What is the integral from 0 to 8 of (8-x)^2 ?

The integral from 0 to 8 of (8-x)^2 is 512/3

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integral from 0 to 8 of (8-x)^2

\int_{0}^{8} (8 - x)^{2} d x

\frac{512}{3}

Solution steps

\int_{0}^{8} (8 - x)^{2} d x

Expand

(8 - x)^{2} : 64 - 16 x + x^{2}

= \int_{0}^{8} 64 - 16 x + x^{2} 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}^{8} 64 d x - \int_{0}^{8} 16 x d x + \int_{0}^{8} x^{2} d x

\int_{0}^{8} 64 d x = 512

\int_{0}^{8} 16 x d x = 512

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

= 512 - 512 + \frac{512}{3}

Simplify

= \frac{512}{3}

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What is the integral from 0 to 8 of (8-x)^2 ?
The integral from 0 to 8 of (8-x)^2 is 512/3