- #1
- 9,623
- 9
The question is to find the following intergal:
[tex]\int x\cdot u^{\frac{1}{2} [/tex] where [itex]u = 2x -1[/itex].
[tex]= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du [/tex]
[tex] u = 2x -1 \Rightarrow x = \frac{u+1}{2} [/tex]
[tex]= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du [/tex]
[tex] = \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} [/tex]
But this isn't going to give me the correct answer. Can anybody see where I've gone wrong? Thank's
[tex]\int x\cdot u^{\frac{1}{2} [/tex] where [itex]u = 2x -1[/itex].
[tex]= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du [/tex]
[tex] u = 2x -1 \Rightarrow x = \frac{u+1}{2} [/tex]
[tex]= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du [/tex]
[tex] = \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} [/tex]
But this isn't going to give me the correct answer. Can anybody see where I've gone wrong? Thank's