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Calculating Definate Integrals

  1. Jul 10, 2012 #1
    I made a pdf so that the equation would be more clear
     

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  3. Jul 10, 2012 #2
    Calculating the antiderivative and evaluating it on the bounds is almost always how definite integrals are done. The fundamental theorem of calculus gives an exact result in this way. Are you sure you didn't make some mistake either in finding the antiderivative or the Riemann sum?

    Also, did you use LaTeX for that? If so, you can use inline and display math between itex or tex tags. Quote this post to see an example:

    [tex]\int_a^b f(x) \; dx = F(b) - F(a)[/tex]

    Edit: though I should say there's an entire field concerning numerical integration--using Riemann sums or improvements on those techniques--to deal with functions whose antidervatives can't be found analytically.
     
    Last edited: Jul 10, 2012
  4. Jul 10, 2012 #3

    HallsofIvy

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    The only purposes of Riemann sums are to give a basic definition of the Riemann integral and, sometimes, to reason out how to set up an integral. You never use Riemann sums to actually do an integral. Find an anti-derivative, evaluate it at the upper and lower bounds, and subtract.
     
  5. Jul 10, 2012 #4

    arildno

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    Eeh, Riemann sums give the rationale for many numerical integration schemes..
     
  6. Jul 10, 2012 #5

    mathwonk

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    There are two ways to compute an integral "exactly". Either take the actual limit of the Riemann sums, which is usually quite hard, or do as Halls emphasizes is much more usual, namely find an antiderivative. (His "always" means 99.9% of the time.)

    If the antiderivative is elusive or totally unavailable, but usually not otherwise, one may use actual explicit Riemann sums to approximate the integral, as arildno says.

    I.e. in almost all textbook problems, like integrating x^18, or cos^4(x) and so on, do as Halls said to get an "exact" answer. In the odd case where the integrand is 1/sqrt(x^5-x+1), or cos(x^2), do as arildno says to get an approximate answer.

    I believe Halls thought you meant to get an exact answer, so his answer was chosen accordingly.
     
  7. Jul 10, 2012 #6

    HallsofIvy

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    Yes, thank you. I didn't think of that. However, I would be inclined to say that most methods, such as Simpson's rule, can be interpreted as approximating the function, over sections of the interval rather than Riemann sums.
     
  8. Jul 10, 2012 #7

    lurflurf

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    ^The initially surprising thing is that approximating a function over an interval by an interpolating polynomial is equal to a Riemann sum.
     
  9. Jul 10, 2012 #8
    correct me if I'm wrong (I've just started definite integration a few days ago): given an integral

    [itex]\int_a^b[/itex] f(x) dx = F(b) - F(a)

    Also, what If im trying to find the area under a curve, when the curve is completly random, and is based on no function.

    Lastly: do i have to do anything special when some where from point a to point b, there is a negative area.

    Thanks
     
    Last edited: Jul 10, 2012
  10. Jul 10, 2012 #9
    You don't have to do anything special about the function taking on negative values. This is accounted for by the antiderivative.

    "based on no function" makes no sense. If you mean you can't write the function in terms of known elementary functions, that's one thing, but I assume that this curve is still well-behaved--continuous, for instance. Just because you can't write it in terms of polynomials, sines, cosines, or anything like that doesn't make it not a function.
     
  11. Jul 10, 2012 #10
    can you give me an example please?
     
  12. Jul 10, 2012 #11
    wait a second, using

    [itex]\int_a^b f(x) dx = F(b)-F(a)[/itex]

    I don't even need to know the function, just the values at point a and point b, am i correct?

    One more question: given a function, say: f(x)=x2
    to find the area under a curve (a to b) would I need to differentiate:

    [itex]\int_a^b[/itex] 2x dx

    or would I simply plug in the function?

    [itex]\int_a^b[/itex] x2 dx
     
    Last edited: Jul 10, 2012
  13. Jul 11, 2012 #12

    arildno

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    True enough.
    The calculative efficiency of the "strict" Riemann sum is low, relative to more advanced schemes that incorporate function approximation over the intervals as well.
    (Of course, Riemann sums ALSO "approximate" functions over intervals, but usually only well when the intervals are extremely tiny)

    The force of Riemann sums, as I see it, is the ease by which proofs in full generality may be given by means of them.

    They are a bit like Cramer's Rule, dreadful as a computational tool, but optimal for some basic theoretical problems of how to prove something.
     
  14. Jul 11, 2012 #13

    micromass

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    No, Riemann sums are a horrible thing to use in proofs. Many proofs turn out to be very technical and ugly. Proving things like Fubini's theorem is (in my opinion) dreadful. Furthermore, Riemann sums are very restrictive things. Many nice results don't hold for them.

    The good thing about Riemann sums is that the definition is easy and intuitive, but that's basically it.
     
  15. Jul 11, 2012 #14
    If I could ask about three questions:

    1: if I want to calculate the definite integral of say

    f(x)=x2

    would i need to differentiate before i find the integral or would i just plug in my function into the integral?

    [itex]\int_a^b[/itex] x2 dx

    or

    [itex]\int_a^b[/itex] 2x dx

    2: when I use certain rules such as the trapazoid rule:

    [itex]\int_a^b[/itex] f(x) dx = (b-a) [itex]\frac{f(a)+f(b)}{2}[/itex]

    does f(x) equal the derivative of F(x) (given that F'(x)=F(x)), or the function value?
     
  16. Jul 12, 2012 #15

    HallsofIvy

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    What you are doing is integrating not differentiating.

    After integrating, you would not still use the integral sign- that's been done. What you are doing is evaluating [itex]x^2[/itex].
    [itex]\int_a^b 2x dx= \left[x^2\right]_a^b= b^2- a^2[/itex]
     
  17. Jul 12, 2012 #16
    Ok, and that is the area of the function?
     
  18. Jul 12, 2012 #17

    HallsofIvy

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    A "function" is not a geometric object and so does not have an "area". Do you mean the area below a graph of a function?
     
  19. Jul 12, 2012 #18
    Yes, I thought that was one of the main points of definite integration?
     
  20. Jul 12, 2012 #19
    Careful, I think you confused HallsOfIvy. I'm pretty sure you want to integrate [itex]x^2[/itex], do you not? There's no differentiation involved. You just do [itex]\int_a^b x^2 \; dx[/itex].

    You can find the area beneath a graph using a definite integral, but that is not the definition of a definite integral. This is a frequent point of confusion when people try to integrate, say, functions in polar coordinates, where instead of approximating an area as a series of rectangles, one has to use triangles instead. The integral is a tool, but the form that integral takes depends on the geometry of the problem.
     
  21. Jul 12, 2012 #20
    I'm not sure if this is right but, lets say I wanted to find the total acceleration from t1 to t2.

    I could integrate the velocity from t1 to t2? v(t2)-v(t1)

    but if i wanted to find the total work done finding the area under the graph of force from t1 to t2, would give the total work, woulnt it?

    because using F(b)-F(a) would give me Δx(?)

    and finding the area under a graph would give me the average value multiplied by time(?)
     
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