What is the integral from a to b of x(b-x)^2 ?

The integral from a to b of x(b-x)^2 is (b^4-3a^4+8ba^3-6b^2a^2)/(12)

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integral from a to b of x(b-x)^2

\int_{a}^{b} x (b - x)^{2} d x

\frac{b ^{4} - 3 a ^{4} + 8 b a ^{3} - 6 b ^{2} a ^{2}}{12}

Solution steps

\int_{a}^{b} x (b - x)^{2} d x

Apply Integration By Parts

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

\int - x^{2} (b - x) d x = - \frac{b x ^{3}}{3} + \frac{x ^{4}}{4}

= [\frac{1}{2} x^{2} (b - x)^{2} - (- \frac{b x ^{3}}{3} + \frac{x ^{4}}{4})]_{a}^{b}

Simplify

[\frac{1}{2} x^{2} (b - x)^{2} - (- \frac{b x ^{3}}{3} + \frac{x ^{4}}{4})]_{a}^{b} : [\frac{1}{12} (3 x^{4} - 8 b x^{3} + 6 b^{2} x^{2})]_{a}^{b}

= [\frac{1}{12} (3 x^{4} - 8 b x^{3} + 6 b^{2} x^{2})]_{a}^{b}

Compute the boundaries:

\frac{b ^{4}}{12} - \frac{1}{12} (3 a^{4} - 8 b a^{3} + 6 b^{2} a^{2})

= \frac{b ^{4}}{12} - \frac{1}{12} (3 a^{4} - 8 b a^{3} + 6 b^{2} a^{2})

Simplify

= \frac{b ^{4} - 3 a ^{4} + 8 b a ^{3} - 6 b ^{2} a ^{2}}{12}

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What is the integral from a to b of x(b-x)^2 ?
The integral from a to b of x(b-x)^2 is (b^4-3a^4+8ba^3-6b^2a^2)/(12)