# Integration problem

1. May 21, 2007

### stanners

Sorry this is the first time using this forum.. so I'm not too good with writing out the equations

1. The problem statement, all variables and given/known data
Find the area of the region between the curves f(x) = 3x^2 and g(x) = sqrt(x/3) for 0 <= x <= 1.

2. Relevant equations

3. The attempt at a solution
So.. since g(x) is greater than f(x) from 0 to 1/3, and less from 1/3 to 1.

I set up the integrals

[int. from 0 to 1/3 of g(x) - int. from 0 to 1/3 of f(x)] + [int. from 1/3 to 1 of f(x) - int. from 1/3 to 1 of g(x)]

Is that set up correctly?

Next is where my problem is...
integral of 3x^2 is x^3, but how do I integrate sqrt(x/3) ?

I used the power rule and got [2(x/3)^3/2]/3, but that doesn't seem right, please help me out with this.

Thanks!

2. May 21, 2007

### Kurdt

Staff Emeritus
Your method seems fine but your integration of the square root is slightly out. Try integrating the following and see why you are slightly out.

$$\frac{1}{\sqrt{3}}\times x^{\frac{1}{2}}$$

Last edited: May 21, 2007
3. May 21, 2007

### orb

Yeah, just factor out the 1/sqrt[3] and treat the sqrt[x] on its own.

4. May 22, 2007

### stanners

So would I end up with 1/sqrt(3) * int. (x^3/2) / (3/2) ?

5. May 22, 2007

### VietDao29

Yes, of course, but without the integral sign thingy. :)