What is the integral from 0 to 1 of x(a+bx^2) ?

The integral from 0 to 1 of x(a+bx^2) is 1/4 (2a+b)

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integral from 0 to 1 of x(a+bx^2)

\int_{0}^{1} x (a + b x^{2}) d x

\frac{1}{4} (2 a + b)

Solution steps

\int_{0}^{1} x (a + b x^{2}) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (a + b x^{2}) - \int b x^{3} d x]_{0}^{1}

= [\frac{1}{2} x^{2} (a + b x^{2}) - \int b x^{3} d x]_{0}^{1}

\int b x^{3} d x = \frac{b x ^{4}}{4}

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

Simplify

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

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

Compute the boundaries:

\frac{1}{4} (2 a + b)

= \frac{1}{4} (2 a + b)

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What is the integral from 0 to 1 of x(a+bx^2) ?
The integral from 0 to 1 of x(a+bx^2) is 1/4 (2a+b)