What is the integral from 0 to 4 of (16-x^2)^2 ?

The integral from 0 to 4 of (16-x^2)^2 is 8192/15

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integral from 0 to 4 of (16-x^2)^2

\int_{0}^{4} (16 - x^{2})^{2} d x

\frac{8192}{15}

Decimal

546.13333 \dots

Solution steps

\int_{0}^{4} (16 - x^{2})^{2} d x

Expand

(16 - x^{2})^{2} : 256 - 32 x^{2} + x^{4}

= \int_{0}^{4} 256 - 32 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} 256 d x - \int_{0}^{4} 32 x^{2} d x + \int_{0}^{4} x^{4} d x

\int_{0}^{4} 256 d x = 1024

\int_{0}^{4} 32 x^{2} d x = \frac{2048}{3}

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

= 1024 - \frac{2048}{3} + \frac{1024}{5}

Simplify

1024 - \frac{2048}{3} + \frac{1024}{5} : \frac{8192}{15}

= \frac{8192}{15}

What is the integral from 0 to 4 of (16-x^2)^2 ?
The integral from 0 to 4 of (16-x^2)^2 is 8192/15