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

The integral from 0 to 2pi of (2-2r^2)r is (-16pi^4+8pi^2)/2

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integral from 0 to 2pi of (2-2r^2)r

\int_{0}^{2 π} (2 - 2 r^{2}) r d r

\frac{- 16 π ^{4} + 8 π ^{2}}{2}

Decimal Notation

- 739.79431 \dots

Solution steps

\int_{0}^{2 π} (2 - 2 r^{2}) r d r

Apply Integration By Parts

: [r^{2} (- r^{2} + 1) - \int - 2 r^{3} d r]_{0}^{2 π}

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{- 16 π ^{4} + 8 π ^{2}}{2}

= \frac{- 16 π ^{4} + 8 π ^{2}}{2}

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What is the integral from 0 to 2pi of (2-2r^2)r ?
The integral from 0 to 2pi of (2-2r^2)r is (-16pi^4+8pi^2)/2