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

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

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

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

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

Solution steps

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

\int_{0}^{2} r 4 - r^{2} d r

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

\int_{0}^{y} 2 e^{- (x + y)} d x

\int_{0.5}^{1} cos (x) d x

\int_{- 1}^{5} (3 x^{4} + 2 x^{2} + 3) d x

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