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

## Homework Statement

The curves are:

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

and

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

## Homework 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..

## The Attempt at a Solution

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

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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?

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?

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.

@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

@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
Perhaps this would be useful.

xm*xn=xm+n

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

@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).

Yes, that is the correct approach.