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Math Help - Integration on an Interval

  1. #1
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    Integration on an Interval



    I really don't know where to start with this since theere is no f(t) equation. Please help!
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  2. #2
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    Quote Originally Posted by rawkstar View Post


    I really don't know where to start with this since theere is no f(t) equation. Please help!
    The graph of f(t) is the same as the graph of f(x). They had to change the variable to define G(x)
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  3. #3
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    I suspect it was not the difference between "f(x)" and "f(t)" that was rawkstar's problem. I suspect he simply means that he is not given a formula for f(x).

    And that's the problem with memorizing formulas rather than learning concepts. You don't need to know the formula if you really understand what is being asked- you read the answers off the graph.

    For example, the "concavity" tells how the slope of the graph is changing. A function is "concave upward" as long as the first derivative is increasing.

    a) By the "fundamental theorem of calculus", the first derivative of G(x)= \int_2^x f(t)dt is, simply, f(x). Looking at the graph, f(x) is increasing between -2 and -1, decreasing for x between -1 and 3/2, then increasing again for x between 3/2 and 3.

    b) If the equation of the tangent line to y= G(x) at x= 0 is y= mx+ b, then m is the derivative of G(x) at x= 0 which means it is the value of f(x) at x= 0 which can be read off the graph.

    Of course, the "b" is G(0) and that is the area under the graph of f(x) from -2 to 0. And that is the area of a rectangle of height 2 and base 2 plus the area of a semi- circle of radius 1.

    c) The "average value" of a function f(x) on [a, b] is \frac{\int_a^b f(x) dx}{b-a}.

    You are told that, on 0 to 3, the average value of f(x) is 0- that tells you that \frac{\int_0^3 f(x)dx}{3}= 0 so that \int_0^3 f(x)dx= 0. And \int_0^3 f(x)dx= G(3)- G(0).

    And you have already found G(0) in (b).
    Last edited by HallsofIvy; March 13th 2010 at 12:52 PM.
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    Quote Originally Posted by HallsofIvy View Post

    b) If the equation of the tangent line to y= G(x) at x= 0 is y= mx+ b, then m is the derivative of G(x) at x= 0 which means it is the value of f(x) at x= 0 which can be read off the graph.

    Of course, the "b" is G(0) and that is the area under the graph of f(x) from -2 to 0. And that is the area of a rectangle of height 2 and base 2 plus the area of a semi- circle of radius 1.
    That was a really good explanation but would you please help me understand the second part of b.

    We are given that the equation of the line tangent to the graph of G(x) at the point where x=0 is y = mx + 7. So shouldn't G(0) be 7?
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