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

The integral from 1 to 2 of (y^2-y^3) is -17/12

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

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

- \frac{17}{12}

Solution steps

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

\int_{1}^{2} y^{2} d y = \frac{7}{3}

\int_{1}^{2} y^{3} d y = \frac{15}{4}

= \frac{7}{3} - \frac{15}{4}

Simplify

\frac{7}{3} - \frac{15}{4} : - \frac{17}{12}

= - \frac{17}{12}

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What is the integral from 1 to 2 of (y^2-y^3) ?
The integral from 1 to 2 of (y^2-y^3) is -17/12