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AmTran_Bus
 one year ago
Can someone help me with this integration by parts?
AmTran_Bus
 one year ago
Can someone help me with this integration by parts?

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AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433873040803:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433873069974:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1That is supposed to say du= 2r dr

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1you forgot the sqrt around the "4+r^2" actually i dont think you need int by parts here .... just the substitution > u = 4+r^2 du = 2r dr

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Hum. This problem is from the integration by parts chapter in the Stewart book :(

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1But ok. Lets do it your way! Can you guide me through it?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1sure ... double check you copied the question correctly though

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1after substitution you get \[\frac{(u4) r}{\sqrt{u}} * \frac{du}{2r}\]

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1the "r" cancels leaving \[\frac{1}{2} \int\limits \frac{u4}{\sqrt{u}} du\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Ok, can you explain the after substitution step? Is u=4+r^2?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1yes so since u = 4 +r^2 , if you solve for r^2 you get r^2 = u4 > r^3 = r^2 *r = (u4)r

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Don't we have to solve for out limits?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1yes but first you must find integral .... then plug in limits at the end

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.1can you finish the integral? \[\frac{u4}{\sqrt{u}} = u^{1/2} 4u^{1/2}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1FYI if you want to use integration by parts, notice that to integrate \[ (4+r^2)^{\frac{1}{2} } \] Taking that as a hint, write the problem as \[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)\ r\ dr \] and let \[ u= r^2 \\ dv = (4+r^2)\ r\ dr \]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Ok. So the (4+r^2)^(1/2) was basically put on top now, right?

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, only because I don't like square roots.

phi
 one year ago
Best ResponseYou've already chosen the best response.1to integrate \[ (4+r^2)^{\frac{1}{2} } \] we need 2 r dr

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1So we havaedw:1433877516461:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Im just trying to get to what you did to be sure I get it.

phi
 one year ago
Best ResponseYou've already chosen the best response.1so split the r^3 into r^2 and r (and put the r next to the dr)

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes. so it should look like this \[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)^{\frac{1}{2}}\ r\ dr\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Very nice. So now I can just do the regular U and dv stuff.

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, u= r^2 dv = what is left over

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Thank you so very much. I still have just a small issue with how you gotdw:1433877820455:dw

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Sorry for that mess. And for asking you so much.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1I get how you brought the sqrt up top.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433877903541:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1you don't have to change the square root to an exponent, but I find it easier to think aboutdw:1433877934816:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{1}{x}= x^{1} \] \[ \sqrt{x} = x^\frac{1}{2} \] \[ \frac{1}{\sqrt{x}}=\frac{1}{x^\frac{1}{2}}=x^{\frac{1}{2}} \]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1Thanks so much phi. I am really glad you showed me how to do it this way.

phi
 one year ago
Best ResponseYou've already chosen the best response.1You still have to do the integral

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.1I know, from all the U= and DV= stuff. But I'm pretty good with that.
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