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Some antiderivative questions

  1. May 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Q1.
    [​IMG]

    Q2.
    [​IMG]

    3. The attempt at a solution

    I really don't know how to go about either of the two.
    In the second one I know that g(0)=100, g'(0)=-10 and g'(15)=0, but have no idea how you would be able to use these go get g(15)

    :confused: :confused:
     
  2. jcsd
  3. May 5, 2007 #2
    Q1 Simplify the integral using all those basic rules. The integral of a constant times a function is equal to a constant times the integral of the function. And the sum of the integrals is equal to the integral of the sum.
    [tex]
    $ \int_{-1}^{1}(3f(x)-g(x))dx = 3\int_{-1}^{1}f(x) dx - \int_{-1}^{1} g(x) dx
    = 6-\int_{-1}^{1} g(x)\\


    6-\int_{-1}^{1} g(x) = 12\\
    \int_{-1}^{1} g(x) = -6\\ $ [/tex]
     
    Last edited: May 5, 2007
  4. May 5, 2007 #3
    whoops! I posted the wrong question. Thanks anyway for the help!

    Here's the one I was supposed to post..

    [​IMG]
     
  5. May 5, 2007 #4
    [tex] let $ u=x^2,\frac{du}{dx}=2x $\\
    $ xdx =\frac{1}{2}du $\\
    when x =0, u=0\\
    when x =2, u=4\\

    $ \int_0^2 xg(x^2)dx=\int_0^4 \frac{1}{2} g(u) du = \frac{1}{2}*16=8$[/tex]
     
  6. May 5, 2007 #5
    Thanks alot of sorting that one out! :approve:

    Now for that dreaded antidiv. graph..:grumpy:
     
  7. May 5, 2007 #6
    I'll take a guess and say that the answer is 25, the curved section of the graph is basically a distraction. You can find the total change between 0 and 15 by finding the area under the derivative graph for this interval. It's equal to -75. As g(0) = 100 it follows that g(15) = 100 -75 =25.
     
  8. May 5, 2007 #7
    hmmm, I'm really clueless, but the theory behind your answer seems like the most logical/correct understanding I have heard so far.
     
  9. May 5, 2007 #8

    HallsofIvy

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    As gunstar-red said, the "looping" part at the end is irrelevant since you are only asked to go up to x= 15. The graph before that is a straight line from (0, -10) to (15, 0). Although it is harder than what he suggested, you can fairly easily write out the equation for g'(x) and integrate to find g(x).
     
  10. May 5, 2007 #9
    I still get 25 doing it your way HallsofIvy.
    g'(x)=2/3x-10
    g(x)=x^3/3-10x+c
    g(0)=100
    .:100=c
    g(x)=x^3/3-10x+100
    g(15)=75-150+100=25

    I think this is because they are in essence the same process.

    g'(x)=f(x)
    g(x)=F(x)+c when x = 0, g(0)=100
    100=F(0)+c
    c=100-F(0)
    so g(x)=F(x) - F(0) +100
    and g(15) =F(15) - F(0) +100
    [tex] $ g(15) = \int_0^{15} g'(x) dx +100 = -75 +100 = 25$ [/tex]
     
  11. May 5, 2007 #10

    HallsofIvy

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    Staff Emeritus
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    Yes, of course, you get the same thing! And, yes, they are basically the same.
     
  12. May 5, 2007 #11
    HallsofIvy, your method makes alot more sense to me solely because theres more math and less theory!
     
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