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Fundamental Theorm of Calc 1 Help Please

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the The Fundamental Theorm of Calculus to Find the following answers

    f(x) =
    0 IF x < -4
    2 IF -4<=x<0
    -3 IF 0<=x<5
    0 IF x>= 5

    g(x) = Integral of f(t) dt Between 4 and x


    Determine the value of each of the following:
    (a) g(−6)=
    (b) g(−3)=
    (c) g(1)=
    (d) g(6)=
    (e) The absolute maximum of g(x) occurs when x= and is the value =





    I get that the derivative of g(x) will be f(x) but I'm stuck after that?
     
  2. jcsd
  3. Jan 19, 2013 #2

    SammyS

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    Re: Fundemental Theorm of Calc 1 Help Please

    That's not supposed to be ... between -4 and x, is it?
    Yes, the derivative of g(x) is f(x). How can you use that information ?

    What do you suppose g(4) is ?

    Can you state The Fundamental Theorem of Calculus ?
     
  4. Jan 19, 2013 #3
    Re: Fundemental Theorm of Calc 1 Help Please

    Yes, sorry my bad its between -4 and x

    g(4) would just be the integral of f(x) and using x = 4 as a substitution

    In this case though, I'm not really sure what f(x) is exactly since it has multiple values or I would just integrate it and sub directly.

    The Fund Theorm states that if f be a continuous real-valued function defined on a closed interval [a, b] then f is differentiable on the open interval(a,b).

    In this case, I realize thats it simple g'(x) = f(x)


    EDIT : I made a graph of f(x) and got it from there, I forgot that the integral of f(x) would simply be the area of the function aswell so it became quite clear on the graph. Thanks, your questions helped me realize it on my own.!
     
    Last edited: Jan 19, 2013
  5. Jan 19, 2013 #4

    SammyS

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    Re: Fundemental Theorm of Calc 1 Help Please

    With the above correction, I would have asked you, 'What is g(-4) ? ' ... with the answer being g(-4) = 0 .
    That should say, g is differentiable on the open interval(a,b).
    Just in case someone comes across this later :

    On the open interval (-∞, -4) , g'(x) = f(x) = 0, and g(-4) = 0 . Therefore, g(x) = 0 for x ≤ -4 .

    On the interval (-4, 0) , g'(x) = f(x) = 2, and g(-4) = 0 . Therefore, g(x) = 2(x - -4) + 0 = 2x + 8 for -4 ≤ x ≤ 0 .

    On the interval (0, 5) , g'(x) = f(x) = -3, and limx→0- g(x) = 8 . Therefore, g(x) = ...
     
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