Finding intersection points of x^{2/3} and x^{3/2} for area between curves

[Asphyxiated]

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.

 

[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?

 

[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?

 

[VeeEight]

Your reasoning is correct. Just subtract the two terms so the difference is equal to 0 and factor out x^2/3. Then find the roots.

 

[Asphyxiated]

@VeeEight

I think you read the problem wrong because I can't 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

 

[Squeezebox]

@VeeEight

I think you read the problem wrong because I can't 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.

x^m*xⁿ=x^m+n

 

[VeeEight]

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

 

[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).

 

[VeeEight]

Yes, that is the correct approach.