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

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

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

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

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

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

= 6 - (- \frac{3}{2}) - 3

Simplify

6 - (- \frac{3}{2}) - 3 : \frac{9}{2}

= \frac{9}{2}

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