What is the integral from 0 to 4 of (4-x^2)^2 ?

The integral from 0 to 4 of (4-x^2)^2 is 1472/15

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integral from 0 to 4 of (4-x^2)^2

\int_{0}^{4} (4 - x^{2})^{2} d x

\frac{1472}{15}

Decimal

98.13333 \dots

Solution steps

\int_{0}^{4} (4 - x^{2})^{2} d x

Expand

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

= \int_{0}^{4} 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

= \int_{0}^{4} 16 d x - \int_{0}^{4} 8 x^{2} d x + \int_{0}^{4} x^{4} d x

\int_{0}^{4} 16 d x = 64

\int_{0}^{4} 8 x^{2} d x = \frac{512}{3}

\int_{0}^{4} x^{4} d x = \frac{1024}{5}

= 64 - \frac{512}{3} + \frac{1024}{5}

Simplify

64 - \frac{512}{3} + \frac{1024}{5} : \frac{1472}{15}

= \frac{1472}{15}

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What is the integral from 0 to 4 of (4-x^2)^2 ?
The integral from 0 to 4 of (4-x^2)^2 is 1472/15