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Math Help - Fundamental Theorem of Calculus

  1. #1
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    Fundamental Theorem of Calculus

    We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.
    I've seen different parts labeled as part one and part two, which are you referring to specifically? how about a specific problem to work through?

    The proof of the theorem should be in your texts, as well as all over the internet. you can look it up. that is really something you go into detail about with advanced calculus though...
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    Quote Originally Posted by JonathanEyoon View Post
    We just went over this in class and I am really confused by it. I know that there are two parts to the theorem and it is part 1 that I am confused by. Can someone please help me understand it? I can easily do the homework problems in the book by simply mimicking the examples but I'm not knowing the WHY which I think is far more important. Any help is much appreciated.
    Is it this one

    \frac{d}{dx}\bigg[\int_a^{x}f(t)dt\bigg]=f(x)

    Or is it

    \int_a^{b}f(x)dx=F(b)-F(a)

    Where F'(x)=f(x)?
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    MmMm... It is this one

    Fundamental Theorems of Calculus -- from Wolfram MathWorld

    where it is stating the second part. Sorry didn't know it gets referred as the second as well.

    I don't really have a problem as such to work with it, it's just I'm not getting what it's saying and simply doing problems by mimicking the examples in my textbook.

    For instance, i don't understand from beginning to end of what the second part is stating.
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  5. #5
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    The first one
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    Quote Originally Posted by JonathanEyoon View Post
    MmMm... It is this one

    Fundamental Theorems of Calculus -- from Wolfram MathWorld

    where it is stating the second part. Sorry didn't know it gets referred as the second as well.

    I don't really have a problem as such to work with it, it's just I'm not getting what it's saying and simply doing problems by mimicking the examples in my textbook.

    For instance, i don't understand from beginning to end of what the second part is stating.
    Ok
    Well its basically saying that once you find the antiderivative of a function, then the definite integral is just the antiderivative evaluated at the upper limit minus the the antiderivative evaluated at the lower limit (assuming hte integral is not improper, don't worry about this part, I jsut wanted a disclaimer)
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    I understand the part of the FTC theorem where it is similar to the Evaluation and Net Change Theorem but I don't understand the other one where f(t) is involved.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I understand the part of the FTC theorem where it is similar to the Evaluation and Net Change Theorem but I don't understand the other one where f(t) is involved.
    That is the second fundamental theorem

    Ok

    The basic reasoning for it is this

    we have that \int_a^{x}f(t)dt where a is a constant and x is a variable

    Then from that we have

    \int_a^{x}f(t)dt=F(x)-F(a) where F'(t)=f(t)

    So now taking the deriative and remebering that snice a is a constant F(a) is also we get

    \frac{d}{dx}\bigg[F(x)-F(a)\bigg]=F'(x)

    and from the earlier assertion

    F'(x)=f(x)


    \therefore\frac{d}{dx}\bigg[\int_a^{x}f(t)dt\bigg]=f(x)
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    Thanks alot fellas! Before you guys disappear, do you mind helping me work through 1 problem using that part of the theorem?


    g(x) = from 0 = lower to x = upper (1 + t^(1/2))dt


    so before I do anything, according to the theorem. This should = 1 + x^(1/2) right?

    So basically we are taking the anti derivative of the function and integrating from 0 to x and then finding the derivative of whatever I get from there? Right?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Thanks alot fellas! Before you guys disappear, do you mind helping me work through 1 problem using that part of the theorem?


    g(x) = from 0 = lower to x = upper (1 + t^(1/2))dt


    so before I do anything, according to the theorem. This should = 1 + x^(1/2) right?

    So basically we are taking the anti derivative of the function and integrating from 0 to x and then finding the derivative of whatever I get from there? Right?
    Your answer is correct, but remember the point of the second fundamental theorem is that you DO NOT have to integrate and differentiate, you can automatically write down the answer
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  11. #11
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    Let f(t) = 1 + t^{\frac{1}{2}}:

    If you're asked to find g'(x) ...

    g(x) = \int_{0}^{x} f(t) dt
    g'(x) = {\color{blue}\frac{d}{dx} \int_{0}^{x} f(t) dt}
    g'(x) = {\color{blue}f(x)} (By the Fundamental Theorem of Calculus).
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  12. #12
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    Thanks alot everyone!
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