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The concept of Integrate

  1. Apr 21, 2007 #1
    I am studying 'Integration'
    These are the questions I have been thinking of, but i still did not get it.

    1.I understood the basic concept of dx and dy
    But I don't know what exactly does dx stand for in the definite and indefinite Integrals.
    [tex]\int x^3\, dx\right)[/tex]

    2. I have read a proof in book about integration product
    [tex]y = u*v [/tex]
    [tex]dy = du*v + u*dv[/tex]
    [tex]\int\, dy\right = \int v\, du\right + \int u\, dv\right[/tex]
    [tex]\int u\, dv\right = uv - \int v\, du\right[/tex]

    I think it might be not serious enought.
    - Why is it without dy? I think it should be [tex]dy/dx = v*du/dx+u*dv/dx[/tex] first.
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2

    Gib Z

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    I like to explain integrals in this manner - you know from your differential calculus class that [tex]\lim_{\Delta x \to 0} \Delta x = dx[/tex]. Basically its a really really tiny change in the x value.

    Now If we had a graph, of say x^3. We graph a section, and cut out a rectangle. The rectangles width will be [itex]\Delta x[/itex] and height will be x^3. But we want an infinitely thin strip so that when we add the strips up, the area is exact. To get an infinitely thin strip, we take the limit of the width to zero, making the width dx.

    So for every strip, the height is x^3, the width is dx, and the integral sign means add them all up.

    For indefinite integrals it just shows what you find the anti derivative with respect to. Eg [itex]\int kx dx[/itex] That tells me k is some constant and I do the integration to the x. If it were a dk, other way around.

    Concerning the proof of the Integration by Parts Formula, you are correct but its easier this way. Start off with the product rule:

    [tex]d(uv)/dx = v*du/dx+u*dv/dx[/tex]

    Take the v*du/dx to the other side.

    [tex]d(uv)/dx - v\frac{du}{dx} = u\frac{dv}{dx}[/tex]

    Now Integrate both sides with respect to x. Basically I mean put an integral sign in front of, and a dx behind, all the terms.

    [tex]\int u\frac{dv}{dx}dx = \int \frac{d(uv)}{dx} dx - \int v\frac{du}{dx} dx[/tex].

    Now the dx's in the denominator and at the end all cancel out.

    [tex]\int u dv = \int 1 d(uv) - \int v du[/tex]

    The Integral of 1, with respect to uv, is just uv. And we have our Rule:

    [tex]\int u dv = uv - \int v du[/tex]
    Last edited: Apr 21, 2007
  4. Apr 21, 2007 #3


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    integrals are imits of areas of rectangles. each rectangle has a heigt and a base lebgth. the f(x) tells you how to measure the height and the dx tells you how to measure the base length.
  5. Apr 21, 2007 #4
    I got it! Thanks so much! :smile:

    But I have two more questions!

    That is why we can solve a problem like that?
    [tex]\int sin(3x)\, dx\right)[/tex]
    [tex]= 1/3 \int sin(3x)\, (3dx)\right)[/tex]

    What does [tex] (3dx)[/tex] tell us? Three times base length? One time base length correlate to [tex]1x [/tex]only?

    And is it right? [tex] \int sin(3x)\, (3dx)\right) = \int sin(3x)\, d3x\right)[/tex]

    What does it mean if it is [tex] d(x^2) [/tex] ?
    Then why am I wrong that if I do that:
    [tex] \int x^3\, dx\right)[/tex]
    [tex] \int x^2\, (xdx)\right)[/tex]
    [tex] \int x^2\, (dx^2)\right)[/tex]
    Last edited: Apr 21, 2007
  6. Apr 21, 2007 #5


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    Yes, because "3" is a constant and can be moved in or out of the integral at will, those are equal. And one reason you might want to do that is to make the subsitution y= 3x. Then dy= 3dx so the integral becomes
    [tex]\frac{1}{3}\int sin(u)du[/itex]

    On the other hand some people like to do it as: u= 3x so du= 3dx and then (1/3)du= dx. Now we have
    [tex]\int sin(u)((1/3)du)= \frac{1}{3}\int sin(u)du[/tex]
  7. Apr 21, 2007 #6

    Gib Z

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    The mistake you are making is that you are thinking that in dx, d is something we are multiplying x by. However that is incorrect, dx just means change in x, limiting to zero. When you did xdx = d(x^2) you are make that mistake.
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