What is the integral of x^4(1-x)^3 ?

The integral of x^4(1-x)^3 is (x^5)/5-(x^6)/2+(3x^7)/7-(x^8)/8+C

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integral of x^4(1-x)^3

\int x^{4} (1 - x)^{3} d x

\frac{x ^{5}}{5} - \frac{x ^{6}}{2} + \frac{3 x ^{7}}{7} - \frac{x ^{8}}{8} + C

Solution steps

\int x^{4} (1 - x)^{3} d x

Expand

x^{4} (1 - x)^{3} : x^{4} - 3 x^{5} + 3 x^{6} - x^{7}

= \int x^{4} - 3 x^{5} + 3 x^{6} - x^{7} 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 x^{4} d x - \int 3 x^{5} d x + \int 3 x^{6} d x - \int x^{7} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 3 x^{6} d x = \frac{3 x ^{7}}{7}

\int x^{7} d x = \frac{x ^{8}}{8}

= \frac{x ^{5}}{5} - \frac{x ^{6}}{2} + \frac{3 x ^{7}}{7} - \frac{x ^{8}}{8}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{6}}{2} + \frac{3 x ^{7}}{7} - \frac{x ^{8}}{8} + C

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What is the integral of x^4(1-x)^3 ?
The integral of x^4(1-x)^3 is (x^5)/5-(x^6)/2+(3x^7)/7-(x^8)/8+C