What is the integral of x^2(1-x)^3 ?

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

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integral of x^2(1-x)^3

\int x^{2} (1 - x)^{3} d x

\frac{x ^{3}}{3} - \frac{3 x ^{4}}{4} + \frac{3 x ^{5}}{5} - \frac{x ^{6}}{6} + C

Solution steps

\int x^{2} (1 - x)^{3} d x

Expand

x^{2} (1 - x)^{3} : x^{2} - 3 x^{3} + 3 x^{4} - x^{5}

= \int x^{2} - 3 x^{3} + 3 x^{4} - x^{5} 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^{2} d x - \int 3 x^{3} d x + \int 3 x^{4} d x - \int x^{5} d x

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

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

= \frac{x ^{3}}{3} - \frac{3 x ^{4}}{4} + \frac{3 x ^{5}}{5} - \frac{x ^{6}}{6}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{3 x ^{4}}{4} + \frac{3 x ^{5}}{5} - \frac{x ^{6}}{6} + C

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