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

[erik05]

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

$$\int 3x \sqrt{5-2x} dx$$

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.

 

[shmoe]

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

 

[dextercioby]

Part integration,maybe...?

Daniel.

 

[Jameson]

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?

$$\int udv = uv - \int vdu$$

Jameson

 

[Brad Barker]

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.

 

[shmoe]

Carrying out his substitution works just fine. By parts also work of course, but it's not necessary.

 

[Jameson]

Fine, we'll do things your way.

$$\int 3x\sqrt{5-2x}dx$$

$$u = 5 - 2x$$

$$du = -2dx$$

$$-\frac{du}{2} = dx$$

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

$$u = 5 - 2x$$

$$2x = 5 - u$$

$$x = \frac{5-u}{2}$$

So now we have...

$$\int 3x\sqrt{5-2x}dx = -\frac{3}{4} \int (5-u) \sqrt{u} du$$

I don't call that easier than integration by parts...

 

[shmoe]

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.

 

[Jameson]

Ah, I fixed them both. Now it's correct :)

 

[shmoe]

Brackets around 5-u in the integral.

 

[Jameson]

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.

 

[shmoe]

No more substitution is needed to integrate that. Distribute over those brackets and go to town.

 

[Jameson]

Ah, crap. Good point. Thought I had you. Alas, substitution will work. Can't win all of them.

Jameson

 

[dextercioby]

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.

Daniel.

 

[shmoe]

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

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).