Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrals with no limits. (Backwards differientation)

  1. Dec 4, 2004 #1
    Alright, I understand clearly what a integral with no limits is, what it does, etc.

    It is simply backwards differentation (differentation deals with the instanteous slope of a parabola)

    And has no limit (no a to b), thus there must be a constant of integration since two (or more) of the same functions can look the same.

    It looks like:

    I understand that these integrals are backwards differentation, thus backwards slopes?

    This doesn't make any sense, what exactly is their use then?

    I see alot of integrals without limits (They do not deal with area like the other type of integrals) being written down about the paradoxes of space (like on the discovery channel/science channel)

    This makes me even more interested on their use of them! Because I am hopefully going to be a professor in astrophysics! So far I'm 16, and I have a great start! I even tutor kids after school in math, its great.

    I understand the use of integrals with limits , from a to b like used in area, it makes sense! However I don't see the use of integrals without limits (thus with the C constant!)


  2. jcsd
  3. Dec 4, 2004 #2
    Consider this problem: given a function f(x), we wish to find a function F(x) such that F'(x) = f(x). The process is called antidifferentiation and F(x) is called an antiderivative of f(x). This "general" problem arises in various applications.

    Consider this problem:

    I have a particle moving along an x-axis such that its velocity at time t is v(t) = 2t. What is its position function, x(t)? We know from physics that x'(t) = v(t). So we want to find a function x(t) such that x'(t) = 2t. This is an antidifferentiation problem.

    How do we solve it? We use educated guesswork. If you're familiar with differentiation then you'll quickly realize that if x(t) = t2, x'(t) = 2t.

    So, we're finished right?

    Not really. Note that x(t) = t2 + 2, x(t) = t2 - 1, and even x(t) = t2 +1000 qualifies as a solution (since in all of those cases, x'(t) = v(t) ).

    Since we have no additional information about x(t), we write the general solution x(t) = t2 + C.

    Now that took a lot of time! The notation to describe what we did is

    [tex]x(t) = \int v(t)\;dt = \int 2t\;dt = t^2 + C[/tex]

    The notation [itex]\int 2t\;dt[/itex] is used to denote all functions x(t) such that x'(t) = 2t.

    Now, if we have additional information about x(t) we can find the value of C. For example, if we have the information that the particle is at x = 5 when t = 0, then we can quickly conclude that x(t) = t2 + 5.

    Other uses of indefinite integral (the one with upper and lower limit is called definite integral):

    - Suppose we know r(t), the rate of change of a city population at time t. Find the function p(t) that describe the population of the city at time t! (note that p'(t) = r(t) )
    - Suppose we know a(t), the acceleration of a particle moving along x-axis at time t. Find its position function x(t)! (note that x'(t) = v(t) and v'(t) = a(t))

    As a comparison: a definite integral (the one you use to find area) yields a number (in that case the area), while an indefinite integral yields a family of functions (as shown earliear, this family has the form of F(x) + C).

    Oh, and btw, the definite and indefinite itegrals are nicely connected to each other. Let f(x) be continuous on [a, b]. Then it can be proved that:

    [tex]\int_a^b f(x)\;dx = \left[\int f(x)\;dx\right]_a^b[/tex]

    I hope that clears things up :)...
    Last edited: Dec 4, 2004
  4. Dec 5, 2004 #3
    So basically, [tex]\int_a^b f(x)\;dx = \left[\int f(x)\;dx\right]_a^b[/tex]

    Is saying that F(x) dx = [tex]\int_a^b f(x)\;dx[/tex]?

    The first integral with limits, is the area under a curve of [tex]f(x)[/tex] of each piece of the area of the curve ([tex]dx[/tex]).

    But, you can solve the area under the curve with the antiderivative of f(x) dx?


    Is this right?
  5. Dec 5, 2004 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yep. This connection between antidifferentiation and integration is so important, it's called the fundamental theorem of calculus.

    \int_a^b f'(t) \, dt = f(b) - f(a)

    and if f is continuous,

    \frac{d}{dx} \int_a^x f(t) \, dt = f(x)
    Last edited: Dec 5, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook