What is the integral from-1 to 3 of (8-4y^2) ?

The integral from-1 to 3 of (8-4y^2) is -16/3

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integral from-1 to 3 of (8-4y^2)

\int_{- 1}^{3} (8 - 4 y^{2}) d y

- \frac{16}{3}

Decimal Notation

- 5.33333 \dots

Solution steps

\int_{- 1}^{3} 8 - 4 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} 4 y^{2} d y

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

\int_{- 1}^{3} 4 y^{2} d y = \frac{112}{3}

= 32 - \frac{112}{3}

Simplify

32 - \frac{112}{3} : - \frac{16}{3}

= - \frac{16}{3}

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What is the integral from-1 to 3 of (8-4y^2) ?
The integral from-1 to 3 of (8-4y^2) is -16/3