What is the integral from 0 to 3 of-te^{-t^2} ?

The integral from 0 to 3 of-te^{-t^2} is -(e^9-1)/(2e^9)

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integral from 0 to 3 of-te^{-t^2}

\int_{0}^{3} - t e^{- t^{2}} d t

- \frac{e ^{9} - 1}{2 e ^{9}}

Decimal

- 0.49993 \dots

Solution steps

\int_{0}^{3} - t e^{- t^{2}} d t

Take the constant out:

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

= - \int_{0}^{3} t e^{- t^{2}} d t

Apply u-substitution

: \int_{0}^{- 9} - \frac{e ^{u}}{2} d u

= - \int_{0}^{- 9} - \frac{e ^{u}}{2} d u

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

= - (- \int_{- 9}^{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_{- 9}^{0} e^{u} d u))

Use the common integral:

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

= - (- (- \frac{1}{2} [e^{u}]_{- 9}^{0}))

Simplify

= - \frac{1}{2} [e^{u}]_{- 9}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{9}}

= - \frac{1}{2} (1 - \frac{1}{e ^{9}})

Simplify

= - \frac{e ^{9} - 1}{2 e ^{9}}

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What is the integral from 0 to 3 of-te^{-t^2} ?
The integral from 0 to 3 of-te^{-t^2} is -(e^9-1)/(2e^9)