How can I solve this integral using substitution and integration by parts?

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Homework Help Overview

The discussion revolves around solving the integral \(\int 3x \sqrt{5-2x} dx\) using substitution and integration by parts. Participants explore different methods to approach the problem, particularly focusing on substitution techniques.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to use substitution with \(u = 5 - 2x\) but encounters difficulties in expressing \(3xdx\) in terms of \(du\). Some participants suggest expressing \(x\) in terms of \(u\) to facilitate the substitution.
  • Others propose considering integration by parts as an alternative method, questioning whether it is necessary given the substitution approach.
  • There is a discussion about the relative difficulty of the two methods, with some participants expressing differing opinions on which is easier.
  • Concerns are raised about potential errors in the algebraic manipulation during the substitution process.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to proceed with the substitution. There is no explicit consensus on the preferred method, as both substitution and integration by parts are being explored. Some participants have noted that further substitution may not be necessary to evaluate the integral.

Contextual Notes

Participants are navigating the complexities of the integral and discussing the implications of their chosen methods. There is an acknowledgment that the integral may require more than one step to evaluate, and the discussion reflects varying levels of familiarity with integration techniques.

erik05
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I seem to be having trouble on this one integral.Any help would be much appreciated.

[tex]\int 3x \sqrt{5-2x} dx[/tex]

I'm assuming the substitution rule applies to this so I have u=5-2x.
Then du=-2dx

And now I'm stuck. How can I get 3xdx to equal -2dx? Please and thanks.
 
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Your substitution is good, try writing x in terms of u to deal with the "x" in 3xdx
 
U-substitution won't work in the normal way of " u =, du = " ... Like Daniel said, you'll need to do integration by parts. Are you familiar with this method?

[tex]\int udv = uv - \int vdu[/tex]

Jameson
 
shmoe said:
Your substitution is good, try writing x in terms of u to deal with the "x" in 3xdx

yep, do this.

it'll be MUCH easier than integration by parts, which you may not even know yet.
 
Carrying out his substitution works just fine. By parts also work of course, but it's not necessary.
 
Fine, we'll do things your way.

[tex]\int 3x\sqrt{5-2x}dx[/tex]

[tex]u = 5 - 2x[/tex]

[tex]du = -2dx[/tex]

[tex]-\frac{du}{2} = dx[/tex]

That takes care of the second term, now for the first.

[tex]u = 5 - 2x[/tex]

[tex]2x = 5 - u[/tex]

[tex]x = \frac{5-u}{2}[/tex]

So now we have...

[tex]\int 3x\sqrt{5-2x}dx = -\frac{3}{4} \int (5-u) \sqrt{u} du[/tex]

I don't call that easier than integration by parts...
 
Last edited by a moderator:
Jameson said:
I don't call that easier than integration by parts...

The relative difficulty is a subjective thing. I object to telling the OP that he'd need integration by parts when his substitution idea is good.

By the way, you lost a pair of brackets in your last line and also a minus sign went missing when you solved for x.
 
Ah, I fixed them both. Now it's correct :)
 
  • #10
Brackets around 5-u in the integral.
 
  • #11
Yeah, fixed 'em. Thanks.

But my point is, even after all of that subsitution, the integral still can't be evaluated in one step. More substitution is needed.
 
  • #12
No more substitution is needed to integrate that. Distribute over those brackets and go to town.
 
  • #13
Ah, crap. Good point. Thought I had you. Alas, substitution will work. Can't win all of them.

Jameson
 
  • #14
shmoe said:
Carrying out his substitution works just fine. By parts also work of course, but it's not necessary.

I didn't claim it was necessary for this example,just gave a suggestion,an advice.He could take or not. :wink:

Daniel.
 
  • #15
dextercioby said:
I didn't claim it was necessary for this example,just gave a suggestion,an advice.He could take or not. :wink:

Of course! The necessary part wasn't aimed at you. Both methods are worth a stab, and proving they give the same answer is a nice bit of algebra practice (they look a bit different).
 

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