i know this is integration by parts but not sure how to actually do it.. $\displaystyle integrate......... 2x sqrt(x) dx$
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Originally Posted by b00yeah05 i know this is integration by parts but not sure how to actually do it.. $\displaystyle integrate......... 2x sqrt(x) dx$ Why do you want to use integration by parts? $\displaystyle \int (2x\sqrt{x}) dx = \int (2 \sqrt{x^5}) dx = \int (2 \cdot x^{\frac52}) dx $ I'll leave the rest for you!
where did you get the 5 from??????
Originally Posted by b00yeah05 where did you get the 5 from?????? Sorry - it's too early in the morning here. Instead of 5 write 3. You should have then: $\displaystyle \int (2 \cdot x^{\frac32}) dx$
sorry..would you mind explaining how you got that?? im assuming you did... 2 . x . sqrt(x) = 2 . sqrt(x^2) then 2 . (x^2)^1/2 correct?
Originally Posted by b00yeah05 sorry..would you mind explaining how you got that?? im assuming you did... 2 . x . sqrt(x) = 2 . sqrt(x^2) then 2 . (x^2)^1/2 correct? no. $\displaystyle x \sqrt{x}$ is not $\displaystyle \sqrt{x^2}$ (how did you get that??) $\displaystyle x \sqrt{x} = x \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}$
Originally Posted by b00yeah05 sorry..would you mind explaining how you got that?? im assuming you did... 2 . x . sqrt(x) = 2 . sqrt(x^2) then 2 . (x^2)^1/2 correct? Not quite: $\displaystyle x = \sqrt{x^2}$ Therefore $\displaystyle x \sqrt{x} = \sqrt{x^2} \cdot \sqrt{x} = \sqrt{x^2 \cdot x} = \sqrt{x^3} = x^{\frac32}$
i see so...the x at the front becomes x^1 = x . x^1/2 therefore......x^1+1/2 = x^3/2
Originally Posted by b00yeah05 i see so...the x at the front becomes x^1 = x . x^1/2 therefore......x^(1+1/2) = x^3/2 Good job - and much easier than the calculations I did. But: Use parantheses!
Originally Posted by Krizalid I think this topic it's pretty useless having this one which is also solved. indeed!
Originally Posted by Krizalid I think this topic it's pretty useless having this one which is also solved. And, in a broader context, there's this one, where the elephant in the room actually gets pointed out.
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