What is the integral from 0 to r of (br+c)2pir ?

The integral from 0 to r of (br+c)2pir is (2pibr^3)/3+picr^2

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integral from 0 to r of (br+c)2pir

\int_{0}^{r} (b r + c) 2 π r d r

\frac{2 πb r ^{3}}{3} + π c r^{2}

Solution steps

\int_{0}^{r} (b r + c) \cdot 2 π r d r

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{r} (b r + c) r d r

Apply Integration By Parts

: [\frac{1}{2} r^{2} (b r + c) - \int \frac{b r ^{2}}{2} d r]_{0}^{r}

= 2 π [\frac{1}{2} r^{2} (b r + c) - \int \frac{b r ^{2}}{2} d r]_{0}^{r}

\int \frac{b r ^{2}}{2} d r = \frac{b r ^{3}}{6}

= 2 π [\frac{1}{2} r^{2} (b r + c) - \frac{b r ^{3}}{6}]_{0}^{r}

Simplify

2 π [\frac{1}{2} r^{2} (b r + c) - \frac{b r ^{3}}{6}]_{0}^{r} : 2 π [\frac{1}{6} (2 b r^{3} + 3 c r^{2})]_{0}^{r}

= 2 π [\frac{1}{6} (2 b r^{3} + 3 c r^{2})]_{0}^{r}

Compute the boundaries:

\frac{1}{6} (2 b r^{3} + 3 c r^{2})

= 2 π \frac{1}{6} (2 b r^{3} + 3 c r^{2})

Simplify

= \frac{2 πb r ^{3}}{3} + π c r^{2}

\int_{1}^{3} \frac{x}{5 - 2 x + x ^{2}} d x

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\int_{0}^{1} (x^{2} - x) e^{- x} d x

\int_{1}^{e} ln^{5} (x) \frac{1}{x} d x

\int_{1}^{2} 2 x + 3 d x

What is the integral from 0 to r of (br+c)2pir ?
The integral from 0 to r of (br+c)2pir is (2pibr^3)/3+picr^2