What is the integral of (1-y)y^n ?

The integral of (1-y)y^n is (y^{n+1})/(n+1)-(y^{n+2})/(n+2)+C

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integral of (1-y)y^n

\int (1 - y) y^{n} d y

\frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2} + C

Solution steps

\int (1 - y) y^{n} d y

Expand

(1 - y) y^{n} : y^{n} - y^{1 + n}

= \int y^{n} - y^{1 + n} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{n} d y - \int y^{1 + n} d y

\int y^{n} d y = \frac{y ^{n + 1}}{n + 1}

\int y^{1 + n} d y = \frac{y ^{n + 2}}{n + 2}

= \frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2}

Add a constant to the solution

= \frac{y ^{n + 1}}{n + 1} - \frac{y ^{n + 2}}{n + 2} + C

What is the integral of (1-y)y^n ?
The integral of (1-y)y^n is (y^{n+1})/(n+1)-(y^{n+2})/(n+2)+C