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

  • #1
264
0

Homework Statement



The curves are:

[tex] f(x)= x^{2/3} [/tex]

and

[tex] g(x)=x^{3/2} [/tex]

Homework Equations



I am assuming that:

[tex] x^{2/3} = x^{3/2} [/tex]

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:

[tex] \sqrt[3]{x^{2}} = \sqrt{x^{3}} [/tex]

but that doesn't help me much..

The Attempt at a Solution



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

Answers and Replies

  • #2
57
0
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
264
0
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
614
0
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
264
0
@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
57
0
@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
 
  • #7
614
0
If you have the equation x2/3 - x3/2 = 0, write this as x2/3(something - 1)
 
  • #8
264
0
@VeeEight

So with:

[tex] x^{2/3}(x^{5/6}-1) = 0 [/tex]

can it be solved for x like so:

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

and

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

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

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
614
0
Yes, that is the correct approach.
 

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