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Math Help - Integral.

  1. #1
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    Integral.

    Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
    below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

    The graph of the function f(x)=(x^2+1) ln⁡(1/2x) is positive above the x-axis and negative below the x-axis.

    Use this fact to find the area enclosed by the graph and the x-axis
    between x = 1 and x = 3, giving your answer to five decimal
    places.

    Can anyone help me.
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  2. #2
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    Re: Integral.

    I can't really read what your question is asking. From the context, I am assuming this is f(x):

    f(x) = (x^2+1)\ln\left(\dfrac{x}{2}\right)

    Is that correct?

    Assuming it is, then \ln 1 = 0 since e^0 = 1. So, for all \dfrac{1}{2} < \dfrac{x}{2} < 1, \ln\left(\dfrac{x}{2}\right)<0 and for all 1<\dfrac{x}{2}<\dfrac{3}{2}, \ln\left(\dfrac{x}{2}\right)>0.

    Also, x^2+1>0 for all real values of x. So, you have a positive times a negative for all values of x where 1 < x < 2 and you have a positive times a positive for all 2<x<3.

    Do you know how to use integration by parts? What method are you using to find the area enclosed by the graph and the x-axis? Are you treating area below the x-axis as "negative" area?
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  3. #3
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    Re: Integral.

    Quote Originally Posted by thomasthetankengine View Post
    Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
    below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

    The graph of the function f(x)=(x^2+1) ln⁡(1/2x) is positive above the x-axis and negative below the x-axis.

    Use this fact to find the area enclosed by the graph and the x-axis
    between x = 1 and x = 3, giving your answer to five decimal
    places.

    Can anyone help me.
    First of all, since you are asked for the AREA, which is a QUANTITY, you need to split the region up into two parts (one where you are above the x axis and one where you are below), and from there set up two integrals. The area is the sum of the MAGNITUDES of these amounts.
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    Re: Integral.

    I understand what i'm being asked now. I'm struggling, however, integrating ln(x/2).
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    Re: Integral.

    Quote Originally Posted by thomasthetankengine View Post
    I understand what i'm being asked now. I'm struggling, however, integrating ln(x/2).
    Let u = x/2. Then \int ln(x/2)~dx = \int ln(u) \cdot (2 du)

    -Dan
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    Re: Integral.

    can anybody help with this?
    I acnt even find any similar examples online. I cant find the area of the curve if there are 2 'functions ' multiplying.

    ln(1/2 x)(x^2+1). I have no idea how to go about this. I can do this type of problem if it was only (x^2 +1) on its own.

    Please help.
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    Re: Integral.

    Do you not have a text book? Every Calculus text I know shows that \int ln(x)dx= (x- 1)ln(x)+ C as a clever example of "integration by parts". And integrating \int ln(x/2)(x^2+ 1)dx can also be done using integration by parts.
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    Re: Integral.

    \int ln(x/2)(x^2+1) = ln(1/2x) * 2x - \int 1/x * 2x
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    Re: Integral.

    my text book is rubbish.
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    Forum Admin topsquark's Avatar
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    Re: Integral.

    An example of integration by parts.

    The underlying idea here is that we may write \int u~dv = uv - \int v~du.

    So let's look at \int (x^2)~ln(x)~dx

    Letting u = ln(x) and dv = x^2~dx (the LHS of the template in u, v above), we may then say that du = \frac{1}{x}~dx and v = \int x^2~dx = \frac{1}{3}x^3

    \int (x^2)~ln(x)~dx = \frac{1}{3} x^3~ln(x) - \int \frac{1}{3}~x^3 \cdot \frac{1}{x}~dx

    The last integral should be easy.

    Sorry about all the reposts. I finally fixed it!

    -Dan
    Last edited by topsquark; June 25th 2014 at 05:43 PM. Reason: Gaah! Not awake right now.
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    Re: Integral.

    Thanks for thAt again. Is the last integral x^3/9 ???
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    Re: Integral.

    Quote Originally Posted by NED1 View Post
    Thanks for thAt again. Is the last integral x^3/9 ???
    Yup.

    If you want some practice, I'd recommend \int x~sin(x)~dx and \int x^2~e^x~dx. (But not in this thread.)

    -Dan
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    Re: Integral.

    I'm really stuck. I have done what you said and this is the result but can't go from here. ∫▒〖(x^2+1)(ln⁡(x/2) )=〗 〖(x〗^2+1)(x-1) ln⁡(x)-∫▒2x xln(x)=〖(x〗^2+1)(x-1) ln⁡(x)-2∫▒〖x(x-1) ln⁡(x) 〗I don't see where i'm going wrong.
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  14. #14
    Forum Admin topsquark's Avatar
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    Re: Integral.

    Quote Originally Posted by thomasthetankengine View Post
    I'm really stuck. I have done what you said and this is the result but can't go from here. ∫▒〖(x^2+1)(ln⁡(x/2) )=〗 〖(x〗^2+1)(x-1) ln⁡(x)-∫▒2x xln(x)=〖(x〗^2+1)(x-1) ln⁡(x)-2∫▒〖x(x-1) ln⁡(x) 〗I don't see where i'm going wrong.
    \int (x^2 - 1)~ln \left ( \frac{x}{2} \right )~dx

    = \int x^2~ln \left ( \frac{x}{2} \right )~dx - \int ln \left ( \frac{x}{2} \right )~dx

    For both integrals, let u = x/2 \implies dx = 2 du. Thus
    = \int (2u)^2~ln(u)~(2~du) - \int ln(u)~(2~du)

    = 8 \int u^2~ln(u)~du - 2 \int ln(u)~du

    Can you take it from here?

    -Dan
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    Re: Integral.

    Quote Originally Posted by topsquark View Post
    \int (x^2 - 1)~ln \left ( \frac{x}{2} \right )~dx

    = \int x^2~ln \left ( \frac{x}{2} \right )~dx - \int ln \left ( \frac{x}{2} \right )~dx

    For both integrals, let u = x/2 \implies dx = 2 du. Thus
    = \int (2u)^2~ln(u)~(2~du) - \int ln(u)~(2~du)


    = 8 \int u^2~ln(u)~du - 2 \int ln(u)~du

    Can you take it from here?

    -Dan
    The integral is that I need to solve is (x^2+1)(ln⁡(x/2)). I have tried to solve it using integration by parts. Let f(x)=x^2+1 and g' (x)=ln⁡(x/2); then f'(x)=2x and g(x) can be solved by the method of integration by substitution. Let u=x/2 then ∫ln⁡(x/2)dx=∫ln⁡(u)(2du)=(x-1) ln⁡(x). I then used the integration by parts and got ∫(x^2+1)(ln⁡(x/2) )=(x^2+1)(x-1) ln⁡(x)-∫2x xln(x)=(x^2+1)(x-1) ln⁡(x)-2∫x(x-1)ln⁡(x)=
    I don't get where the two integrals come in. Is that because you used integration by parts twice? I'm really sorry, I'm getting really confused.
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