What is the integral from 0 to 1 of x^2e^{5x} ?

The integral from 0 to 1 of x^2e^{5x} is (17e^5-2)/(125)

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integral from 0 to 1 of x^2e^{5x}

\int_{0}^{1} x^{2} e^{5 x} d x

\frac{17 e ^{5} - 2}{125}

Decimal

20.16818 \dots

Solution steps

\int_{0}^{1} x^{2} e^{5 x} d x

Apply Integration By Parts

: [\frac{1}{5} e^{5 x} x^{2} - \int \frac{2}{5} e^{5 x} x d x]_{0}^{1}

= [\frac{1}{5} e^{5 x} x^{2} - \int \frac{2}{5} e^{5 x} x d x]_{0}^{1}

\int \frac{2}{5} e^{5 x} x d x = \frac{2}{125} (5 e^{5 x} x - e^{5 x})

= [\frac{1}{5} e^{5 x} x^{2} - \frac{2}{125} (5 e^{5 x} x - e^{5 x})]_{0}^{1}

Compute the boundaries:

\frac{17 e ^{5}}{125} - \frac{2}{125}

= \frac{17 e ^{5}}{125} - \frac{2}{125}

Simplify

= \frac{17 e ^{5} - 2}{125}

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What is the integral from 0 to 1 of x^2e^{5x} ?
The integral from 0 to 1 of x^2e^{5x} is (17e^5-2)/(125)