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Math Help - Area under graph question. Integral of ln.

  1. #1
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    Area under graph question. Integral of ln.

    Can't seem to solve this question:

    Sketch the graph of ln(x-3), then, without evaluating the integral, prove that

    3/2ln10 < Integral of lnlx-3l, upper limit = 8, lower limit = 5 < ln125


    Sorry I can't type the inequalities out in proper form ><, and by lx-3l I mean the modulus/absolute value.

    This question falls under the topic of 'area under curve by integration', so I know it has something to do with that, but I just can't figure it out.

    This is the graph with y = ln(x-3), y = 3/2ln10 and y = ln124

    Area under graph question. Integral of ln.-lnx-3.png
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Area under graph question. Integral of ln.

    Hint:

    Area of trapezoid: A_T=\frac{1}{2}\left(B+b \right)h=\frac{3}{2}\left(\ln(2)+\ln(5) \right)=\frac{3}{2}\ln(10)

    Area of rectangle: A_R=bh=3\ln(5)=\ln(125)
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  3. #3
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    Re: Area under graph question. Integral of ln.

    Quote Originally Posted by MarkFL2 View Post
    Hint:

    Area of trapezoid: A_T=\frac{1}{2}\left(B+b \right)h=\frac{3}{2}\left(\ln(2)+\ln(5) \right)=\frac{3}{2}\ln(10)

    Area of rectangle: A_R=bh=3\ln(5)=\ln(125)
    Woah, how did you see that so quickly

    Ahh got it after drawing the whole graph out on paper. Thanks a lot
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Area under graph question. Integral of ln.

    Glad to help. Rectangles and trapezoids are commonly used to approximate definite integrals, so those were the shapes I considered first.
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