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

The integral from 0 to 1 of x(ax+bx^2) is 1/12 (4a+3b)

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

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

\frac{1}{12} (4 a + 3 b)

Solution steps

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

Multiply by the conjugate of

a x + b x^{2} : \frac{a ^{2} x ^{2} - b ^{2} x ^{4}}{a x - b x ^{2}}

= \int_{0}^{1} x \frac{a ^{2} x ^{2} - b ^{2} x ^{4}}{a x - b x ^{2}} d x

Simplify

x \frac{a ^{2} x ^{2} - b ^{2} x ^{4}}{a x - b x ^{2}} : x^{2} (b x + a)

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

Apply Integration By Parts

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

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

\int \frac{b x ^{3}}{3} d x = \frac{b x ^{4}}{12}

= [\frac{1}{3} x^{3} (b x + a) - \frac{b x ^{4}}{12}]_{0}^{1}

Simplify

[\frac{1}{3} x^{3} (b x + a) - \frac{b x ^{4}}{12}]_{0}^{1} : [\frac{1}{12} (3 b x^{4} + 4 a x^{3})]_{0}^{1}

= [\frac{1}{12} (3 b x^{4} + 4 a x^{3})]_{0}^{1}

Compute the boundaries:

\frac{1}{12} (4 a + 3 b)

= \frac{1}{12} (4 a + 3 b)

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What is the integral from 0 to 1 of x(ax+bx^2) ?
The integral from 0 to 1 of x(ax+bx^2) is 1/12 (4a+3b)