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Math Help - Area bounded by given curves

  1. #1
    DBA
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    Area bounded by given curves

    Find the area of the region by the given curves.
    y=5ln(x) and y=xln(x)

    This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
    <br />
\int\! u\,dv = uv - \int\! v\,du<br />

    I have no idea how to start here. I am not even sure how the regon should look like.

    Thanks for any help.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by DBA View Post
    Find the area of the region by the given curves.
    y=5ln(x) and y=xln(x)

    This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
    <br />
\int\! u\,dv = uv - \int\! v\,du<br />

    I have no idea how to start here. I am not even sure how the regon should look like.

    Thanks for any help.
    FYI that is the integration by parts formula: Integration by parts - Wikipedia, the free encyclopedia and you're expected to use it on both equations

    5\int ln(x)

    To find the integral of ln(x) note that \ln(x) = 1 \times \ln(x)

    Now to integrate by parts. Since we don't know the integral of ln(x) we must say that u = \ln(x) and so du = \frac{1}{x} dx

    Therefore dv =\, dx giving v = x. Using the formula you posted

    \int \ln(x)\,dx = x\ln(x) - \int \left(x \times \frac{1}{x}\right)\, dx = x\ln(x)-x+k = x(ln(x)-1)+k where k is a constant

    Reintroducing the 5:

    5x[ln(x)-1]+C where C=5k


    Can you do the second integral? It's much the same but with x instead of 1
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  3. #3
    DBA
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    I still do not understand what I am doing. I am looking for one area and if I use the formula on both equations I get 2 answers.
    What does these answers represent?
    And what are the limits of the integrals?

    I tried to do the second integral
    Sorry I am not that good with this LaTex stuff yet, so I write an S for integral
    I set u=lnx, du = 1/x dx and dv = x, v = 1/2x^2
    S xln(x) = 1/2 x^2 * ln(x) - S 1/2 x^2 * 1/x dx

    But I still have a product in my integral? So what do I do now?
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  4. #4
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    Quote Originally Posted by DBA View Post
    Find the area of the region by the given curves.
    y=5ln(x) and y=xln(x)

    This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
    \int\! u\,dv = uv - \int\! v\,du
    " alt="
    \int\! u\,dv = uv - \int\! v\,du
    " />

    I have no idea how to start here. I am not even sure how the regon should look like.

    Thanks for any help.
    First, find the intersection points by setting the two curves equal to each other
    5ln(x)=xln(x)
    Solving, you get x=1,5
    The area of the region = the integral from 1 to 5 for the bigger curve - the smaller curve.
    To know which the is bigger, take any value for x in (1,5) and substitute in both of them.
    Let take x=e
    substitute this in y=5ln(x) you got 5
    substitute this in y=xln(x) you got e
    since 5>e
    Hence, 5ln(x)>xln(x) for all x in (1,5)
    So the bigger curve is y=5ln(x) and the smaller curve is y=xln(x)

    The area = \int_1^5 ( 5ln(x) - xln(x) ) dx = 5 \int_1^5 ln(x) dx - \int_1^5 xln(x) dx

    Now, Solve this by using the integration by parts.
    The final answer = the desired area.

    Here is the final answer:
    http://www.wolframalpha.com/input/?i...x%3D1+to+x%3D5
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