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Are integrals approximations?

  1. Apr 5, 2007 #1
    Hey, I'm a math novice. As far as I can tell, we use integrals to solve such problems as the area beneath a curve. If I understand it correctly, integrals use a method where an infinite series of rectangular columns are shot up at the curve to probe its varying slope. These columns, because they are flat on top, have their corners sliced off by the varying curved slope, and integrals are designed to sum and cancel all these extra sliced off column shavings, so that we get a true picture of the varying slope.

    If this is wrong already, please correct me.

    So I was wondering, are integrals perfect at finding the area beneath a curved slope, or are they approximations?

  2. jcsd
  3. Apr 5, 2007 #2
    well I'm no mathemetician but as I understand it, it's an exact value. Since your using an infinite amount of rectangles the error that you get from those rectangle corners are infinitely small.
    Also, you perhaps haven't learned it but, you don't have to even use rectangles with flat tops to get the approximate integral. your rectangle can be a trapezoid to best fit the curve your integrating, or a parabola which will fit very closely to the curve and minimize the error of an approximate integral significantly. But either way you integrate it, when you're doing true integration, the error is infinately small and infact and exact value.
    It's a lot easier to explain with diagrams. hopefully it made sense though.
  4. Apr 5, 2007 #3


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    Integrals are exact. The stuff you're talking about, Riemann sums of tiny rectangles or trapezoids or whatever, are approximations of integrals, not integrals proper.

    - Warren
  5. Apr 5, 2007 #4
    Integrals are exact because we define them as such. The integral shouldn't be understood as "the area under a curve". In effect, it has an analytical meaning before a geometrical one, even though that's contrary to its development. We simply define the area under a curve as the integral, not the opposite. Hence, the integral is only a limit of a summation before anything else. Also, what you are talking about is kind of ambiguous. What is the exact area under a curve? If you are talking about drawing the curve and then use some means to calculate the area physically, then this is not mathematical at all. In mathematics we define singular points, and lines that have thickness 0 and perfect right triangles and etc. In other words, we define areas, and the integral is what we use to do so.
    Last edited: Apr 5, 2007
  6. Apr 5, 2007 #5
    I don't know how else to define area enclosed by general curves but as a definite integral. From this point of view the integral is exact by definition.

    If you are asking from a physics point of view, then every experiment has an error tolerance [tex] \epsilon [/tex]. Then using finitely many rectangles we can approximate the area to within [tex] \epsilon [/tex] of some value; exact has no meaning in physics.
  7. Apr 6, 2007 #6
    All excellent replies, thank you. I can see the folly in my question now. To help me further understand integrals, perhaps you could please provide me of some applications of integrals in the physical world. If you could give me 3 examples in different disciplines or areas, I would be thankful.

  8. Apr 6, 2007 #7
    The integral is used in every natural science because most functions in nature aren't linear. For example, to calculate the distance covered by a uniformly accelerated body, we integrate the acceleration in respect to time twice - once to give us the velocity and twice to give us the covered distance.

    [tex]v = \int {a} dt [/tex]

    [tex]d = \int v dt [/tex]

    or simply

    [tex]d =\int \int a { dt} [/tex]

    Even in simple mechanics, the integral is indispensable. Any definition can be expressed as an integral. For example, work, [tex]W = Fd[/tex] is

    [tex] W = \int F { dd} [/tex]
    Last edited: Apr 6, 2007
  9. Apr 6, 2007 #8


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    Or more precisely;

    [tex]W = \int\vec{F}\cdot d\vec{s}[/tex]

  10. Apr 6, 2007 #9
    There are integrals everywhere in physics. Examples are available in the tens of thousands. Here are some very simple ones:

    One of the (many) representations of Maxwell's laws, which completely determine the electromagnetic field in classical electromagnetism, is

    [tex]\oint_S D \cdot \mbox{d}\vec{a} = q_{f_{\mbox{encl}}}[/tex]

    [tex]\oint_S B \cdot \mbox{d}\vec{a} = 0,[/tex]

    [tex]\oint_C E \cdot \mbox{d}\vec{l} = -\frac{\partial \Phi_B}{\partial t},[/tex]

    [tex]\oint_C H\cdot \mbox{d}\vec{l} = I_{f_{\mbox{encl}}} + \frac{\partial \Phi_D}{\partial t},[/tex]

    where the first two are surface integrals and the last two are contour integrals.

    In Schrodinger's formulation of nonrelativistic quantum mechanics, the normalization condition for a wavefunction [itex]\psi[/itex] is

    [tex]1 = \langle \psi | \psi \rangle = \int_{\mbox{allspace}} \overline{\psi(\vec{r})} \psi(\vec{r}) \mbox{d}^3\vec{r},[/tex]

    and the expectation value of any observable (or, more generally, any operator) [itex]\Omega[/itex] is

    [tex]\langle \psi | \Omega | \psi \rangle = \int_{\mbox{allspace}} \overline{\psi(\vec{r})}(\Omega \psi(\vec{r})) \mbox{d}^3\vec{r}.[/tex]

    For any continuous, real function [itex]f[/itex] defined on the reals an antiderivative of [itex]f[/itex] is given by the definite integral [itex]\int_0^x f(x^\prime) dx^\prime[/itex].
    Last edited: Apr 6, 2007
  11. Apr 6, 2007 #10


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    In order to answer this question, we should first have a look at what we want "area" to be!

    I'll take the Riemann version for non-negative functions over some interval:

    1. The area under some curve should be a real, non-negative number
    2. Areas should have an additive property, i.e, the area over the whole interval ought to equal the sum of the areas of an arbitrary partition of the interval into sub-intervals.
    3. In the special case of a constant function, the area under its curve should be reducible to the area of a rectangle.

    We might perhaps set up other properties of "area" than these, but I'll limit myself to them.

    Now, the first requirement in order for an "area" to exist in the Riemann sense would be that any sequence of rectangle partitions, or Riemann sums converges to the SAME number.
    Functions whose Riemann sums DO have this property is called INTEGRABLE.
    The unique number they converge to is called the "integral" of f over the interval.

    The finite partitions themselves have property 2, and it can be proven that for integrable functions, the integral also has this property, and also property 3 is fulfilled.

    This is the justification for calling the integral of f for the area beneath the graph of f, since the integral is shown to have the properties we desire areas to have.
  12. Apr 6, 2007 #11

    Gib Z

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    I won't go through the exact maths, but heres some useful examples of what the integral could help with.

    1. Finding the area under a curve, as you know.

    2. Given a particles function of velocity, you can work out how far its traveled in a certain amount of time. Given acceleration, velocity and/or displacement as well.

    3. I have a function which gives me the population density of a country at particular places. But I want to know the total population of the country, I integrate and sum all the tiny bits up :)

    4. I might have a continuous curve and I want the average value between A nd B. I can't just take some points and average them, but what I would do is find the area under the curve between the points I want to find the average of, A and B. Now, say I make a rectangle with the same area and same width as the interval, or B-A, then the height of this rectangle is the average value!

    5. Data, who posted above, gave you some very complex integrals in relation to physics. I would recommend you ignore them for now, a bit advanced.
  13. Apr 7, 2007 #12
    While I agree that Chaos shouldn't, at this point, spend any time at all trying to understand what the facts I posted mean, I always think it's fun to see what you could understand given some time and work :wink:.

    If he continues in mathematics and/or physics, none of those will seem nearly so daunting in two or three years.
    Last edited: Apr 7, 2007
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