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

The integral from-1 to 3 of 2y-(y^2-3) is 32/3

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

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

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

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

Simplify

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

\int_{- 1}^{3} 2 y d y = 8

\int_{- 1}^{3} y^{2} d y = \frac{28}{3}

\int_{- 1}^{3} 3 d y = 12

= 8 - \frac{28}{3} + 12

Simplify

8 - \frac{28}{3} + 12 : \frac{32}{3}

= \frac{32}{3}

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What is the integral from-1 to 3 of 2y-(y^2-3) ?
The integral from-1 to 3 of 2y-(y^2-3) is 32/3