What is the integral of (1-r)sqrt(r) ?

The integral of (1-r)sqrt(r) is 2/3 r^{3/2}-2/5 r^{5/2}+C

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integral of (1-r)sqrt(r)

\int (1 - r) r d r

\frac{2}{3} r^{\frac{3}{2}} - \frac{2}{5} r^{\frac{5}{2}} + C

Solution steps

\int (1 - r) r d r

Expand

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

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

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

\int r^{\frac{3}{2}} d r = \frac{2}{5} r^{\frac{5}{2}}

= \frac{2}{3} r^{\frac{3}{2}} - \frac{2}{5} r^{\frac{5}{2}}

Add a constant to the solution

= \frac{2}{3} r^{\frac{3}{2}} - \frac{2}{5} r^{\frac{5}{2}} + C

\frac{\partial}{\partial y} (- 2 x)

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\int 6^{s i n (x)} cos (x) d x

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 5 x}{x + 1})

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

What is the integral of (1-r)sqrt(r) ?
The integral of (1-r)sqrt(r) is 2/3 r^{3/2}-2/5 r^{5/2}+C