What is the integral from 0 to 1 of (y^2-y^3) ?

The integral from 0 to 1 of (y^2-y^3) is 1/12

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integral from 0 to 1 of (y^2-y^3)

\int_{0}^{1} (y^{2} - y^{3}) d y

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

\int_{0}^{1} 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_{0}^{1} y^{2} d y - \int_{0}^{1} y^{3} d y

\int_{0}^{1} y^{2} d y = \frac{1}{3}

\int_{0}^{1} y^{3} d y = \frac{1}{4}

= \frac{1}{3} - \frac{1}{4}

Simplify

\frac{1}{3} - \frac{1}{4} : \frac{1}{12}

= \frac{1}{12}

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What is the integral from 0 to 1 of (y^2-y^3) ?
The integral from 0 to 1 of (y^2-y^3) is 1/12