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Homework Help: My nightly integral thread:

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm sitting down and doing somewhere around 50 problems this evening for practice, and I just got stuck. I will probably use this one thread to post multiple questions if that is okay.

    2. Relevant equations

    3. The attempt at a solution
    Edited out this problem, it was my mistake.

    Edit: Oh man, I see my mistake..... ( I wrote the problem down wrong)
    Last edited: Jul 7, 2011
  2. jcsd
  3. Jul 7, 2011 #2


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    Where you given a hint to use that substitution?

    If you were integrating x3(2+x4)5, then that would be a good substitution.
  4. Jul 7, 2011 #3
    Yeah, that was how it was supposed to be.

    I u-sub the 2+x^4 and then the dx = du/4x^3 cancels out the other x^3. I read the problem wrong and didn't bother to recheck it lol. Yes it told me to use that substitution. I figured that one out, thank you.
  5. Jul 7, 2011 #4


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    I started a post pursuing the m = u-2 idea, just to see, but it quickly became a nightmare.

    LOL !
  6. Jul 7, 2011 #5
    Yeah, I found that out on my own, then I glanced over and "Oh I wrote a wrong exponent" lol.

    Now I am messing with this one:

    [tex]\int sec(2x)tan(2x)dx[/tex]

    I'm thinking that I should re-write this one in terms of sine/cosine so that:

    [tex]\int \frac{sin(2x)}{cos^{2}(2x)}dx[/tex]
    let u=cos(2x)
    [tex]\int \frac{sin(2x)}{u^{2}(-sin(2x))}du[/tex]
    [tex]- \int \frac{1}{u^{2}}du[/tex]
    [tex]- \int u^{-2} du[/tex]
    and so on.

    edit: I should have just realized that the antiderivative of sectan is sec. Duh!
    Last edited: Jul 7, 2011
  7. Jul 7, 2011 #6


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    Just use u=2x.

    What is the derivative of sec(θ) ?

    Otherwise, let u = cos(2x), that will work too.
  8. Jul 7, 2011 #7

    I see there wasn't really any work involved in this one lol, but I used the u-substitution anyway without thinking.
  9. Jul 7, 2011 #8


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    Sorry, I kept editing my previous post.

    Once you change to sines & cosines, use u = cos(2x).
  10. Jul 7, 2011 #9
    Okay, would someone mind checking this one please?

    [tex]\int x^{3} \sqrt{x^{2}+1} dx[/tex]

    I let [itex]u=x^{2}+1[/itex], such that:

    [tex]\int \frac{x^{3}u^{\frac{1}{2}}}{2x}du[/tex]
    [tex]\int \frac{x^{2}u^{\frac{1}{2}}}{2}du[/tex]

    And so, [itex]x^{2} = u-1[/itex], such that:
    [tex]\int \frac{(u-1)u^{\frac{1}{2}}}{2}du[/tex]

    Which simplifies to:
    [tex]\frac{1}{2} \int u^{\frac{3}{2}} du - \frac{1}{2} \int u^{\frac{1}{2}} du[/tex]

    Taking the indefinite integral yields:


    I believe the solution to be the following:
    [tex]\frac{\sqrt{(x^{2}+1)^{5}}}{5} - \frac{\sqrt{(x^{2}+1)^{3}}}{3} + C[/tex]
    Is it correct, and should I simplify more somehow?
  11. Jul 7, 2011 #10


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    You missed a factor of 1/2.
  12. Jul 7, 2011 #11
    Ah, I see what you are saying. I do that alot in differentiation, I forget to that they are composition functions and don't chain-rule out the argument. I'll re-do that problem, thank you for spotting that! In that case, which should I substitute U for? The argument? u=2x ?
  13. Jul 7, 2011 #12


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    A nice thing about finding an indefinite integral (anti-derivative) is that you can check you answer by taking the derivative of the result.

    Not much of a way to simplify this. You could factor out (1/15)√[(x2+1)3], but why?
  14. Jul 7, 2011 #13


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    When you said that u = cos(2x), then you should get du = -2sin(2x) dx.
  15. Jul 8, 2011 #14
    Thank you sammy, I corrected the above mistake. I am working through some definite integrals at the moment, and I may want to post some questions (I have one about constants and the fundamental theorem), but I am not sure how to make a specific symbol in latex.

    How do you make this symbol (in latex), and what is it actually called?
    Sorry, I'm trying to use my constant PF homework help questions to my advantage by learning latex as I go, and I could not figure out how to find this symbol on the internet, and it's not in the Advanced Edit drop-downs.
    Thank you.
    Last edited: Jul 8, 2011
  16. Jul 8, 2011 #15


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    It should be \right|. It is in the latex reference button under boundaries I believe.
  17. Jul 8, 2011 #16

    I like Serena

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    I think we simply call it "vertical bar" and you can type it as a vertical bar: |. :)

    If you follow rf's suggestion, it will come out bigger, which looks nicer.
    But you will have to put \left. on the left side of your expresion to get proper latex.

    To get the boundaries, you'll type a subscript/superscript combination.
    Like: \right|_a^b.
  18. Jul 8, 2011 #17
    I understand how to do the sub/super script for it. What do you mean when you say "I have to put a \left|" on the left of my equation. Can you please show me an example?
  19. Jul 8, 2011 #18
    \left. f(x) \right|_a^b


    [tex]\left. f(x)\right|_a^b[/tex]
  20. Jul 8, 2011 #19
    Thank you, that example helped.
    Why is one \left. (period) and the other bound \left| (pipe) ?

    Anyway, here is my question for tonight. I have thought about it all day for the most part. I learned about the riemann sum definition of a definite integral, did a bunch of practice problems on the subject, and watched MIT and Khan academy videos on the topic. I am finally satisfied that taking the middle/left/right to be the height of the rectangle makes no difference as n approaches infinity (squeeze theorem et al.).

    Then they throw this one on me.

    [tex]\int_{a}^{b}f(x)dx = \left. F(x) \right|_{a}^{b} = F(b)-F(a)[/tex]

    What? How the heck? This one is simply not intuitive at all for me at all.

    [tex]F(x+\Delta x)=\int_{a}^{x+\Delta x}f(t)dt[/tex]
    Okay sure.

    [tex]F(x+\Delta x) - F(x) = \int_{a}^{x+\Delta x}f(t)dt - \int_{a}^{x+\Delta x}f(t)dt[/tex]

    then they claim:
    "It can be shown that"
    [tex]\int_{a}^{x}f(t)dt + \int_{x}^{x+\Delta x}f(t)dt = \int_{a}^{x+\Delta x}f(t)dt[/tex]
    (The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

    Okay, here is where it breaks down for me. How are they determining that these integrals are the area? I understand that the sum of the areas from the start to the mid point, and from the midpoint to the end, is equal to the area from the start to the end. How does the integral mean area exactly?? I must have missed the part (not really) where they show how the riemann sum definition relates to the integral? I thought that is what they were basically proving here??
  21. Jul 8, 2011 #20
    For every \right there must be a left. Usually, we do something like

    \left( x^2+x \right)

    which yields


    But here we don't want any left-bracket, we want no symbol. That is what \left. is for. So

    \left. x^2+x \right)


    [tex]\left. x^2+x\right)[/tex]

    You see, there is no left bracket there, there is nothing. So, if we want to type

    [tex]\left. f(x)\right|_a^b[/tex]

    we can't just type f(x)\right_a^b since there must be a left delimiter. So we type

    \left. f(x)\right|_a^b

    But we know that


    denotes the (oriented) area under f from a to b. This was the point of the entire Riemann-sum definition. The integral is supposed to be the area!!!!
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