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

The integral of (y^2-y^3) is (y^3)/3-(y^4)/4+C

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integral of (y^2-y^3)

\int (y^{2} - y^{3}) d y

\frac{y ^{3}}{3} - \frac{y ^{4}}{4} + C

Solution steps

\int y^{2} - y^{3} 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^{2} d y - \int y^{3} d y

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

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

= \frac{y ^{3}}{3} - \frac{y ^{4}}{4}

Add a constant to the solution

= \frac{y ^{3}}{3} - \frac{y ^{4}}{4} + C

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sec^{2} (\frac{x}{2})

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