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

The integral from 0 to r of (br^2+cr)(2pir) is pi/6 (3br^4+4cr^3)

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

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

\frac{π}{6} (3 b r^{4} + 4 c r^{3})

Solution steps

\int_{0}^{r} (b r^{2} + cr) \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^{2} + cr) r d r

Multiply by the conjugate of

b r^{2} + cr : \frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr}

= 2 π \cdot \int_{0}^{r} \frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr} r d r

Simplify

\frac{b ^{2} r ^{4} - c ^{2} r ^{2}}{b r ^{2} - cr} r : r^{2} (b r + c)

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

Apply Integration By Parts

: [\frac{1}{3} r^{3} (b r + c) - \int \frac{b r ^{3}}{3} d r]_{0}^{r}

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

\int \frac{b r ^{3}}{3} d r = \frac{b r ^{4}}{12}

= 2 π [\frac{1}{3} r^{3} (b r + c) - \frac{b r ^{4}}{12}]_{0}^{r}

Simplify

2 π [\frac{1}{3} r^{3} (b r + c) - \frac{b r ^{4}}{12}]_{0}^{r} : 2 π [\frac{1}{12} (3 b r^{4} + 4 c r^{3})]_{0}^{r}

= 2 π [\frac{1}{12} (3 b r^{4} + 4 c r^{3})]_{0}^{r}

Compute the boundaries:

\frac{1}{12} (3 b r^{4} + 4 c r^{3})

= 2 π \frac{1}{12} (3 b r^{4} + 4 c r^{3})

Simplify

= \frac{π}{6} (3 b r^{4} + 4 c r^{3})

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What is the integral from 0 to r of (br^2+cr)(2pir) ?
The integral from 0 to r of (br^2+cr)(2pir) is pi/6 (3br^4+4cr^3)