What is the integral from-2 to 2 of (4+y^2)-(2y^2) ?

The integral from-2 to 2 of (4+y^2)-(2y^2) is 32/3

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integral from-2 to 2 of (4+y^2)-(2y^2)

\int_{- 2}^{2} (4 + y^{2}) - (2 y^{2}) d y

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{- 2}^{2} 4 + y^{2} - 2 y^{2} d y

Simplify

= \int_{- 2}^{2} 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_{- 2}^{2} 4 d y - \int_{- 2}^{2} y^{2} d y

\int_{- 2}^{2} 4 d y = 16

\int_{- 2}^{2} y^{2} d y = \frac{16}{3}

= 16 - \frac{16}{3}

Simplify

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

= \frac{32}{3}

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What is the integral from-2 to 2 of (4+y^2)-(2y^2) ?
The integral from-2 to 2 of (4+y^2)-(2y^2) is 32/3