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Integration help

  1. Nov 19, 2015 #1
    Int.png Hi

    Pls can anyone explain how the attached picture was worked out?

    Thanks
     
    Last edited by a moderator: Nov 19, 2015
  2. jcsd
  3. Nov 19, 2015 #2

    SteamKing

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    You should be able to work this integral out by using what you learned in Calc I.

    Treat TC as a constant and take TB to be the variable of integration.

    After you find the antiderivative, substitute the limits and use the rules of logarithms to obtain the final result. That's all there is to it.

    BTW, if you haven't learned this, ##\int \frac{dx}{x}= ln\,x + C##
     
  4. Nov 21, 2015 #3

    HallsofIvy

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    Perhaps you have seen [itex]\int_a^b \frac{dx}{c- x}[/itex]. To integrate that let u= c- x so du= -dx.
     
  5. Nov 24, 2015 #4
    Bit confused as the equation isn;t in the format ##\int \frac{dx}{x}= ln\,x + C## ??
     
  6. Nov 24, 2015 #5
    Let's look at what happens on [itex]\int_a^b \frac{dx}{c- x}[/itex], and you'll do the necessary substitutions later.

    1. If ## c \notin [a,b]##, then you are integrating a continuous function on ##[a,b]##. In this case, there is no discussion needed, and the theory says your integral is equal to ##F(b) - F(a)##, where ##F## is an antiderivative of ##\frac{1}{c-x}##.
      If ## b < c ## then you can choose ##F(x) = - \ln(c-x) ##
      If ## c < a ## then you can choose ##F(x) = - \ln(x - c) ##
      In any case, you can choose ##F(x) = -\ln |c-x| ## and [itex]\int_a^b \frac{dx}{c- x} = F(b) - F(a) = \ln |\frac{ c-a }{c-b}| = \ln \frac{ c-a }{c-b}[/itex]
    2. If ##c\in[a,b]##, then there is a discontinuity at ##x=c##, and it can be shown that ##\frac{1}{c-x}## is not integrable. That's why I think you are in case (1) given the answer.
     
  7. Nov 24, 2015 #6

    SteamKing

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    Just how much integral calculus have you studied?
     
  8. Nov 24, 2015 #7

    Ssnow

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    A primitive for the inverse is the natural logaritm, it is important understand how the module work in the argument of logaritm in relation of your physical quantities...
     
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