# Integration problem

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

## Homework Statement

Find the area of the region between the curves f(x) = 3x^2 and g(x) = sqrt(x/3) for 0 <= x <= 1.

## 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!

Kurdt
Staff Emeritus
Gold Member
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:
Yeah, just factor out the 1/sqrt and treat the sqrt[x] on its own.

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

VietDao29
Homework Helper
So would I end up with 1/sqrt(3) * int. (x^3/2) / (3/2) ?

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