What is the integral from-1 to 3 of (8+2y-y^2) ?

The integral from-1 to 3 of (8+2y-y^2) is 92/3

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integral from-1 to 3 of (8+2y-y^2)

\int_{- 1}^{3} (8 + 2 y - y^{2}) d y

\frac{92}{3}

Decimal

30.66666 \dots

Solution steps

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

\int_{- 1}^{3} 8 d y = 32

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

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

= 32 + 8 - \frac{28}{3}

Simplify

32 + 8 - \frac{28}{3} : \frac{92}{3}

= \frac{92}{3}

\int_{1}^{2} 12 x^{2} d x

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What is the integral from-1 to 3 of (8+2y-y^2) ?
The integral from-1 to 3 of (8+2y-y^2) is 92/3