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Thread: 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
    $\displaystyle
    \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.
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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
    $\displaystyle
    \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.
    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

    $\displaystyle 5\int ln(x)$

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

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

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

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

    Reintroducing the 5:

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


    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
    $\displaystyle
    \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
    $\displaystyle 5ln(x)=xln(x)$
    Solving, you get $\displaystyle x=1,5$
    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 $\displaystyle (1,5)$ and substitute in both of them.
    Let take $\displaystyle x=e$
    substitute this in $\displaystyle y=5ln(x)$ you got 5
    substitute this in $\displaystyle y=xln(x)$ you got e
    since $\displaystyle 5>e$
    Hence, $\displaystyle 5ln(x)>xln(x)$ for all x in $\displaystyle (1,5)$
    So the bigger curve is $\displaystyle y=5ln(x)$ and the smaller curve is $\displaystyle y=xln(x)$

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

    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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