Limit Definition of Indefinite Integrals?

  • #1

Main Question or Discussion Point

Hello,

I was just wondering, we have what could be called the indefinite derivative in the form of d/dx x^2=2x & evaluating at a particular x to get the definite derivative at that x. But with derivation, we can algebraically manipulate the limit definition of a derivative to actually evaluate to 2x from x^2. Is there a similar process available to algebraically manipulate a limit definition of an indefinite integral to get the expected result?

Integration is normally just taught as the reverse of derivation, and while that works of course, I was just curious if there was a way to directly determine the indefinite integral of a function by using limits or differentials.
 

Answers and Replies

  • #2
Stephen Tashi
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A function (if it has a derivative) has a unique derivative. The "indefinite integral" (antiderivative) of a function is not unique. You can apply a definition of integration and obtain a particular indefinite integral of a function f(x). Pick an arbitrary value x = a and compute the area under the curve from x = a to x = b using whatever definition of integration you learned (for example, as the limit of a Riemann sums). The result of this computation gives you a value F(b) that is a function of b. Considering "b" to be a variable itself , F(b) is is an antiderivative of f(x). You could just as well write F(b) as F(x) since the names of variables don't matter when there no relation between the two variables.

Definite Integration is not introduced as the reverse of integration in most calculus texts. After you become familiar with The Fundamental Theorem Of Calculus, you may get the impression that integration is just the reverse of differentiation and forget how definite integration was actually defined.
 
  • #3
I hope I haven't missed anything, but the examples you gave all seem to deal with definite integrals, or if they obtain the indefinite integral, it is implied that derivation rules were applied in reverse (i.e. power rule, d/dx of ln(x)=1/x, etc.). As for the definite integral, if I recall correctly, it was taught starting with Riemann sums so I do get that. I just mean when the indefinite integral was introduced, usually it's just taught as the reverse process of obtaining the "indefinite derivative", just with the addition of a constant.

To add clarity, when I say being able to algebraically manipulate a limit definition to get the indefinite integral, I mean doing so without resorting to the reverse power rule, and familiar reversed rules based on the derivatives of ln(x), sin(x), etc. Am I making any sense here?
 
  • #4
Stephen Tashi
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To add clarity, when I say being able to algebraically manipulate a limit definition to get the indefinite integral, I mean doing so without resorting to the reverse power rule, and familiar reversed rules based on the derivatives of ln(x), sin(x), etc. Am I making any sense here?
If you are asking whether you can begin with the expression [itex] lim_{h\rightarrow 0} \frac{ F(x+h) - F(x)} {h} = f(x) [/itex] and solve it to find [itex] F(x) [/itex] by taking limits with out considering the values of [itex] f(x) [/itex] over an entire interval then I'd say the answer is no, you can't.

You can define [itex] F(x) [/itex] as [itex] F(x) = \int_a^x {f(x) dx} [/itex] and apply the definition [itex] F'(x) = lim_{h\rightarrow 0} \frac{F(x+h) - F(x)}{h} [/itex] You'd be applying a limit definition to a function defined by a definite integral. So the limit is affected the values [itex] f(x) [/itex] takes on an interval.
 
  • #5
If you are asking whether you can...with out considering the values of f(x) f(x) over an entire interval then I'd say the answer is no, you can't.
Yes this is generally what I'm talking about, though I didn't mean to imply any sort of restriction such as not considering values over the entire interval. It just so happens that I kept running into the fact that I know of no way of determining directly the indefinite integral without applying derivation in reverse. I was imagining some sort of sum, maybe using sigma notation, where summing all possible infinitesimal intervals of x times the value of the integrand at that x would yield the indefinite integral.

You can define F(x) F(x) as F(x)=∫xaf(x)dx F(x) = \int_a^x {f(x) dx} and apply the definition F′(x)=limh→0F(x+h)−F(x)h F'(x) = lim_{h\rightarrow 0} \frac{F(x+h) - F(x)}{h} You'd be applying a limit definition to a function defined by a definite integral. So the limit is affected the values f(x) f(x) takes on an interval.
Yeah I remember that now that you mentioned it, but as you mentioned, that defines F(x) as a function of a definite integral so would still implicitly be using reverse derivation to determine the indefinite integral in order to actually evaluate F(x).
 
  • #6
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It just so happens that I kept running into the fact that I know of no way of determining directly the indefinite integral without applying derivation in reverse.
Although it doesn't seem to make sense, we don't call it derivation when we take the derivative. We call it differentiation.

In contrast, completion of the square is used in the derivation of the quadratic formula.
 
  • #7
Stephen Tashi
Science Advisor
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Yeah I remember that now that you mentioned it, but as you mentioned, that defines F(x) as a function of a definite integral so would still implicitly be using reverse derivation to determine the indefinite integral in order to actually evaluate F(x).
No, it wouldn't. The definition of a definite integral does not depend on using reverse diffentiation. You're confusing the technique of doing a definite integral with the definition of a definite integral.
 

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