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

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

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

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

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

Decimal Notation

- 14.39953 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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