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

The integral of rsqrt(4-r^2)-r^2 is -1/3 (4-r^2)^{3/2}-(r^3)/3+C

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integral of rsqrt(4-r^2)-r^2

\int r 4 - r^{2} - r^{2} d r

- \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3} + C

Solution steps

\int 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 r 4 - r^{2} d r - \int r^{2} d r

\int r 4 - r^{2} d r = - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}}

\int r^{2} d r = \frac{r ^{3}}{3}

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3}

Add a constant to the solution

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3} + C

\frac{1}{( s )}

sin (sin (x))

\frac{d}{d x} ((x - 1)^{2} (x - 2))

f (x) = sin (x^{3})

\frac{\partial}{\partial x} (Q (x) 0. 5^{\frac{x}{35}})

What is the integral of rsqrt(4-r^2)-r^2 ?
The integral of rsqrt(4-r^2)-r^2 is -1/3 (4-r^2)^{3/2}-(r^3)/3+C