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

The integral from 0 to 1 of 2y^2-2y^3 is 1/6

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

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

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

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

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

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

= \frac{2}{3} - \frac{1}{2}

Simplify

\frac{2}{3} - \frac{1}{2} : \frac{1}{6}

= \frac{1}{6}

What is the integral from 0 to 1 of 2y^2-2y^3 ?
The integral from 0 to 1 of 2y^2-2y^3 is 1/6