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

The integral from 0 to 4 of (sqrt(4-x))^2 is 8

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

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

8

Solution steps

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

Simplify

(4 - x)^{2} : 4 - x

= \int_{0}^{4} 4 - 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}^{4} 4 d x - \int_{0}^{4} x d x

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

\int_{0}^{4} x d x = 8

= 16 - 8

Simplify

= 8

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What is the integral from 0 to 4 of (sqrt(4-x))^2 ?
The integral from 0 to 4 of (sqrt(4-x))^2 is 8