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Area under y=x^2

  1. Sep 15, 2008 #1
    This is out of my textbook:

    EXAMPLE 6:
    Find the area under the parabola y=x2 from 0 to 1.

    SOLUTION:
    An antiderivative of f(x) = x2 is F(x) = 1/3x3. The required area is found using Part 2 of the Fundamental Theorem....

    My question is: how was the antiderivative obtained?
     
  2. jcsd
  3. Sep 15, 2008 #2
    The second part of the fundamental theorem of calculus says, essentially, that if we have a continuous function f over some interval, [a,b]. Then let F be an antiderivative of f.

    So it follows that:
    f(x) = F'(x)

    -----

    Take the derivative of F(x) and you will indeed obtain f(x). All that happened was the derivative in reverse.
     
  4. Sep 15, 2008 #3
    When you have a variable (let's use x) you can use this formula:

    (1/n+1)x^n+1.
    Read as: "One over n plus one, times the variable, that variable raised to n plus one. "n" is the original power from the problem.
    Example: x^4 would be 1/5 x^5. To prove this, you could find the derivative of my solution. (which would be 5*(1/5) x^5-1 = x^4.
    Hope this helped a little.
     
  5. Sep 15, 2008 #4
    Basically, to find an antiderivative, you have to think up a function that, if you take its derivative, you would get your original function back again.

    Now, the Power Rule of derivative says that if [tex]f(x) = ax^n[/tex], then [tex]f'(x) = anx^{n-1}[/tex]. So use this backwards:

    You have [tex]f(x) = x^2[/tex]. Your task is to find [tex]F(x)[/tex].

    In this case, you the exponent in [tex]f(x)[/tex] is [tex]n-1 = 2[/tex], so the exponent in [tex]F(x)[/tex] is [tex]n = 3[/tex].

    You also know that in [tex]f(x)[/tex], [tex]an = 1[/tex], so [tex]a = 1/3[/tex].

    Now, you can write [tex]F(x) = ax^n = 1/3x^3[/tex].
     
  6. Sep 15, 2008 #5
    Thanks everyone. I actually do understand how to get this particular example's anitderivative. I wanted to know if there was a more general way of achieving this. I can only imagine once f(x) gets a little more complicated, finding the antiderivative could get quite painful (although finding the derivative can also be quite painful :wink:).
     
  7. Sep 15, 2008 #6
    there is four way to antiderivative (as i have known ) ...

    first one is subtution
    second using the table of intgration
    third is by part
    fourth is friction



    excuse my splling
     
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