2} d u

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

= - \int_{- 4 π^{2}}^{0} - \frac{e ^{u}}{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_{- 4 π^{2}}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 4 π^{2}}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 4 π^{2}}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4 π^{2}}}

= \frac{1}{2} (1 - \frac{1}{e ^{4 π^{2}}})

Simplify

= \frac{e ^{4 π^{2}} - 1}{2 e ^{4 π^{2}}}

\int 16 sin^{2} (x) cos^{3} (x) d x

\frac{1}{4 x + 3}

\frac{\partial}{\partial x} (7 x - 3 y + 4)

\int_{0}^{1} x^{6} - x d x

y^{^{'}} + y \cdot cos (x) = sin (x) \cdot cos (x)

What is the integral from 0 to 2pi of xe^{-x^2} ?
The integral from 0 to 2pi of xe^{-x^2} is (e^{4pi^2}-1)/(2e^{4pi^2)}