What is the integral from 0 to 9 of 8-sqrt((64x)/9) ?

The integral from 0 to 9 of 8-sqrt((64x)/9) is 24

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integral from 0 to 9 of 8-sqrt((64x)/9)

\int_{0}^{9} 8 - \frac{64 x}{9} d x

24

Solution steps

\int_{0}^{9} 8 - \frac{64 x}{9} d x

Simplify

8 - \frac{64 x}{9} : 8 - \frac{8}{3} x

= \int_{0}^{9} 8 - \frac{8}{3} x 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}^{9} 8 d x - \int_{0}^{9} \frac{8}{3} x d x

\int_{0}^{9} 8 d x = 72

\int_{0}^{9} \frac{8}{3} x d x = 48

= 72 - 48

Simplify

= 24

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What is the integral from 0 to 9 of 8-sqrt((64x)/9) ?
The integral from 0 to 9 of 8-sqrt((64x)/9) is 24