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

The integral from-2 to 1 of (3-y)-(y^2+1) is 9/2

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

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

\frac{9}{2}

Decimal

4.5

Solution steps

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

Simplify

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

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

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

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

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

Simplify

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

= \frac{9}{2}

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What is the integral from-2 to 1 of (3-y)-(y^2+1) ?
The integral from-2 to 1 of (3-y)-(y^2+1) is 9/2