What is the integral of x(4-x)^2 ?

The integral of x(4-x)^2 is 8x^2-(8x^3)/3+(x^4)/4+C

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integral of x(4-x)^2

\int x (4 - x)^{2} d x

8 x^{2} - \frac{8 x ^{3}}{3} + \frac{x ^{4}}{4} + C

Solution steps

\int x (4 - x)^{2} d x

Expand

x (4 - x)^{2} : 16 x - 8 x^{2} + x^{3}

= \int 16 x - 8 x^{2} + x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 16 x d x - \int 8 x^{2} d x + \int x^{3} d x

\int 16 x d x = 8 x^{2}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

= 8 x^{2} - \frac{8 x ^{3}}{3} + \frac{x ^{4}}{4}

Add a constant to the solution

= 8 x^{2} - \frac{8 x ^{3}}{3} + \frac{x ^{4}}{4} + C

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What is the integral of x(4-x)^2 ?
The integral of x(4-x)^2 is 8x^2-(8x^3)/3+(x^4)/4+C