# Homework Help: Find the area between two curves - Help finding the limits of integration

1. Apr 30, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

The curves are:

$$f(x)= x^{2/3}$$

and

$$g(x)=x^{3/2}$$

2. Relevant equations

I am assuming that:

$$x^{2/3} = x^{3/2}$$

is going to give me the limits of integration but I don't know how to solve for x on this equation.

Could also put it this way:

$$\sqrt[3]{x^{2}} = \sqrt{x^{3}}$$

but that doesn't help me much..

3. The attempt at a solution

no real attempt because I don't know what to do.

2. Apr 30, 2010

### Squeezebox

I don't know how to do with algebra but, here's a hint:

What two numbers can be raised to any power and still equal themselves?

3. Apr 30, 2010

### Asphyxiated

I would say, 0 and 1, right? because -1 will change signs if raised to an even power so that doesn't fit. If that's right what was the clue that that would be the answer?

4. Apr 30, 2010

### VeeEight

Your reasoning is correct. Just subtract the two terms so the difference is equal to 0 and factor out x2/3. Then find the roots.

5. Apr 30, 2010

### Asphyxiated

@VeeEight

I think you read the problem wrong because I cant factor out the x^(2/3) as the lines are x^3/2 and x^2/3, not the same, or else I could easily solve for x

6. Apr 30, 2010

### Squeezebox

Perhaps this would be useful.

xm*xn=xm+n

7. Apr 30, 2010

### VeeEight

If you have the equation x2/3 - x3/2 = 0, write this as x2/3(something - 1)

8. Apr 30, 2010

### Asphyxiated

@VeeEight

So with:

$$x^{2/3}(x^{5/6}-1) = 0$$

can it be solved for x like so:

$$\sqrt[3]{x^{2}} = 0 \rightarrow x = 0$$

and

$$\sqrt[6]{x^{5}}-1 = 0$$

$$\sqrt[6]{x^{5}}=1 \rightarrow x =1$$

so the limits will be 0 to 1? (I know those are the right limits because when the integral is evaluated from 0 to 1 the answer checks out in the back of the book).

9. Apr 30, 2010

### VeeEight

Yes, that is the correct approach.