AmTran_Bus
  • AmTran_Bus
Can someone help me with this integration by parts?
Mathematics
jamiebookeater
  • jamiebookeater
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AmTran_Bus
  • AmTran_Bus
|dw:1433873040803:dw|
AmTran_Bus
  • AmTran_Bus
|dw:1433873069974:dw|
AmTran_Bus
  • AmTran_Bus
That is supposed to say du= 2r dr

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dumbcow
  • dumbcow
you 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
  • AmTran_Bus
Hum. This problem is from the integration by parts chapter in the Stewart book :(
AmTran_Bus
  • AmTran_Bus
But ok. Lets do it your way! Can you guide me through it?
dumbcow
  • dumbcow
sure ... double check you copied the question correctly though
AmTran_Bus
  • AmTran_Bus
Ok. I did.
dumbcow
  • dumbcow
after substitution you get \[\frac{(u-4) r}{\sqrt{u}} * \frac{du}{2r}\]
dumbcow
  • dumbcow
the "r" cancels leaving \[\frac{1}{2} \int\limits \frac{u-4}{\sqrt{u}} du\]
AmTran_Bus
  • AmTran_Bus
Ok, can you explain the after substitution step? Is u=4+r^2?
dumbcow
  • dumbcow
yes so since u = 4 +r^2 , if you solve for r^2 you get r^2 = u-4 --> r^3 = r^2 *r = (u-4)r
AmTran_Bus
  • AmTran_Bus
Got ya. Thanks.
AmTran_Bus
  • AmTran_Bus
Don't we have to solve for out limits?
AmTran_Bus
  • AmTran_Bus
*our
dumbcow
  • dumbcow
yes but first you must find integral .... then plug in limits at the end
AmTran_Bus
  • AmTran_Bus
dumbcow
  • dumbcow
can you finish the integral? \[\frac{u-4}{\sqrt{u}} = u^{1/2} -4u^{-1/2}\]
phi
  • phi
FYI 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
  • AmTran_Bus
Ok. So the (4+r^2)^(-1/2) was basically put on top now, right?
phi
  • phi
yes, only because I don't like square roots.
phi
  • phi
to integrate \[ (4+r^2)^{-\frac{1}{2} } \] we need 2 r dr
AmTran_Bus
  • AmTran_Bus
So we havae|dw:1433877516461:dw|
AmTran_Bus
  • AmTran_Bus
Im just trying to get to what you did to be sure I get it.
phi
  • phi
so split the r^3 into r^2 and r (and put the r next to the dr)
AmTran_Bus
  • AmTran_Bus
Ok. r^2 *r
phi
  • phi
yes. 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
  • AmTran_Bus
Very nice. So now I can just do the regular U and dv stuff.
phi
  • phi
yes, u= r^2 dv = what is left over
AmTran_Bus
  • AmTran_Bus
Thank you so very much. I still have just a small issue with how you got|dw:1433877820455:dw|
AmTran_Bus
  • AmTran_Bus
Sorry for that mess. And for asking you so much.
AmTran_Bus
  • AmTran_Bus
I get how you brought the sqrt up top.
AmTran_Bus
  • AmTran_Bus
|dw:1433877903541:dw|
phi
  • phi
you don't have to change the square root to an exponent, but I find it easier to think about|dw:1433877934816:dw|
phi
  • phi
\[ \frac{1}{x}= x^{-1} \] \[ \sqrt{x} = x^\frac{1}{2} \] \[ \frac{1}{\sqrt{x}}=\frac{1}{x^\frac{1}{2}}=x^{-\frac{1}{2}} \]
AmTran_Bus
  • AmTran_Bus
Thanks so much phi. I am really glad you showed me how to do it this way.
phi
  • phi
You still have to do the integral
AmTran_Bus
  • AmTran_Bus
I know, from all the U= and DV= stuff. But I'm pretty good with that.

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