What is the integral from-4 to 4 of sqrt(3)(16-y^2) ?

The integral from-4 to 4 of sqrt(3)(16-y^2) is (256)/(sqrt(3))

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integral from-4 to 4 of sqrt(3)(16-y^2)

\int_{- 4}^{4} 3 (16 - y^{2}) d y

\frac{256}{3}

Decimal

147.80166 \dots

Solution steps

\int_{- 4}^{4} 3 (16 - y^{2}) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{- 4}^{4} 16 - 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

= 3 (\int_{- 4}^{4} 16 d y - \int_{- 4}^{4} y^{2} d y)

\int_{- 4}^{4} 16 d y = 128

\int_{- 4}^{4} y^{2} d y = \frac{128}{3}

= 3 (128 - \frac{128}{3})

Simplify

3 (128 - \frac{128}{3}) : \frac{256}{3}

= \frac{256}{3}

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What is the integral from-4 to 4 of sqrt(3)(16-y^2) ?
The integral from-4 to 4 of sqrt(3)(16-y^2) is (256)/(sqrt(3))