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Popular Calculus >

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

Solution

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

Solution

- \frac{1}{3}

+1

Decimal

- 0.33333 \dots

Solution steps

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

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

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

= \frac{2}{3} - 1

Simplify

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

= - \frac{1}{3}

Graph

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Frequently Asked Questions (FAQ)

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