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

The integral from 0 to 2 of te^{3t} is (5e^6+1)/9

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

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

\frac{5 e ^{6} + 1}{9}

Decimal Notation

224.23821 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{6} \frac{e ^{u} u}{9} d u

= \int_{0}^{6} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{6} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{6}

= \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{6}

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

= \frac{1}{9} [e^{u} u - e^{u}]_{0}^{6}

Compute the boundaries:

5 e^{6} + 1

= \frac{1}{9} (5 e^{6} + 1)

Simplify

= \frac{5 e ^{6} + 1}{9}

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What is the integral from 0 to 2 of te^{3t} ?
The integral from 0 to 2 of te^{3t} is (5e^6+1)/9