7.1 Integration By Parts/30: Difference between revisions
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Now, we need to substitute u back | Now, we need to substitute u back | ||
<math>\frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} + C </math> | <math>\frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} + C </math> | ||
<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr ~~~ = ~~~ \left [ \frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} \right ] </math> | <math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr ~~~ = ~~~ \left [ \frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} \right ] </math> | ||
Revision as of 12:39, 29 November 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= 4+r^{2} \\[2ex] r^{2} &= u-4 \\[2ex] 2r\cdot dr &= du \\[2ex] \end{align} }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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 ~~~ = ~~~ }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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}} }
Now, we need to substitute u back
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} + C }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr ~~~ = ~~~ \left [ \frac{\left ( r^{2}+4 \right )^{\frac{3}{2}}}{3} - 4\left ( r^{2}+4 \right )^{\frac{1}{2}} \right ] }