7.1 Integration By Parts/30: Difference between revisions

Revision as of 11:46, 29 November 2022

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

${\begin{aligned}u&=4+r^{2}{x}\\[2ex]du&={\frac {1}{\sqrt {1-x^{2}}}}\;dx\\[2ex]\end{aligned}}$

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

@@ Line 1: / Line 1: @@
 <math> \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}r^{3}\cdot \frac{1}{\sqrt{4+r^{2}}}\cdot dr  </math>
 <math>
 \begin{align}
-u &= \arcsin {x}
+u &= 4+r^{2}{x}
    \\[2ex]
 du &= \frac{1}{\sqrt{1-x^2}}\;dx \\[2ex]
@@ Line 12: / Line 10: @@
 \end{align}
 </math>
+<math> \int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\cdot dr  ~~~ = ~~~  \int_{0}^{1}r^{3}\cdot \frac{1}{\sqrt{4+r^{2}}}\cdot dr  </math>