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Mathematics 16 Online
OpenStudy (anonymous):

integral of 1/x * (2/3)x^(3/2)

OpenStudy (mimi_x3):

Is it: \[\large \int\limits\frac{1}{x} *\frac{2}{3} x^{\frac{3}{2}} \] ?

OpenStudy (anonymous):

yes

OpenStudy (turingtest):

\[\frac1x\cdot\frac23x^{3/2}=\frac23x^{1/2}\]

OpenStudy (anonymous):

and how exactly did you get that answer? sorry i just needa know for future reference

OpenStudy (mimi_x3):

\[\large \int\limits\frac{2}{3} \sqrt{x} =>\int\limits\frac{2\sqrt{x}}{3} =>\frac{2}{3} \int\limits\sqrt{x} \] The integrate it

OpenStudy (mimi_x3):

Then*

OpenStudy (turingtest):

...with the rule\[\large \int x^ndx=\frac{x^{n+1}}{n+1}+C\]

OpenStudy (anonymous):

ok but i still have a question if you dont mind answering...how did you get from 1/x*(2/3)x^(2/3) to 2/3(x^1/2)??

OpenStudy (anonymous):

like do you just take the intergral of them both seperately or what? im so confused

OpenStudy (turingtest):

algebra\[\large\frac{x^a}{x^b}=x^{a-b}\]\[\frac1x\cdot\frac23x^{3/2}=\frac23\frac {x^{3/2}}x=\frac23x^{3/2-1}=\frac23x^{1/2}\]now integrate\[\frac23\int x^{1/2}dx=\frac23\frac{x^{\frac12+1}}{\frac12+1}=x^{3/2}+C\]

OpenStudy (anonymous):

ohhhhh ok i see now i see! yayy thank you

OpenStudy (turingtest):

welcome :D

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