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Integral Calc. question(s)

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello,

    This is a 2 part question I guess. In general math, you form a sort of equality by doing to one side what you do to the other, meaning if you add 3 to one side, you add 3 to the other. In this example, you have to put a 2 on the right side, but then you put a half on the other, why is that? Why isn't it 2 on both?

    Second question is where did the one fourth come from? (1/4th) on the second line. I thought that the derivative of the function (2 dx) later drops out of the final equation, but here it drops out and a fourth replaces it apparently, how/why did this happen?

    2. Relevant equations



    3. The attempt at a solution

    Its already solved, I just need an explanation of how it came to be.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hello Anon481516! Welcome to PF! :smile:
    no, you do on one side of the = symbol what you do to the other

    in this case, both the 1/2 and the 2 are on the same side of the = symbol :wink:
    from the fourth power :smile:
     
  4. Feb 21, 2013 #3

    SteamKing

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    If you look carefully at the first line, you will see that the left hand integral has been re-written on the right hand side so that you have (1/2)*integral [(2x-5)^3] * 2 dx.

    The derivative of (2x-5) is 2*dx. The 1/2 factor is added so that it cancels when multiplied by the 2*dx.

    Since you have integral [(2x-5)^3] *(1/2)*2dx, then by the integration of power rule, where:

    integral [u^n] du = [1/(n+1)]*u^(n+1)

    1/2*integral [(2x-5)^3]*2dx = 1/2*(1/4)*(2x-5)^4

    Are you sure you didn't recognize this already? Calculus problems are going to get much harder.
     
  5. Feb 22, 2013 #4
    To be honest, no I didnt see it. The missing intermediate step from the first line to the second line threw me off. I forgot to see the change in exponent from 3 to 4, and then that term over 4, which resulted in moving the 4 from the denominator beside the 1/2, which would be 1/4th.

    When you say, and they say, "cancel", does that imply a result of 1? I thought a -2 on the other side would cancel it out...But then that would be -4 if multiplied...So when canceling, it should be 1 and not 0? Seems like a basic question, but may as well ask it.
     
  6. Feb 22, 2013 #5

    SteamKing

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    2*(1/2) = 1
     
  7. Feb 22, 2013 #6

    SammyS

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    That's fine for equations.

    Two things you can do to an algebraic expression without changing its value are:
    1. Multiply the expression by 1 . (That's what is done in the case you're asking about.)

    2. Add zero to the expression. ​
     
  8. Feb 23, 2013 #7

    tiny-tim

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    genius!:smile:
     
  9. Feb 23, 2013 #8
    Alright. But I noticed that there are many rules regarding integrating a function. Some rules look similar to one another, but you can only use one rule at a time, correct? From the previous example, you went from the actual function to 1/2 ∫ (2x-5) ... (2dx), then from there you used the n+1/n+1 rule. Some of these rules can seem and are somewhat basic, but how do you know what rules to use? Do People have a sheet of rules by them when doing these? Once functions use logarithms or square roots, they will involve natural logs and other such equivalents when integrating, then they will get rather complex and I dont see how people could do them from memory alone.
     
  10. Feb 23, 2013 #9

    SammyS

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    In my opinion, the most important skill you need for doing integrations is to be very accomplished at taking derivatives, in particular, being able to use the chain rule several levels deep and also being good with the product rule, as well as with the sum and difference rules. You should know the derivatives of all the basic functions. When I say know I mean you should be able to write them down with out looking them up.

    If you are familiar with integral tables, (Does anybody use these any more?) you will notice that rather than being in the form
    [itex]\displaystyle \int\,x^n\,dx=
    \frac{1}{n+1}x^{n+1}+C[/itex]​
    and
    [itex]\displaystyle \int\,\cos(x)\,dx=
    \sin(x)+C[/itex]​
    they are in the form
    [itex]\displaystyle \int\,u^n\,du=\frac{1}{n+1}u^{n+1}+C[/itex]​
    and
    [itex]\displaystyle \int\,\cos(u)\,du=
    \sin(u)+C\ .[/itex]​

    For the integral in your example, you have (2x-5)3 as the integrand. In this case, u = (2x-5), therefore, du = u'(x)dx = 2dx . (That can also be written, du = d(2x - 5) = 2dx .

    So they multiplied dx by (1/2)∙2, kept the 2 with the dx and move the constant multiplier out of the integral. Then, (2x-5)3 (2dx) becomes (2x-5)3d(2x-5) which is u3du. Now you can use the power rule for the anti-derivative (the integral).
     
  11. Feb 23, 2013 #10
    @SammyS, integral tables are useless (at least for me) because my professor expects us to know how to do all integrals without them.
     
  12. Feb 23, 2013 #11

    SammyS

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    That's hardly the point ... although I suppose you could interpret my statement that way.

    The way the integrals are expressed in the tables is such that by their very nature, substitution is suggested.
     
  13. Feb 23, 2013 #12

    SteamKing

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    The goal in doing the practice problems is to give yourself experience in recognizing basic integral forms. For example, if you have an integrand with an expression raised to a power, then one of the integral forms to consider is int u^n = n*u^n+1. Likewise, if you have int (du/u), then a log will probably be in the result.
     
  14. Feb 25, 2013 #13
    Sammy, I'm not sure I followed that last part you said, "Then, (2x-5)3 (2dx) becomes (2x-5)3d(2x-5) which is u3du."

    Are you saying that the derivative of U3 is (2x-5)? I thought it was just 2. I may be confused by the wording.
     
  15. Feb 25, 2013 #14

    tiny-tim

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    no SammyS is saying that the derivative of U = 2x - 5 is dU = d(2x- 5) = 2dx …
     
  16. Feb 26, 2013 #15
    Alright, I think I understood that. Thanks to everyone who helped.

    I'll probably be back soon with more questions.
     
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