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

The integral from 0 to 1/2 of x^2e^{2x} is e/8-1/4

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

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

\frac{e}{8} - \frac{1}{4}

Decimal

0.08978 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{e}{8} - \frac{1}{4}

= \frac{e}{8} - \frac{1}{4}

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What is the integral from 0 to 1/2 of x^2e^{2x} ?
The integral from 0 to 1/2 of x^2e^{2x} is e/8-1/4