7.1 Integration By Parts/30: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 16: Line 16:
<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr  ~~~ = ~~~  \int_{0}^{1}\frac{r}{2\sqrt{u}}\cdot du  ~~~ = ~~~  \int_{0}^{1}\frac{u-4}{2\sqrt{u}}\cdot du  ~~~ = ~~~  \frac{}{}\frac{1}{2} \int_{0}^{1} \left (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}} \right ) \cdot du  ~~~ = ~~~  </math>
<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr  ~~~ = ~~~  \int_{0}^{1}\frac{r}{2\sqrt{u}}\cdot du  ~~~ = ~~~  \int_{0}^{1}\frac{u-4}{2\sqrt{u}}\cdot du  ~~~ = ~~~  \frac{}{}\frac{1}{2} \int_{0}^{1} \left (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}} \right ) \cdot du  ~~~ = ~~~  </math>


<math> \frac{1}{2} \left [ \left (\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right ) - \left ( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right ) \right ]  ~~~ = ~~~  \frac{1}{2} \left [ \left (\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right ) - 4\left ( \frac{u^{\frac{1}{2} }}{\frac{1}{2}} \right )   ~~~ = ~~~  \right ] </math>
<math> \frac{1}{2} \left [ \left (\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right ) - \left ( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right ) \right ]  ~~~ = ~~~  \frac{1}{2} \left [ \left (\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right ) - 4\left ( \frac{u^{\frac{1}{2} }}{\frac{1}{2}} \right )\right ]    ~~~ = ~~~  \frac{u^{\frac{3}{2}}}{3} - 4u^{\frac{1}{2}} </math>

Revision as of 12:25, 29 November 2022