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

The integral from 0 to 2 of rsqrt(4-r^2)-r^2 is 0

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integral from 0 to 2 of rsqrt(4-r^2)-r^2

\int_{0}^{2} r 4 - r^{2} - r^{2} d r

0

Solution steps

\int_{0}^{2} r 4 - r^{2} - r^{2} d r

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}^{2} r 4 - r^{2} d r - \int_{0}^{2} r^{2} d r

\int_{0}^{2} r 4 - r^{2} d r = \frac{8}{3}

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

= \frac{8}{3} - \frac{8}{3}

Simplify

= 0

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What is the integral from 0 to 2 of rsqrt(4-r^2)-r^2 ?
The integral from 0 to 2 of rsqrt(4-r^2)-r^2 is 0