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

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

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

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

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

Decimal

0.21616 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{- 2}^{0} e^{u} d u)

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

(- 29) (- 2 - 5)

x \to - \infty lim (3 x^{6})

\frac{4 s + 20}{s ^{2} + 8 s + 16}

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 t ^{6}})

\frac{\partial}{\partial y} (x^{4})

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