# Integrate sqrt problem

1. Sep 6, 2005

### MathGnome

Ok, I've been doing work for about 4 hours straight and I think my brain is fried. I know this is easy, it is just not working in my head.

Anyway, the problem is this:

Integrate the sqrt(4x) + sqrt(4x) on the interval 0 to 1

I get, (8^3/2)/3 + (8^3/2)/3 but apparently this is not right. I'm probably forgetting something I'll hit myself in the head for . Any help though?

Thx,
MathGnome

PS: This is not homework.

2. Sep 6, 2005

### amcavoy

I'm sorry, but I'm having trouble seeing what the problem is. Did you mean $\int_{0}^{1}\left(\sqrt{4x}+\sqrt{4x}\right)dx$? That's what it looks like. If so, you can simplify before integration.

3. Sep 6, 2005

### MathGnome

Yep, that's the one.

$2\int_{0}^{1}\left(\sqrt{4x}\right)dx$ (Is this the simplified form?)

Last edited: Sep 6, 2005
4. Sep 6, 2005

### amcavoy

Alright, you can take the square root of 4. Now you have the following:

$$4\int_0^1\sqrt{x}dx=\frac{8{\sqrt{x}}^3}{3}\right|_0^1=\boxed{\frac{8}{3}}$$

Yes, you simplified correctly.

Last edited: Sep 7, 2005
5. Sep 6, 2005

### MathGnome

Ahh, I see what I did I think. I think I tried to just throw out the 1 and ended up with (8^(3/2))/3

Thanks for the help!
It's hard to stop from taking shortcuts! Argh!

Last edited: Sep 6, 2005