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

The integral from 0 to 1 of (2x+1)e^{3x} is (7e^3-1)/9

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

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

\frac{7 e ^{3} - 1}{9}

Decimal

15.51097 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2}{3} e^{3 x} d x = \frac{2}{9} e^{3 x}

= [\frac{1}{3} e^{3 x} (2 x + 1) - \frac{2}{9} e^{3 x}]_{0}^{1}

Simplify

[\frac{1}{3} e^{3 x} (2 x + 1) - \frac{2}{9} e^{3 x}]_{0}^{1} : [\frac{1}{9} (6 e^{3 x} x + e^{3 x})]_{0}^{1}

= [\frac{1}{9} (6 e^{3 x} x + e^{3 x})]_{0}^{1}

Compute the boundaries:

\frac{7 e ^{3}}{9} - \frac{1}{9}

= \frac{7 e ^{3}}{9} - \frac{1}{9}

Simplify

= \frac{7 e ^{3} - 1}{9}

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What is the integral from 0 to 1 of (2x+1)e^{3x} ?
The integral from 0 to 1 of (2x+1)e^{3x} is (7e^3-1)/9