7.1 Integration By Parts/30: Difference between revisions

Revision as of 11:57, 29 November 2022

$\int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr$

${\begin{aligned}u&=4+r^{2}\\[2ex]r^{2}&=u-4\\[2ex]2r\cdot dr&=du\\[2ex]\end{aligned}}$

$\int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr~~~=~~~\int _{0}^{1}{\frac {r}{\sqrt {4+r^{2}}}}\cdot dr$

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	</math>		</math>

	<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr ~~~ = ~~~ \int_{0}^{1}\frac{r~~^{3}~~}{\sqrt{4+r^{2}}}\cdot dr </math>		<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr ~~~ = ~~~ \int_{0}^{1}\frac{r}{\sqrt{4+r^{2}}}\cdot dr </math>