What is the integral from 0 to 1 of xe^{-5x^2} ?

The integral from 0 to 1 of xe^{-5x^2} is (e^5-1)/(10e^5)

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integral from 0 to 1 of xe^{-5x^2}

\int_{0}^{1} x e^{- 5 x^{2}} d x

\frac{e ^{5} - 1}{10 e ^{5}}

Decimal Notation

0.09932 \dots

Solution steps

\int_{0}^{1} x e^{- 5 x^{2}} d x

Apply u-substitution

: \int_{0}^{- 5} - \frac{e ^{u}}{10} d u

= \int_{0}^{- 5} - \frac{e ^{u}}{10} d u

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

= - \int_{- 5}^{0} - \frac{e ^{u}}{10} d u

Take the constant out:

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

= - (- \frac{1}{10} \cdot \int_{- 5}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{10} [e^{u}]_{- 5}^{0})

Simplify

= \frac{1}{10} [e^{u}]_{- 5}^{0}

Compute the boundaries:

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

= \frac{1}{10} (1 - \frac{1}{e ^{5}})

Simplify

= \frac{e ^{5} - 1}{10 e ^{5}}

\int_{0}^{1} \frac{5 y}{e ^{2 y}} d y

\frac{d}{d x} (ln (x^{2} + y^{2}))

\frac{\partial}{\partial x} (x^{2} e^{x^{2}})

\frac{\partial}{\partial x} (e^{x} sin (x y))

\frac{\partial}{\partial x} (x + y + z)

What is the integral from 0 to 1 of xe^{-5x^2} ?
The integral from 0 to 1 of xe^{-5x^2} is (e^5-1)/(10e^5)