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Integration check

  1. Apr 27, 2005 #1

    I need to solve for y

    [tex]\int\frac{dy}{y}=k \int dt[/tex]
    [tex]y=e^{kt} +C[/tex]

    is this correct? also, I've always wondered why the integral of [itex]dy[/itex] goes away? the integration of [itex]dx[/itex] becomes x, but what about dy? Also, if I were to integrate x^2 dx... the dx goes away after that integratin as well, why is that?
  2. jcsd
  3. Apr 27, 2005 #2


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    No, you need to add the constant of integration as soon as you have integrated, you will see why.
  4. Apr 27, 2005 #3


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    Could u reformulate your question...?I don't understand "the 'dy' goes away"...:confused:

    It's justa plain simple integration,

  5. Apr 27, 2005 #4
    [tex] \int dy [/tex]

    dy represents an infinitesimally small size of y's. the integral sums an infinite amount of these infinitesimally small y's. This evaluates to y.

    In incorrect algebra this translates to

    [tex] \infty * \left(\frac{y}{\infty}\right) = y [/tex]
  6. Apr 27, 2005 #5
    okay.. here's the question that's bothering me...


    [tex]\int k dx=kx[/tex]
    [tex]\int x^2 dx = \frac{1}{3} x^3 [/tex]

    does this mean that.... [tex]\int x^2 dx =\frac{1}{3} x^2 * x = \frac{1}{3} x^3 [/tex]

    so that the dx turned into the x so that it could multiply the x^2?

    I was never explained why...

    [tex]\int x^n dx = \frac{x^{n+1}}{n+1} [/tex]
    as in... when you add the n+1... I was never told where the dx went.. did the dx become the 1 added to the n? also, why do you have to divide by n+1?
  7. Apr 27, 2005 #6
    then you mean...
    [tex]\int\frac{dy}{y}=k \int dt[/tex]

    also.. for [tex]\int\frac{dy}{y}=k \int dt[/tex]

    why doesnt it become

    since [tex]\int dy = y[/tex] just as [tex]\int dt = t[/tex]
  8. Apr 27, 2005 #7
    This is correct pretty much. Remember the Riemann sum? It's something close to

    [tex] \sum_{i=1}^{\infty} f(x_i)\Delta x_i [/tex]

    One element draws a rectangle of height f(x) and width delta x. I'm sure you remember the approximation techniques drawing several rectangles gives a better approximation, and the sum of an infinite amount of rectangles is exact, well the infinite amount of rectangles becomes the integral. The small width's of [itex] \Delta x [/itex] turn the dx into x.

    [tex] x^{n+1} = x*x^n [/tex]

    Do you see why?

    Take [itex] f(x) = x^n [/tex]

    [tex] \int f(x) dx = \sum_{i=1}^{\infty}(x_i^n)\Delta x_i [/tex]

    The [itex] \Delta x_i [/itex] turn into x, and when multiplied by [itex] x^{n} \mbox{ gives } x^{n+1} [/itex] I dont remember the exact reasoning for the n+1 in the denominator, but it follows from some kind of triange ratio I think. Someone else would have to take that one.
  9. Apr 27, 2005 #8


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    That "dx","dy","dt" is what we call "integration measure".Maybe you should check again your notes on Riemann integration,i think you're unclear what this means...

  10. Apr 27, 2005 #9
    like I said, I was given the formulas but never explained why they do what they do... so I'm clueless and just accepted it for a while now, but I am really going to have a hard time understanding advanced math if I dont understand the basic concept. Could someone please explain the 'dy' and 'dx' concepts? btw, I have not learned summation and will not for another half a year (sry whozum)
  11. Apr 27, 2005 #10
    This is correct. Anytime you see something in the form of [tex]\frac{dy}{dt} = ky[/tex] you should spot that this will integrate to the form of [tex] y = Ce^{kt}[/tex]. This is the general model for exponential growth / decay.

    When you integrate, the "dy" will kind of go away. You would call this integrating with respect to y.

    Do a simple integral like [tex]\int2xdx[/tex] You get x^2 + C, not x*x^2 + C.
    Try your integral again. [tex]\int\frac{dy}{y} = \int{y}^{-1}dy = \ln |y|+ C[/tex]

    Last edited: Apr 27, 2005
  12. Apr 27, 2005 #11
    You can delete your own posts. It's the first box on the 'edit' screen.

    Almost. The only thing that you need to be careful about is to make sure that you understand that the C in the first line is not the same as the C in the second line, and you should typically label them differently (call one [itex]C_1[/itex], for example).

    As well, keep in mind that

    [tex]\int \frac{1}{x} \ dx = \ln x+C[/tex]

    is not quite true. It's really

    [tex]\int \frac{1}{x} \ dx = \ln |x|+C,[/tex]

    and this changes things sometimes.

    Now, to your other questions. For now, when you see

    [tex]\int f(x) \ dx,[/tex]

    you should interpret it just to mean the most general form of the antiderivative of [itex]f(x)[/itex], and the [itex]\int[/itex] and [itex]dx[/itex] as nothing more than notational conveniences (the [itex]dx[/itex] tells you precisely what variable you want an antiderivative with respect to). In particular,

    [tex]\int x^n \ dx = \frac{x^{n+1}}{n+1} + C, \ n \geq 0[/tex]

    is true precisely because the right side is the most general form of an antiderivative of [itex]x^n[/itex] (read: every antiderivative of [itex]x^n[/itex] has this form for some [itex]C[/itex], and every function with this form is an antiderivative).

    Now, there certainly is reasoning behind using this particular notation and you will learn about it when you study differentials in more detail.
    Last edited: Apr 27, 2005
  13. Apr 27, 2005 #12
    you say this... but what about my example....

    the dx turned into an x, it did not just dissapear. Also, Whozum agreed:

    why does the dx turn into a x and the dy goes away?
  14. Apr 27, 2005 #13
    edit: Come to think of it, theres alot more to this than I can properly explain so I'll abstain.
    Last edited: Apr 27, 2005
  15. Apr 27, 2005 #14
    It's my understanding that dy, dx, etc. have no meaning in and of themselves they are merely parts of the symbol [tex]\int\dx[/tex]. But it turns out they can be manipulated algebraicly anyways--that's why the Leibnitz notation is better than the f', f'' stuff. Anyhow, you aren't integrating dx you are integrating 1. [tex]\int1dx[/tex]
  16. Apr 27, 2005 #15
    okay, that makes more sense now...
  17. Apr 27, 2005 #16

    [tex]\int dy = y?[/tex]
  18. Apr 29, 2005 #17

    [tex]\int d(insert anything here) = (insert anything here)[/tex]
  19. Apr 29, 2005 #18

    [tex]\int \ d y = y + K[/tex]
  20. Apr 30, 2005 #19
    The d(some variable) does not turn into that variable and then multiply with the other arguments.

    The way you seem to be doing some integrals is in the form of:

    [tex]\int ax^ndx = \frac{ax^n}{n+1}*x + C[/tex]

    This will correct evaluate any integral in that form where x is not -1, but you may be using the wrong reasoning. The constant turning dy into y or dx into x and then multiplying may work out for integrals using the general power rule, but it does not always work.

    Look at my earlier example of [tex]\int \frac{1}{y} dy[/tex]

    Using your logic, I would do this [tex] \frac{y^{(-1+1)}}{(-1+1)}*y + C[/tex]
    which in this case is incorrect.

    Does that make sense? I hope this helps.

    Last edited: Apr 30, 2005
  21. Apr 30, 2005 #20
    well, a simpler example is just

    [tex]\int e^x \ dx = e^x + C.[/tex]

    Explain that one using this reasoning!

    The reason that the integral I put above is true is precisely because the right side is the most general form of the antiderivative of the integrand, [itex]e^x[/itex] - which is precisely what an indefinite integral is defined to be!
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