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

The integral from 1 to 0 of x^3e^{2x} is -(e^2+3)/8

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

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

- \frac{e ^{2} + 3}{8}

Decimal Notation

- 1.29863 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} x^{3} e^{2 x} d x

Apply Integration By Parts

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

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

\int \frac{3}{2} e^{2 x} x^{2} d x = \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

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

Compute the boundaries:

\frac{e ^{2}}{8} + \frac{3}{8}

= - (\frac{e ^{2}}{8} + \frac{3}{8})

Simplify

= - \frac{e ^{2} + 3}{8}

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What is the integral from 1 to 0 of x^3e^{2x} ?
The integral from 1 to 0 of x^3e^{2x} is -(e^2+3)/8