# Integral Question

1. May 30, 2005

### 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.

2. May 30, 2005

### shmoe

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

3. May 30, 2005

### dextercioby

Part integration,maybe...?

Daniel.

4. May 30, 2005

### 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

5. May 30, 2005

### Brad Barker

yep, do this.

it'll be MUCH easier than integration by parts, which you may not even know yet.

6. May 30, 2005

### shmoe

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

7. May 30, 2005

### 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...

Last edited: May 30, 2005
8. May 30, 2005

### shmoe

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.

9. May 30, 2005

### Jameson

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

10. May 30, 2005

### shmoe

Brackets around 5-u in the integral.

11. May 30, 2005

### 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.

12. May 30, 2005

### shmoe

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

13. May 30, 2005

### Jameson

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

Jameson

14. May 30, 2005

### dextercioby

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

Daniel.

15. May 30, 2005

### shmoe

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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