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

The integral of ((1-x)^3)/3 is 1/3 x-(x^2)/2+1/3 x^3-(x^4)/(12)+C

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

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

\frac{1}{3} x - \frac{x ^{2}}{2} + \frac{1}{3} x^{3} - \frac{x ^{4}}{12} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int (1 - x)^{3} d x

Expand

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

= \frac{1}{3} \cdot \int 1 - 3 x + 3 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

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

\int 1 d x = x

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

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

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

= \frac{1}{3} (x - \frac{3 x ^{2}}{2} + x^{3} - \frac{x ^{4}}{4})

Simplify

\frac{1}{3} (x - \frac{3 x ^{2}}{2} + x^{3} - \frac{x ^{4}}{4}) : \frac{1}{3} x - \frac{x ^{2}}{2} + \frac{1}{3} x^{3} - \frac{x ^{4}}{12}

= \frac{1}{3} x - \frac{x ^{2}}{2} + \frac{1}{3} x^{3} - \frac{x ^{4}}{12}

Add a constant to the solution

= \frac{1}{3} x - \frac{x ^{2}}{2} + \frac{1}{3} x^{3} - \frac{x ^{4}}{12} + C

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What is the integral of ((1-x)^3)/3 ?
The integral of ((1-x)^3)/3 is 1/3 x-(x^2)/2+1/3 x^3-(x^4)/(12)+C