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Math Help - fundamental theorem of cal

  1. #1
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    fundamental theorem of cal

    I'm not sure how to begin/approach/do this problem or apply the theorem.
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  2. #2
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    Quote Originally Posted by frog09 View Post
    I'm not sure how to begin/approach/do this problem or apply the theorem.
    You must deduce an equation for f(x) from the given graph. It is the graph of sin(x), shifted 1 place along positive x, and shifted 2 placed along positive y, and squished/expanded by a factor of... something! You must then inject that equation into the integral:

     g(x) = \int_{-3}^{2x-1}f(t)dt

    And carry out the integration to get a function in x for g(x).
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  3. #3
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    Quote Originally Posted by frog09 View Post
    I'm not sure how to begin/approach/do this problem or apply the theorem.
    The second fundemental theorem is

    \frac{d}{dx} \int_a^x f(t)\, dt = f(x)
    or if

    g(x) = \int_a^x f(t)\, dt then g'(x) = f(x)

    However to use this theorem, it must look exactly of this form. What you have is

    g(x) = \int_{-3}^{2x-1} f(t)\, dt

    so use the chain rule. Let u = 2x-1

     \frac{d}{dx} \int_{-3}^{u} f(t)\, dt = \frac{d}{du} \int_{-3}^{u} f(t)\, dt \cdot \frac{d u}{dx}
    Can you continue?
    Last edited by Jester; January 3rd 2009 at 11:46 AM.
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    would  f'(x)=g(x) be true in this case too?

    and how do you use this theorem to find domain of g?
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  5. #5
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    Quote Originally Posted by frog09 View Post
    would  f'(x)=g(x) be true in this case too?

    and how do you use this theorem to find domain of g?
    If g(x)=\int_{u(x)}^{v(x)} f~dx then the domain of g would the values of x that make \int_{u(x)}^{v(x)} f~dx defined.

    EDIT: Typo
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  6. #6
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    ok i think i may have got it...
     f'(x)=g(x)
    so by the fundamental theorem
     f'(x)= f(u) du/dx
    f'(x)= g(2x-1)* 2
    which would mean
     g(x)= (1/2) \int_{-3}^u u du

    u=2x-1
    so when x=1/2, u=-1; and when x=-3, u=-7


    changing the limits would change the domain to [-1, -7]?!
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    Quote Originally Posted by frog09 View Post
    ok i think i may have got it...
     f'(x)=g(x)
    so by the fundamental theorem
     f'(x)= f(u) du/dx
    f'(x)= g(2x-1)* 2
    which would mean
     g(x)= (1/2) \int_{-3}^u u du

    u=2x-1
    so when x=1/2, u=-1; and when x=-3, u=-7


    changing the limits would change the domain to [-1, -7]?!
    If g(x) = \int_{-3}^{2x-1} f(t) \, dt then g' (x)= 2f(2x-1), not  f'(x)=g(x)!

    Now we will use g' (x)= 2f(2x-1), to get an idea of the domain of g(x), i.e. places where g is defined. From your graph f(t) is only defined on

    -3 \le t \le 5 so f(2x-1) is only defined on -3 \le 2x-1 \le 5\;\; \Rightarrow\;\; -1 \le x \le 3

    So in this interval f(2x-1) is defined so  g'(x) is defined and since

    g(x) = \int_{-3}^{2x-1} f(t)\,dt then so is  g(x).
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    just one more question... in order to find where g(x) is a maximum, how would i adjust the f graph in order to look for the max?

    i know a maximum at x will be when it changes from positive to negative at x, but on the f graph, there is no such point.
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  9. #9
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    Quote Originally Posted by frog09 View Post
    just one more question... in order to find where g(x) is a maximum, how would i adjust the f graph in order to look for the max?

    i know a maximum at x will be when it changes from positive to negative at x, but on the f graph, there is no such point.
    The maximum (or minimum) will occur at a critical point, a place where  g' = 0 \; \text{or}\; g' DNE. So

    g'(x) = 2 f(2x-1) and so g'(x) = 0 \; \text{when}\;2 f(2x-1) = 0 and from the graph f(0)=0 which means that x = \frac{1}{2}
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