Understanding Indefinite Integrals

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Homework Help Overview

The discussion revolves around understanding the concept of indefinite integrals in calculus, particularly in relation to their interpretation and application in finding areas under curves. The original poster expresses confusion about the meaning of indefinite integrals compared to definite integrals, especially in the context of functions representing physical quantities like velocity and displacement.

Discussion Character

  • Conceptual clarification, Exploratory

Approaches and Questions Raised

  • The original poster attempts to relate the concept of indefinite integrals to their understanding of definite integrals and the physical interpretation of area under curves. They question what it means to take an indefinite integral and how it relates to the area under a function over an infinite interval.

Discussion Status

Some participants provide clarifications regarding the nature of indefinite integrals, explaining that they represent a family of functions that give the area under the original function up to any point. This includes a discussion about the relationship between indefinite and definite integrals, as well as the role of the constant of integration.

Contextual Notes

The original poster indicates that they are seeking a deeper understanding of the concept as part of their homework, suggesting that they may have constraints related to their current learning materials or methods of explanation.

rambo5330
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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 I am 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 let's say m/s.

Now with the integral.. let's 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 let's 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 doesn't make sense.. can someone please explain that too me?
 
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anyone?
 
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 f: [a, b] \to \mathbb{R} is an integrable function, and let F(x) = \int_a^x f(t)\,dt. Then we say F is the indefinite integral of f, and abbreviate this as \int f(x)\,dx = F(x) + C. The constant C comes from the fact that our choice of a as the lower limit in the integral is sort of arbitrary, and \int_a^x f(t)\,dt - \int_{a'}^x f(t)\,dt = \int_a^{a'} f(t)\,dt, where the last integral is a constant depending only on a and a'.

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

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