What is the integral from-2 to 2 of 1/2 (4-x^2)^2 ?

The integral from-2 to 2 of 1/2 (4-x^2)^2 is 256/15

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integral from-2 to 2 of 1/2 (4-x^2)^2

\int_{- 2}^{2} \frac{1}{2} (4 - x^{2})^{2} d x

\frac{256}{15}

Decimal Notation

17.06666 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{2} (4 - x^{2})^{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2}^{2} (4 - x^{2})^{2} d x

Expand

(4 - x^{2})^{2} : 16 - 8 x^{2} + x^{4}

= \frac{1}{2} \cdot \int_{- 2}^{2} 16 - 8 x^{2} + x^{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{- 2}^{2} 16 d x - \int_{- 2}^{2} 8 x^{2} d x + \int_{- 2}^{2} x^{4} d x)

\int_{- 2}^{2} 16 d x = 64

\int_{- 2}^{2} 8 x^{2} d x = \frac{128}{3}

\int_{- 2}^{2} x^{4} d x = \frac{64}{5}

= \frac{1}{2} (64 - \frac{128}{3} + \frac{64}{5})

Simplify

\frac{1}{2} (64 - \frac{128}{3} + \frac{64}{5}) : \frac{256}{15}

= \frac{256}{15}

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What is the integral from-2 to 2 of 1/2 (4-x^2)^2 ?
The integral from-2 to 2 of 1/2 (4-x^2)^2 is 256/15