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Homework Help: Area between curve and axes

  1. Dec 11, 2003 #1
    Find the are between the curve [tex] y=\sqrt{1-x} [/tex] and the coordinate axes
     
  2. jcsd
  3. Dec 11, 2003 #2

    NateTG

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    Do you know integral calculus?
     
  4. Dec 12, 2003 #3

    ShawnD

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    I think it would be f(max) - f(min) where f(x) = (2/3)(1 - x)^(3/2)
     
  5. Dec 12, 2003 #4
    You have to perform the integral
    [tex]\int_{0}^{1}\sqrt{1-x}dx[/tex]
    Try the substitution u=1-x
     
  6. Dec 12, 2003 #5
    First find the domain and range that will give u limits of integration


    Why you need a substitution

    [tex]\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
     
  7. Dec 12, 2003 #6
    I do believe this integral is equivalent to the one I posted. But how does your integral follow from the problem?
     
  8. Dec 12, 2003 #7
    I was just shortening the step which are required for substitutions

    Anyway i will be thinking that way too which u have asked
     
  9. Dec 12, 2003 #8
    In order to do this problem, we usually take the following steps.
    1. Sketch the curve [tex] y=\sqrt{1-x} [/tex] and find out what exactly you need to find.

    2. Find the x-intercept(s) or y-intercept(s).

    3. Write down a definite integral and solve the problem.

    In this case, the x-intercept is 1, so you can find out the area by [tex]\int_{0}^{1}\sqrt{1-x}dx[/tex]

    It should be
    [tex]-\int_{0}^{1}\sqrt{1-x}d(1-x)[/tex]
     
  10. Dec 12, 2003 #9
    Area is positive so in any case it is modulus
    of the integral
     
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