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Integrate sqrt problem

  • Thread starter MathGnome
  • Start date
10
0
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 :cry:. Any help though?

Thx,
MathGnome

PS: This is not homework.
 

Answers and Replies

663
0
I'm sorry, but I'm having trouble seeing what the problem is. Did you mean [itex]\int_{0}^{1}\left(\sqrt{4x}+\sqrt{4x}\right)dx[/itex]? That's what it looks like. If so, you can simplify before integration.
 
10
0
apmcavoy said:
I'm sorry, but I'm having trouble seeing what the problem is. Did you mean [itex]\int_{0}^{1}\left(\sqrt{4x}+\sqrt{4x}\right)dx[/itex]? That's what it looks like. If so, you can simplify before integration.
Yep, that's the one.

[itex]2\int_{0}^{1}\left(\sqrt{4x}\right)dx[/itex] (Is this the simplified form?)
 
Last edited:
663
0
Alright, you can take the square root of 4. Now you have the following:

[tex]4\int_0^1\sqrt{x}dx=\frac{8{\sqrt{x}}^3}{3}\right|_0^1=\boxed{\frac{8}{3}}[/tex]

Yes, you simplified correctly.
 
Last edited:
10
0
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! :redface:
 
Last edited:

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