What is the integral from 0 to 2 of (3x+1)e^{2x} ?

The integral from 0 to 2 of (3x+1)e^{2x} is (11e^4+1)/4

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integral from 0 to 2 of (3x+1)e^{2x}

\int_{0}^{2} (3 x + 1) e^{2 x} d x

\frac{11 e ^{4} + 1}{4}

Decimal Notation

150.39491 \dots

Solution steps

\int_{0}^{2} (3 x + 1) e^{2 x} d x

Apply Integration By Parts

: [\frac{1}{2} e^{2 x} (3 x + 1) - \int \frac{3}{2} e^{2 x} d x]_{0}^{2}

= [\frac{1}{2} e^{2 x} (3 x + 1) - \int \frac{3}{2} e^{2 x} d x]_{0}^{2}

\int \frac{3}{2} e^{2 x} d x = \frac{3}{4} e^{2 x}

= [\frac{1}{2} e^{2 x} (3 x + 1) - \frac{3}{4} e^{2 x}]_{0}^{2}

Simplify

[\frac{1}{2} e^{2 x} (3 x + 1) - \frac{3}{4} e^{2 x}]_{0}^{2} : [\frac{1}{4} (6 e^{2 x} x - e^{2 x})]_{0}^{2}

= [\frac{1}{4} (6 e^{2 x} x - e^{2 x})]_{0}^{2}

Compute the boundaries:

\frac{11 e ^{4}}{4} + \frac{1}{4}

= \frac{11 e ^{4}}{4} + \frac{1}{4}

Simplify

= \frac{11 e ^{4} + 1}{4}

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What is the integral from 0 to 2 of (3x+1)e^{2x} ?
The integral from 0 to 2 of (3x+1)e^{2x} is (11e^4+1)/4