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Math Help - calculate integral approximations

  1. #1
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    calculate integral approximations

    Hi, I need help on solving this problem.

    I tried to find the antiderivative of this equation first and got:

    -ln / 4x^4

    I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

    So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

    I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

    Thank you.
    Attached Thumbnails Attached Thumbnails calculate integral approximations-help.jpg  
    Last edited by abilitiesz; November 1st 2009 at 10:44 PM.
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  2. #2
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    Quote Originally Posted by abilitiesz View Post
    Hi, I need help on solving this problem.

    I tried to find the antiderivative of this equation first and got:

    -ln / 4x^4

    I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

    So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

    I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

    Thank you.
    Someone may be along in a short while who can make a good guess at what you are asking, but in the mean time post the question as asked and explain where you are having difficulties.

    CB
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  3. #3
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    Oh ok, I edited my first post and put the question along with the equations. Basically, all I'm asking for is if my anti derivative is correct. I know how to solve it after that.
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  4. #4
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    Quote Originally Posted by abilitiesz View Post
    Hi, I need help on solving this problem.

    I tried to find the antiderivative of this equation first and got:

    -ln / 4x^4

    I then tried to find the interbals of the width so I did.. 8-2/5 = 6/5 = 1.2

    So, thus I added 1.2 to the endpoints and got.. 2, 3.2, 4.4, 5.6, 6.8, 8.

    I then plugged it into the equation I solved for to try to find T5, but I keep getting the wrong answer. Is my anti derivative even correct?

    Thank you.
    I will guess from the notation that you're expected to get the approximate value of the integral using the Trapeziodal rule (with n = 5) and the Midpoint rule (with n = 5). There are formulas in which you substitute the stuff you've been given to get the answer. It involves nothing more than simple but tedious arithmetic.

    I don't know why you would be trying to get an anti-derivative when the question is clearly not asking you to.
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  5. #5
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    Quote Originally Posted by abilitiesz View Post
    Hi, I need help on solving this problem.

    I tried to find the antiderivative of this equation first and got:

    -ln / 4x^4
    The anti-derivative of [\ln(x)]^5 is not what you have here. Also you are unlikely be able to find the anti-derivative, which is one of the points of asking you to integrate the function numerically.

    WolframAlpha gives:

    \int \log^5(x) dx = -120 x+x \log^5(x)-5 x \log^4(x)+ 20 x \log^3(x)-60 x \log^2(x)+120 x \log(x)+C

    which can be obtained by repeated integration by parts.

    CB
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    The anti-derivative of [\ln(x)]^5 is not what you have here. Also you are unlikely be able to find the anti-derivative, which is one of the points of asking you to integrate the function numerically.

    WolframAlpha gives:

    \int \log^5(x) dx = -120 x+x \log^5(x)-5 x \log^4(x)+ 20 x \log^3(x)-60 x \log^2(x)+120 x \log(x)+C

    which can be obtained by repeated integration by parts.

    CB
    Since the integrand appears to be \frac{1}{(\log x)^5} the business of attempting to get an anti-derivative is even more dire (and even more clearly not intended) ....
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Since the integrand appears to be \frac{1}{(\log x)^5} the business of attempting to get an anti-derivative is even more dire (and even more clearly not intended) ....

    Not really dire-er, the main difference is that the chain of integrations by parts terminates with \text{li}(x) rather than something more elementary

    CB
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    Not really dire-er, the main difference is that the chain of integrations by parts terminates with \text{li}(x) rather than something more elementary

    CB
    Well, it looks like this question has been \text{li}x\text{ed} into shape ....
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