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

The integral from 0 to 2 of te^{2t} is (3e^4+1)/4

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

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

\frac{3 e ^{4} + 1}{4}

Decimal

41.19861 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{4} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{4} e^{u} u d u

Apply Integration By Parts

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{4}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{4}

Compute the boundaries:

3 e^{4} + 1

= \frac{1}{4} (3 e^{4} + 1)

Simplify

= \frac{3 e ^{4} + 1}{4}

What is the integral from 0 to 2 of te^{2t} ?
The integral from 0 to 2 of te^{2t} is (3e^4+1)/4