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Math Help - Calculus Help

  1. #1
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    Calculus Help

    Problem in attachment
    Attached Thumbnails Attached Thumbnails Calculus Help-graph.jpg  
    Last edited by CaptainBlack; February 18th 2006 at 11:44 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by frozenflames
    Graph- horizontal line at y =2 ending at (1,2). Diagonal line from (1,2) to (2,1). Curved line from (2,1) to (3, .5) horizontal line from (3,.5) to (4,.5).

    (EXACT PROBLEM on PROBLEM # 8 http://www.fralibrary.com/teachers/z...ltChoiceII.pdf

    If
    3
    ∫ f (x) dx = 2.3 and F '(x) = f(x), then F(3) - F(0) =
    1

    (A) 0.3 (B) 1.3(C) 3.3(D) 4.3 (E) 5.3
    If I understood what you---and the problem #8--- mean, then,
    ---your description of the graph does not tally with the figure as shown.
    ---the area under the graph from x=1 to x=3 is 2.3 sq.units.
    ---find the area under the graph from x=0 to x=3.
    ---the F'(x) = f(x) is beyond me.

    Okay.
    "INT.(1->3)[f(x)]dx" means the area under the graph of f(x) and above the x-axis from x=1 to x=3.
    It means also F(3) minus F(1).
    And it is equal to 2.3 sq.units as given.

    So if we are to find F(3) minus F(0), which is the area under the graph of f(x) and above the x-axis from x=0 to x=3, then we just need to find the area under the graph from x=0 to x=1 and then add that to 2.3 sq.units.
    Meaning,
    F(3) -F(0) = {F(1) -F(0)} +{F(3) -F(1)}
    F(3) -F(0) = {F(1) -F(0)} +2.3
    F(3) -F(0) = {2*1} +2.3 -------------F(1) -F(0) is a rectangle that is 2units by 1unit.
    F(3) -F(0) = 2 +2.3
    F(3) -F(0) = 4.3 sq.units ----------answer.

    Or, D from the multiple choice is the answer.

    ----------------------------------------------------
    Ooppss, you are in integration now?
    Okay.

    F(1) -F(0) = INT.(0->1)[f(x)]dx
    Per the graph as shown,
    f(x) = 2 -------------(a horizontal line at y=2.)
    from x=0 to x=1,
    So,
    INT.(0->1)[f(x)]dx
    = INT.(0->1)[2]dx
    = [2x](0->1)
    = 2(1) -2(0)
    = 2 -0
    = 2 sq.units.
    Last edited by ticbol; February 18th 2006 at 11:00 AM. Reason: integration way
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  3. #3
    Grand Panjandrum
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    What we need here is the fundamental theorem of calculus, one version
    of which is:

    If F'(x)=f(x), then:

    <br />
\int_a^b f(x) dx=F(b)-F(a)  <br />

    So:

    <br />
F(3)-F(0)=\int_0^3 f(x) dx<br />

    Which can be seen from counting squares is greater than 4 and less than 5
    so the correct answer is (D) 4.3. Another way of arriving at this answer is
    to split the integral into two parts, from 0 to 1 which is clearly 2, and
    from 1 to 3 which we are told is 2.3.

    RonL
    Last edited by CaptainBlack; February 18th 2006 at 11:43 AM.
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