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

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

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

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

\frac{3 e ^{2} - 1}{4}

Decimal

5.29179 \dots

Solution steps

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

Apply Integration By Parts

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

\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}

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

Simplify

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

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

Compute the boundaries:

\frac{3 e ^{2}}{4} - \frac{1}{4}

= \frac{3 e ^{2}}{4} - \frac{1}{4}

Simplify

= \frac{3 e ^{2} - 1}{4}

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