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Visualizing integration

  1. Feb 22, 2012 #1
    Lets say I want to integrate sin from 0 to pi

    The answer is 2

    But how do visualize it in terms of the graph?

    am I summing up the area under the graph?

    So it's like max value of 1 on the y axis
    While the x axis stretches from 0 to 3.14?

    If so, then why does doing the calculus in terms of cos (after integrating) give me the same result?

    What is the reason behind it? What am I doing essentially?

  2. jcsd
  3. Feb 22, 2012 #2


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    Most of what you are saying is almost correct- you can visualize the integral as the signed area below the graph of y= sin(x) and above y= 0 between x= 0 and [itex]x= \pi[/itex]. If the graph is below the y= 0, the "area" is negative.

    But I don't understand your statement "why does doing the calculus in terms of cos (after integrating) give me the same result?". That's not true at all:
    [tex]\int_0^{\pi} sin(x)dx= \left[-\cos(x)\right]_0^{\pi}= -(-1)-(-1)= 2[/tex]

    [tex]\int_0^{\pi} cos(x)dx= \left[sin(x)\right]_0^{\pi}= 0- 0= 0[/tex]

    With y= cos(x), for [itex]\pi/2< \pi< \pi[/itex], one half of the graph is above the y-axis, the other is below and so the two cancel.
  4. Feb 24, 2012 #3
    er no , thats not what i meant

    i mean, if integrating sin from 0 to pi means counting the area underneath the graph, then i will get 2 as the answer right?

    so now my other question is why does doing the same integration, sin from 0 to pi, BUT using the method of

    ∫ sinx = -cos x over 0 to pi, gives me 2 too?

    essentially, why is the integral of sin, -cos?

    and why does summing 0 to pi for -cos = counting the area underneath the graph?
  5. Feb 24, 2012 #4
    Fundamental theorem of calculus.
  6. Feb 24, 2012 #5


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    hi quietrain! :smile:
    because if the area from 0 to x is A(x),

    then A(x + dx) - A(x) is the area of a little almost-rectangle with width dx and height sinx.

    ie approximately A(x + dx) - A(x) = sinx dx,

    or approximately [A(x + dx) - A(x)]/dx = sinx …

    in the limit, dA/dx = sinx :wink:

    (and cos(x+dx) - cosx = 2sin(x + dx/2)sin(dx/2) ~ sinx)
  7. Feb 26, 2012 #6
    "Why" is a question for Religion, not Science.
    "What" and "How" are the province of Science and Mathematics.

    Now your question ..... "What am I doing essentially ?"
    Answer; You are finding the area under the curve by multiplying the equation for the curve by dx and adding all those slim vertical rectangular areas while taking the limit as dx => 0. The limit gives you the most accurate answer possible. So ......

    Integration is Multiplication [in the limit] while one of the multiplicands is changing
    That is why Integration also yields Volume or Work or Distance and thus is so useful.
    Any quantity under change that is calculated by multiplication requires Integration.

    Remember, Calculus is the Mathematics of Change.
  8. Feb 26, 2012 #7


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  9. Feb 26, 2012 #8
    This too:

    http://www.khanacademy.org/video/introduction-to-definite-integrals?topic=calculus [Broken]

    You'll be able to see what it means and its physics application.
    Last edited by a moderator: May 5, 2017
  10. Feb 27, 2012 #9
    alright thanks everyone!
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