Indefinite integrals

  • Thread starter rambo5330
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  • #1
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This is really a full on homework question but it WILL help me to solve my homework... by helping me fully understand the integral.

So im trying to understand exactly what the indefinte integral means?

heres my train of thought...
if our function F(x) = x2 then its derivative is F'(x) = 2x
and this F'(x) will give you the slope of a tangent line to F(x) at any point x .

If the graph was Distance vs. time this slope would be the velocity in lets say m/s.

Now with the integral.. lets say F(x) now represents velocity vs time...
then i understand if i take the DEFINITE integral on an interval of this graph i will get the area under the curve of F(x)= x^2... and lets say again its m/s vs time (s) then (m/s)(s) = m i.e. velocity x time = displacement.....

so what exactly is happening if i take the indefinite integral? i'm having trouble visiualising this... does is it trying to find the area under the entire function? i,e to infinity.. this doesnt make sense.. can someone please explain that too me?
 

Answers and Replies

  • #3
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It means not "trying to find the area under the entire function", but "trying to give a function which gives the area under the original function up to any point". That is, think about the indefinite integral as a way of abbreviating the function you get by letting the value of the upper limit in a definite integral vary.

Suppose [tex]f: [a, b] \to \mathbb{R}[/tex] is an integrable function, and let [tex]F(x) = \int_a^x f(t)\,dt[/tex]. Then we say [tex]F[/tex] is the indefinite integral of [tex]f[/tex], and abbreviate this as [tex]\int f(x)\,dx = F(x) + C[/tex]. The constant [tex]C[/tex] comes from the fact that our choice of [tex]a[/tex] as the lower limit in the integral is sort of arbitrary, and [tex]\int_a^x f(t)\,dt - \int_{a'}^x f(t)\,dt = \int_a^{a'} f(t)\,dt[/tex], where the last integral is a constant depending only on [tex]a[/tex] and [tex]a'[/tex].

In your example, if you have a velocity function [tex]f(t) = 2t[/tex] (corresponding to a constant acceleration of [tex]2\,\mathrm{m}/\mathrm{s}^2[/tex]) then the corresponding distance-traveled function is [tex]F(t) = \int_0^t f(\tau)\,d\tau = t^2[/tex].
 
  • #4
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thank you very much ! very helpful to me! not even my text books could explain that part as clearly as you.
 

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