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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

- 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}

\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}

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