Integration By Parts and Substitution

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[SOLVED] Integration By Parts and Substitution

Short background; Took Calc 1 my senior year in highschool. Got As all 4 quarters and found it quite easy. Freshman year comes around and I sign up for Calc 2. Turns out the only teacher teaching Calculus 2 for my fall and spring semester is a horrible teacher who doesn't care about his students. So I decided to wait until my Sophomore year to take Calc 2.

So fast forward; I am picking up on all the new techniques and what have you, but I keep forgetting all the old, little essential things because of that year off.

So heres the problem;

[tex]\int\pi^{\sqrt{x}}dx[/tex]

Solve using integration by parts. At the bottom of the problem my professor gave a hint saying "You may like to do a substitution first."

And I'm stumped. I haven't used substitution at all in the past 15 months.

The only substitution I can think of is [tex]{\sqrt{x}}=m[/tex], but that is of no use.

Thanks in advance. =P
 

Answers and Replies

  • #2
morphism
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Well,

[tex]\pi^{\sqrt{x}} = e^{\sqrt{x} \, \ln \pi}[/tex]

And maybe the substitution u=\sqrt{x} from here?
 
  • #3
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Well,

[tex]\pi^{\sqrt{x}} = e^{\sqrt{x} \, \ln \pi}[/tex]

And maybe the substitution u=\sqrt{x} from here?

Played around with it for a while, didn't find anything. Not saying your wrong tho, I'm probably wrong.
 
  • #4
morphism
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I decided to use the substitution [itex]u=\ln(\pi)\sqrt{x}[/itex] instead, but that doesn't really matter. This transforms the integral into

[tex] \frac{2}{(\ln{\pi})^2} \int u e^u du. [/tex]

And from here it's an easy application of integration by parts.
 
  • #5
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I decided to use the substitution [itex]u=\ln(\pi)\sqrt{x}[/itex] instead, but that doesn't really matter. This transforms the integral into

[tex] \frac{2}{(\ln{\pi})^2} \int u e^u du. [/tex]

And from here it's an easy application of integration by parts.

lol thanks for the help. But this is the part I don't remember how to do. Can't figure out how to get from
[tex]\int{\pi^\sqrt{x}}\ dx[/tex]
to
[tex] \frac{2}{(\ln{\pi})^2} \int u e^u du. [/tex]
 
  • #6
Dick
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pi^(sqrt(x))=exp(ln(pi)*sqrt(x)). Then u=ln(pi)*sqrt(x), du=ln(pi)*dx/(2*sqrt(x)). Please check this and then try to continue.
 
  • #7
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pi^(sqrt(x))=exp(ln(pi)*sqrt(x)). Then u=ln(pi)*sqrt(x), du=ln(pi)*dx/(2*sqrt(x)). Please check this and then try to continue.

[tex]u=ln\pi\sqrt{x}[/tex]
[tex]du=\frac{ln\pi dx}{2\sqrt{x}}[/tex]

If I'm not wrong, that substitution would require;

[tex]\int\frac{ln\pi}{2\sqrt{x}} e^\sqrt{x}ln\pi[/tex]

I can account for the [tex]\frac{ln\pi}{2}[/tex] but I don't remember how to get the [tex]\frac{1}{\sqrt{x}}[/tex] before the e in the integral so that you can apply the du.
 
  • #8
Dick
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You are wrong. The integral is exp(u)*dx. Would you please solve for dx in terms of u and du so we can get on with this? Sorry to be snappy, it's getting late.
 
  • #9
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Still drawing a blank.

And I've never seen the term exp(u). I was thinking of it as e^(u) but the more I play with the numbers the more I see I'm wrong.
 
  • #10
morphism
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exp(u) is indeed just e^u.

We have:
du=ln(pi)*dx/(2*sqrt(x))

and

u=ln(pi)*sqrt(x).

It shouldn't be hard to write dx in terms of u and du.
 
  • #11
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Okay, pretty sure I got it. Like I said, I was just forgetting simple little substitution rules.

Thanks.
 
  • #12
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Okay, came up with a solution (I think)

Oh yea, the original problem was from 1 to 4. [tex]\int\pi^{\sqrt{x}}[/tex]

u = m dv = e^m dm
du = dm v = e^m

me^m - [tex]\int[/tex] e^m dm

And from there subtract.

I came to 12.27177549 but this little online integral solver is telling me otherwise. (http://www.solvemymath.com/online_math_calculator/integrals/definite_integral_solver.php [Broken])
 
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  • #13
Dick
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Ok, so u*e^u-e^u is ok. But you've lost all of the multiplicative constants like ln(pi) and 2. Where did they go?
 
  • #14
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Ok, so u*e^u-e^u is ok. But you've lost all of the multiplicative constants like ln(pi) and 2. Where did they go?


Ahhhh. Thank you very much.
 

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