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Math Help - graph under intervals

  1. #1
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    graph under intervals

    I don't even know what to call this - and I am bit baffled by the question:
    Q) explain why the graph of the function f(x)=(4-x)In(1/2*x) lies above the x- aixs for (2<x<4) and below the x-axis for 4<x<6.

    does this mean x= 3 and 5 and you substitute this in equation to see what you get?

    Now I know this- so sorry for posting it. But I have to use this fact that graph is positive 2<x<4 and negative for 4<x<6 and find the area enclosed by the graph anf the x- axis between x=2 and x=6
    Last edited by vidhi96; July 2nd 2011 at 08:01 AM. Reason: I figured out the answer myself- but the extension to this question is still unclear
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  2. #2
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    Re: graph under intervals

    Quote Originally Posted by vidhi96 View Post
    I don't even know what to call this - and I am bit baffled by the question:
    Q) explain why the graph of the function f(x)=(4-x)In(1/2*x) lies above the x- aixs for (2<x<4) and below the x-axis for 4<x<6.
    To be clear the function is f(x) = (4 - x)\ln \left( {\frac{x}{2}} \right).

    If 2<x<4 then (4-x)>0~\&~\ln \left( {\frac{x}{2}} \right)>0.

    But if x>4 then (4-x)<0.
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    Re: graph under intervals

    For f(x) to be positive you know that 4-x \text{  and  } \ln \left(\dfrac{x}{2}\right) must have the same sign (and from the domain x > 0)


    For the graph to be above the x axis then both "parts" of f(x) must have the same sign.

    For 0 < x < 2 then 4-x > 0 \text{  and  } \ln \left(\dfrac{x}{2}\right) < 0

    For 2 < x < 4 then 4-x > 0 \text{  and  } \ln \left(\dfrac{x}{2}\right) > 0

    For 4 < x then 4-x < 0 \text{  and  } \ln \left(\dfrac{x}{2}\right) > 0
    Attached Thumbnails Attached Thumbnails graph under intervals-graph1.png  
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    Re: graph under intervals

    thanks but I am not sure how to calculate area of this function between x=2 and x=6 uisng what we just proved
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    Re: graph under intervals

    hey thanks and I am glad I did the same thing- but I don't know how to calculate the area of the function / graph between x=2 and x=6 - any help will do
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    Re: graph under intervals

    You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by \int_a^b f(x)dx.

    To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
    You should be able to integrate each of those by parts.
    Last edited by HallsofIvy; July 2nd 2011 at 09:12 AM.
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    Re: graph under intervals

    I got the part - as I will use integration by substitution hence 2u=x but I am confused over limit of integration so b=4 and a=2 is that correct?


    Quote Originally Posted by HallsofIvy View Post
    You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by \int_a^b f(x)dx.

    To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
    You should be able to integrate each of those by parts.
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    Re: graph under intervals

    Quote Originally Posted by HallsofIvy View Post
    You posted this in the Calculus section so you should understand that the area under the graph of y= f(x), above the x-axis, for a< x< b is given by \int_a^b f(x)dx.

    To integrate (4- x)ln(x/2), let u= x/2 so that x= 2u and you have (4- 2u)ln(u)= 4ln(u)- 2uln(u).
    You should be able to integrate each of those by parts.
    Hi ya I can't figure it out and it seems to complicated can you please show me thanks
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    Re: graph under intervals

    Let u = \ln \left(\dfrac{x}{2}\right) \text{  and  } v' = 4-x

    Integration by parts it \displaystyle \int u\,dv = uv - \int v\,du


    Spoiler:
    u' = \dfrac{1}{x} and v = 4x-\dfrac{x^2}{2}


    \displaystyle \ln \left(\dfrac{x}{2}\right) \cdot \left(4x-\dfrac{x^2}{2}\right) - \int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx
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    Re: graph under intervals

    Many thanks
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  11. #11
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    Re: graph under intervals

    hey do you mind if I complete the whole thing and send it to you for a check.

    Just as a started Integral of (4x-x^2/2)*1/x dx = (7x^2/4)+c is this correct?
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    Re: graph under intervals

    It would be better to update this thread with your working.

    You can cancel an x in the integral though because we know it cannot be 0 and we're not integrating from 0.

    \int \left[\left(4x-\dfrac{x^2}{2}\right) \cdot \dfrac{1}{x} \right]\ dx = \int \left(\dfrac{4x}{x} -\dfrac{x^2}{2x}\right) \ dx

    = \int \left(4 - \dfrac{x}{2}\right) \dx  = \int 4\, dx - \dfrac{1}{2} \int x\ dx

    I'm not sure where 7/4 comes from and you will need to take care with your signs when you subtract it with the first part of the expression
    Last edited by e^(i*pi); July 2nd 2011 at 02:26 PM. Reason: tex rendering
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    Re: graph under intervals

    that was my error- i worked it all out as this integration was part of big question. I iwll upload entire question once I finish typing it - thnaks again.
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