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

The integral from-1 to 2 of (2-y^2)-(-y) is 9/2

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

Simplify

= \int_{- 1}^{2} 2 - y^{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_{- 1}^{2} 2 d y - \int_{- 1}^{2} y^{2} d y + \int_{- 1}^{2} y d y

\int_{- 1}^{2} 2 d y = 6

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

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

= 6 - 3 + \frac{3}{2}

Simplify

6 - 3 + \frac{3}{2} : \frac{9}{2}

= \frac{9}{2}

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What is the integral from-1 to 2 of (2-y^2)-(-y) ?
The integral from-1 to 2 of (2-y^2)-(-y) is 9/2