What is the integral from 0 to 5 of te^{-t^2} ?

The integral from 0 to 5 of te^{-t^2} is (e^{25}-1)/(2e^{25)}

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integral from 0 to 5 of te^{-t^2}

\int_{0}^{5} t e^{- t^{2}} d t

\frac{e ^{25} - 1}{2 e ^{25}}

Decimal

0.49999 \dots

Solution steps

\int_{0}^{5} t e^{- t^{2}} d t

Apply u-substitution

: \int_{0}^{- 25} - \frac{e ^{u}}{2} d u

= \int_{0}^{- 25} - \frac{e ^{u}}{2} d u

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

= - \int_{- 25}^{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_{- 25}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 25}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 25}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{25}}

= \frac{1}{2} (1 - \frac{1}{e ^{25}})

Simplify

= \frac{e ^{25} - 1}{2 e ^{25}}

\int_{0}^{1} [2 x] sin (π x) d x

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\int_{- 1}^{1} sin^{2} (x) d x

\int_{0}^{1} (2 x + 2) d x

\int_{1}^{2} (e^{2 t} + \frac{1}{2 t}) d t

What is the integral from 0 to 5 of te^{-t^2} ?
The integral from 0 to 5 of te^{-t^2} is (e^{25}-1)/(2e^{25)}