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Math Help - Fundamental Theorom of Calculus

  1. #1
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    Fundamental Theorom of Calculus

    Hi, so I got this problem:
    f(x) = x - 2√x, intervals are: [0,2]

    This is what I did:
    First I turned the square root sign into an exponent
    1. x - 2x1/2

    Then I integrated it
    2. 02(x - 2x1/2)dx

    Then I used the antiderivative
    3. (x2/2) - (4/3)x3/2
    Just letting you know in case if you were wondering about the (4/3), since I know that it would be multiplied by (2/3) since it's the inverse of the exponent, then it was multiplied by the 2 on the outside.

    Then I did f(b) - f(a), which means plugging in 2, then plugging in 0, and subtracting it from each other, though I left out the 0 part because it makes everything 0, so I only plugged in the 2, therefore I did not need to subtract
    4. This is what I got:
    (2)2/2 -(4/3)(2)3/2
    (4/2) - (4/3)√23
    2 - (4/3)√(4*2)
    2 - (4/3) * (2√2)
    2 - (8√2)/3

    I don't really know where to go on from there.

    Someone showed me this (and by the way, it's the correct answer in the back of the book as well)
    Fundamental Theorom of Calculus-calculus-fundamental-43.png
    But I am really confused on how they did those next steps!
    I see the left side, which is f(c)(2-0), and I'm confused what that means, and where the 6 came from on the right side.
    And so on.

    Can someone explain to me or help me find a simplier way for me to understand?

    Thanks!
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  2. #2
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    Re: Fundamental Theorom of Calculus

    i suspect you are not telling us the whole problem, just a part of it. your answer for the definite integral is correct, and matches what you were given:

    2 - 8√2/3 = 6/3 - 8√2/3 = (6 - 8√2)/3

    it looks if you're using an integral version of the mean-value theorem, that is, c is a value in (0,2) where the definite integral of f from 0 to 2 matches a rectangle of height f(c) and width 2 (the width of the interval [0,2]).

    if we define F(x) = \int_a^x f(t)\ dt, for x in [a,b] then by the mean-value theorem, there is a c in (a,b) with:

    f(c) = F'(c) = \frac{F(b) - F(a)}{b - a}.

    that is:

    f(c)(b - a) = \int_a^b f(t)\ dt
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  3. #3
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    Re: Fundamental Theorom of Calculus

    Quote Originally Posted by Deveno View Post
    i suspect you are not telling us the whole problem, just a part of it. your answer for the definite integral is correct, and matches what you were given:

    2 - 8√2/3 = 6/3 - 8√2/3 = (6 - 8√2)/3

    it looks if you're using an integral version of the mean-value theorem, that is, c is a value in (0,2) where the definite integral of f from 0 to 2 matches a rectangle of height f(c) and width 2 (the width of the interval [0,2]).

    if we define F(x) = \int_a^x f(t)\ dt, for x in [a,b] then by the mean-value theorem, there is a c in (a,b) with:

    f(c) = F'(c) = \frac{F(b) - F(a)}{b - a}.

    that is:

    f(c)(b - a) = \int_a^b f(t)\ dt
    Oh sorry!
    I forgot to put in the specifics!
    My bad, this is what we're trying to find:

    Find the value of c guaranteed by the Mean Value Theorem for Integrals for the function over the indicated interval.
    So I can see within that Mean Value Theorom, they multiplied the denominator into F'(C), right?

    And I see now, so that 6 came from the 2, which made it into 6/3 so they're all on the same fraction!

    Thanks

    Then after that, they just divide 2 from both side since the left side had the (2-0) thing right?
    Last edited by Chaim; February 9th 2013 at 10:25 AM.
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