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Math Help - Definite integral question

  1. #1
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    Unhappy Definite integral question

    Quote Originally Posted by Chris L T521 View Post
    For this to be a pdf, \int_{-\infty}^{\infty}P\left(x\right)\,dx=1

    Thus, solve the equation \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1 for k.

    Can you take it from here?
    see this question below(integration)They are asking that the integral be evaluated.
    is the procedure just like what we did(integration in the above question?)


    \int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr  ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx
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  2. #2
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    Quote Originally Posted by Pringles View Post
    Quote Originally Posted by Chris L T521 View Post
    For this to be a pdf, \int_{-\infty}^{\infty}P\left(x\right)\,dx=1

    Thus, solve the equation \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1 for k.

    Can you take it from here?
    see this question below(integration)They are asking that the integral be evaluated.
    is the procedure just like what we did(integration in the above question?)


    \int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr  ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx
    It has nothing to do with that.

    You're expected to know and apply the standard form \int \frac{1}{x + a} \, dx = \ln |x + a| (I have omitted the arbitrary constant).
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    It has nothing to do with that.

    You're expected to know and apply the standard form \int \frac{1}{x + a} \, dx = \ln |x + a| (I have omitted the arbitrary constant).
    No disrespect meant, but I am wondering why you don't just include the arbitrary constant in your posts about integration? Surely it's easier and more mathematically rigorous to write "+ C" rather than "(I have omitted the arbitrary constant)".
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  4. #4
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    Quote Originally Posted by Pringles View Post
    Evaluate the integral

    \int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr  ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx


    I would appreciate it if someone would be kind enough to give me the full workings with some explanation.


    Thankyou
    Do you know how expression was split into two?

    \displaystyle \int_{0}^{1}\frac{1}{x^{2}+3x+2}dx=\int_{0}^{1}\fr  ac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx

    The general rule for such integrals is  \displaystyle \int \frac{1}{x+a}dx = \ln{|x+a|} + C . Hence:

    \displaystyle  \int_{0}^{1}\frac{1}{x+1}dx-\int_{0}^{1}\frac{1}{x+2}dx = [\ln{|x+1|}]^1_0 -[\ln{|x+2|}]^1_0

    \displaystyle = [\ln{|1+1|}-\ln{|0+1|}]-[\ln{|1+2|}-\ln{|0+2|}]

    =\displaystyle  \ln{|2|}-\ln{|1|}-\ln{|3|}+\ln{|2|}

    =\displaystyle  2\ln{|2|}-\ln{|3|} since \displaystyle  \ln{|1|} = 0

    =\displaystyle  \ln{|2^2|}-\ln{|3|} since  \displaystyle a\ln{|b|} = \ln{|b^a|}

    =\displaystyle  \ln{|4|}-\ln{|3|}

    =\displaystyle  \ln{|\frac{4}{3}|} since \displaystyle  \ln{|a|}-\ln{|b|}=\ln{|\frac{a}{b}|}
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  5. #5
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    Quote Originally Posted by Mush View Post
    No disrespect meant, but I am wondering why you don't just include the arbitrary constant in your posts about integration? Surely it's easier and more mathematically rigorous to write "+ C" rather than "(I have omitted the arbitrary constant)".
    If the result is to be used by the OP in a definite integral then I omit the constant (and say so).

    If the result is part of an indefinite integral that the OP is trying to solve then I include the constant.

    My reasons for doing this are not too difficult to fathom I think ....
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