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Area of the region S between the graphs of f and g over the interval [0,2]

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the area of the region S between the graph of f and g over the interval [0,2]
    if f(x) = x(x-2) and g(x) = x/2



    2. Relevant equations



    3. The attempt at a solution

    The result is

    a(S) = [tex]\int_0^2 [g(x) - f(x)] dx [/tex] = [tex]\int_0^2 (5x/2 - x^2) dx [/tex] = 5/2 2^2/2 - 2^3/3 = 7/3

    This problem is in page 89 of apostol's vol 1 calculus.

    I dont understad well where did [tex]\int_0^2 (5x/2 - x^2) dx [/tex] come from. Im starting to study integral calculus so this is relatively new to me. Could someone explain to me where did [tex]\int_0^2 (5x/2 - x^2) dx [/tex] come from?, and to explain the rest of the process.

    I would appreciate it thanks in advance.
     
  2. jcsd
  3. Nov 15, 2009 #2
    Did you actually perform the operation g(x)-f(x)? It seems that you are looking at a solution and trying to figure out the problem by following the solution. This isn't the way to do things at all. You need to attempt the problem and struggle with it before you look at the solution.
     
  4. Nov 15, 2009 #3
    I didnt perform the operation im not sure where to start, my guess would be to write

    [tex]\int_0^2 [g(x) - f(x)] dx [/tex] = [tex]\int_0^2 (x/2) - x(x-2) dx [/tex]

    instead of [tex]\int_0^2 (5x/2 - x^2) dx [/tex] then to get the integral of both functions and then to replace the x on each function by 0 and 2 , and then to make the substraction.
     
  5. Nov 15, 2009 #4

    Mark44

    Staff: Mentor

    Did you draw a graph of the region whose area you're trying to find? To get the integrand, you need the typical volume element, whose area [itex]\Delta V[/itex] is [upper y value - lower y value] [itex]\Delta x[/itex]. When you have the graph it's easy to see which function gives you the upper y value and which one gives you the lower y value.
     
  6. Nov 15, 2009 #5
    When someone suggests to do something, just do it!

    g(x)-f(x) = (x/2)-[x(x-2)] = (x/2)-[x2-2x] = (x/2)-x2+2x = 5x/2-x2
     
  7. Nov 15, 2009 #6
    I gotta reread my algebra book though well I got a different result
    (x/2) - x^2 + 2x = [x - 2(x^2 +2x)] /2 = [x - 2x^2 + 4x] / 2 = [5x -2x^2] / 2 ?
     
  8. Nov 15, 2009 #7
    Are you self-studying calculus? Because your algebra skills are a little weird to be studying calculus out of Apostol. Why did you find a common denominator? Can you not notice that your final answer is still 5x/2-x2?
     
  9. Nov 15, 2009 #8
    Well at school I didnt learn a lot of math I dont even recall having done geometry back then. but I have self studied everything from algebra all the way to calculus. I have been having some problems with factorization recently so I guess I will give a look to my algebra book. Im self studying calculus out of Apostol, appart from this I havent needed a lot of helps to get through it.
    Thanks a lot.
     
    Last edited: Nov 15, 2009
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