What is the integral from 0 to 1 of (xe^x)^2 ?

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

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integral from 0 to 1 of (xe^x)^2

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

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

Decimal

1.59726 \dots

Solution steps

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

Simplify

(x e^{x})^{2} : e^{2 x} x^{2}

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

Apply Integration By Parts

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

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

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

= [\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}}{4} - \frac{1}{4}

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

Simplify

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

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