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)/2

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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}

Decimal Notation

4051.04196 \dots

Solution steps

\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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{9} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{9} - 1

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

Simplify

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

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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)/2