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Thread: Fund. Theorem Confusion

  1. #1 Fund. Theorem Confusion 
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    Something has me confused and it's probably pretty simple.

    If I find the area under some function between two points a and b, the Fundamental Theorem says it's equal to the difference of the functional values of the 'parent' function valued at the same two points. Say f(x) = x then I'm calling its parent F(x) = (x^2)/2, as an example.

    If the x and y coordinates refer to lengths, like meters, then the area under the curve would be expressed in square meters. But F(b) - F(a) is also equal to this area. But it's coming from the difference of the functional values found on its y axis.

    So how did this y axis get to become units of square meters?

    And then I could do it all over starting with the area under f(x) = x^2 and get square meters as the dimension of my answer, and then find the same answer from F(b) - F(a) from F(x) = (x^3)/3?

    I must have a flaw in my understanding here!


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  3. #2  
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    this is just simple integration really. What you have in a square is very simple just length*breadth to find the area. Take a line and find its slope if it doesnt have a curvature at all then its just a simple slope. However often a line has a rate of change we define as a derivative. Now if we know the rate of change of the curve and the straight line distance between the two points then its is relatively simple to find the area beneath the curve since it is proportional to the rate of change of distance so you can find the length of the curve you cannot find the area of a rectangle without its length and the same applies here you want your length.here is a simple example of the area bounded by two curvesy=x/squareroot(x^2+1) and y=x^4-x
    now by constructing a graph of the two curves its easy to see the lines intesect at approximately 0 an 1.18from x/squareroot(x^2+1)=x^4-x
    Area=lim{1.18+0}[function above]dx
    integrating the first term we use u=x^2+1.Then du=2xdx
    and when x is 1.18 then u is approximately 2.39
    Therefore A=1/2lim{2.39+1}du/squareroot(u)-lim{1.18+0}(x^4-x)dx
    =squareroot(2.39)-1-((1.18^5)/5)+(1.18)^2/2
    =0.785


    Last edited by fiveworlds; March 1st, 2013 at 04:45 PM.
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  4. #3  
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    Quote Originally Posted by Ledger View Post
    Something has me confused and it's probably pretty simple.

    If I find the area under some function between two points a and b, the Fundamental Theorem says it's equal to the difference of the functional values of the 'parent' function valued at the same two points. Say f(x) = x then I'm calling its parent F(x) = (x^2)/2, as an example.

    If the x and y coordinates refer to lengths, like meters, then the area under the curve would be expressed in square meters.
    So far so good. Continue to keep in mind that x and y are lengths, and that y=f(x), not F(x).

    But F(b) - F(a) is also equal to this area. But it's coming from the difference of the functional values found on its y axis.
    That's where you tripped up. F(x) is the antiderivative of f(x). While f(x) has dimensions of length, its antiderivative F(x) has dimensions of length-squared (area).

    So how did this y axis get to become units of square meters?
    The y-axis of f(x) is still length. F(x) is an area.

    And then I could do it all over starting with the area under f(x) = x^2 and get square meters as the dimension of my answer, and then find the same answer from F(b) - F(a) from F(x) = (x^3)/3?
    Yes, if you keep mistaking the antiderivative for the original function, you will have a persistent "off by one" error and spiral off into the nth dimension. So don't do that. HTH.
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  5. #4  
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    also you should note that anti differentiation is also called integration
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  6. #5  
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    Thanks for the response tk. It wasn't actually a trip up as I recognized F(x) as the 'parent' function (result of integrating f(x) or its antiderivative) all along. What is key to my confusion, however, is what you said in the second sentence that, "...its antiderivative F(x) has dimensions of length-squared (area)." This is not clear to me. I mean, it has to be true because what is pulled off of the y axis by doing F(b) - F(a) equals the result of the integration, which is area.

    This gets right at it. If F(x) has area on the y axis then what, length on the x axis? This where it becomes muddy.
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  7. #6  
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    Quote Originally Posted by Ledger View Post
    This gets right at it. If F(x) has area on the y axis then what, length on the x axis? This where it becomes muddy.
    You've said it right -- the abscissa is still x, which has units of length. so F(x) is a running tally of area. F(b) is the area under the original f(x) up to x=b, and F(a) is the area of f(x) up to x=a. The difference between those two areas is the area of the piece between x=a and x=b.

    Hopefully the mud is clearing a bit...if not, post again.
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  8. #7  
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    Quote Originally Posted by tk421 View Post
    Quote Originally Posted by Ledger View Post
    This gets right at it. If F(x) has area on the y axis then what, length on the x axis? This where it becomes muddy.
    You've said it right -- the abscissa is still x, which has units of length. so F(x) is a running tally of area. F(b) is the area under the original f(x) up to x=b, and F(a) is the area of f(x) up to x=a. The difference between those two areas is the area of the piece between x=a and x=b.

    Hopefully the mud is clearing a bit...if not, post again.
    Thanks, F(x) being a "running tally of area" really helped the mud settle.
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  9. #8  
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    Quote Originally Posted by Ledger View Post
    Quote Originally Posted by tk421 View Post
    Quote Originally Posted by Ledger View Post
    This gets right at it. If F(x) has area on the y axis then what, length on the x axis? This where it becomes muddy.
    You've said it right -- the abscissa is still x, which has units of length. so F(x) is a running tally of area. F(b) is the area under the original f(x) up to x=b, and F(a) is the area of f(x) up to x=a. The difference between those two areas is the area of the piece between x=a and x=b.

    Hopefully the mud is clearing a bit...if not, post again.
    Thanks, F(x) being a "running tally of area" really helped the mud settle.
    Glad to hear it.

    --Cheers,
    tk
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