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

The integral of x^2((1-x)-((1-x)^2)/2) is -(x^5}{10}+\frac{x^3)/6+C

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

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

- \frac{x ^{5}}{10} + \frac{x ^{3}}{6} + C

Solution steps

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

Expand

x^{2} (1 - x - \frac{( 1 - x ) ^{2}}{2}) : - \frac{x ^{4}}{2} + \frac{x ^{2}}{2}

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

\int \frac{x ^{4}}{2} d x = \frac{x ^{5}}{10}

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

= - \frac{x ^{5}}{10} + \frac{x ^{3}}{6}

Add a constant to the solution

= - \frac{x ^{5}}{10} + \frac{x ^{3}}{6} + C

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What is the integral of x^2((1-x)-((1-x)^2)/2) ?
The integral of x^2((1-x)-((1-x)^2)/2) is -(x^5}{10}+\frac{x^3)/6+C