What is the integral of x(4-x)^5 ?

The integral of x(4-x)^5 is 512x^2-(1280x^3)/3+160x^4-32x^5+(10x^6)/3-(x^7)/7+C

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integral of x(4-x)^5

\int x (4 - x)^{5} d x

512 x^{2} - \frac{1280 x ^{3}}{3} + 160 x^{4} - 32 x^{5} + \frac{10 x ^{6}}{3} - \frac{x ^{7}}{7} + C

Solution steps

\int x (4 - x)^{5} d x

Expand

x (4 - x)^{5} : 1024 x - 1280 x^{2} + 640 x^{3} - 160 x^{4} + 20 x^{5} - x^{6}

= \int 1024 x - 1280 x^{2} + 640 x^{3} - 160 x^{4} + 20 x^{5} - x^{6} 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 1024 x d x - \int 1280 x^{2} d x + \int 640 x^{3} d x - \int 160 x^{4} d x + \int 20 x^{5} d x - \int x^{6} d x

\int 1024 x d x = 512 x^{2}

\int 1280 x^{2} d x = \frac{1280 x ^{3}}{3}

\int 640 x^{3} d x = 160 x^{4}

\int 160 x^{4} d x = 32 x^{5}

\int 20 x^{5} d x = \frac{10 x ^{6}}{3}

\int x^{6} d x = \frac{x ^{7}}{7}

= 512 x^{2} - \frac{1280 x ^{3}}{3} + 160 x^{4} - 32 x^{5} + \frac{10 x ^{6}}{3} - \frac{x ^{7}}{7}

Add a constant to the solution

= 512 x^{2} - \frac{1280 x ^{3}}{3} + 160 x^{4} - 32 x^{5} + \frac{10 x ^{6}}{3} - \frac{x ^{7}}{7} + C

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What is the integral of x(4-x)^5 ?
The integral of x(4-x)^5 is 512x^2-(1280x^3)/3+160x^4-32x^5+(10x^6)/3-(x^7)/7+C