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Math Help - Marginal Revenue with definite integral

  1. #1
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    Marginal Revenue with definite integral

    Hello,

    I was given a graph of a marginal revenue function which i have attached. (If for some reason it does not come up, these are main points on the graph (0,100), (500,0) and (1000,-100).)
    The first thing i needed to do is find the equation of the marginal revenue function

    What i have done so far:
    found slope which is -0.2
    y intercept is 100
    Therefore marginal revenue function is R'(x)=--0.2x + 100

    However my prof tells me she wants me to find it as a definite integral with limit 0 to 1000.

    Would I start the integral as
    ∫(100 - 0.2x)dx [0,1000]
    and finish with
    = 100x -0.1x^2 + C [0,1000]
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mellowdano View Post
    Hello,

    I was given a graph of a marginal revenue function which i have attached. (If for some reason it does not come up, these are main points on the graph (0,100), (500,0) and (1000,-100).)
    The first thing i needed to do is find the equation of the marginal revenue function

    What i have done so far:
    found slope which is -0.2
    y intercept is 100
    Therefore marginal revenue function is R'(x)=--0.2x + 100

    However my prof tells me she wants me to find it as a definite integral with limit 0 to 1000.

    Would I start the integral as
    ∫(100 - 0.2x)dx [0,1000]
    and finish with
    = 100x -0.1x^2 + C [0,1000]
    \int_0^{1000}(-0.2x+100)dx=(-\frac{1}{10}x^2+100x)|_0^{1000}

    Now evaluate...

    This being a definte integral, there will be no constant C.
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  3. #3
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    Joined
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    when i evaluates the answer is 0.
    = ∫1000 (100 0.2x) dx
    0
    = (100x 0.1x2)|1000
    0
    = [100(1000) 0.1(1000)2] [100(0) 0.1(0)2]
    =[100000 100000] [0 0]
    = 0-0
    = 0

    Would I conclude from this that the 2 x intercepts are therefore x=0 and x=1000 without graphing it?
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