What is the integral from 0 to 2 of re^{-r^2} ?

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

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integral from 0 to 2 of re^{-r^2}

\int_{0}^{2} r e^{- r^{2}} d r

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

Decimal

0.49084 \dots

Solution steps

\int_{0}^{2} r e^{- r^{2}} d r

Apply u-substitution

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

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

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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