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Finding the Relationship between the Limit Method and Direct Integral Method for Area

by vanmaiden
Tags: area, curve, direct integral, limit
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vanmaiden
#1
Nov30-11, 09:36 PM
P: 101
1. The problem statement, all variables and given/known data
I am in the process of studying integration and finding the areas under curves. So far, I know of two methods of finding the area under a curve: the limit method and the direct integral method. Could someone explain the relationship between these two methods?


2. Relevant equations
[itex]\int[/itex]f(x) dx = F(x)|[itex]^{b}_{a}[/itex] = F(b) - F(a) = Area

[itex]lim_{n→∞}[/itex] [itex]\sum^{n}_{i = 1}[/itex] [itex]f(x_{i})[/itex]Δx = Area

3. The attempt at a solution
I noticed in the direct integration method for finding the area under a curve that the area under the curve is equal to the change in y of a more complicated function: the integral. I graphed it out on my calculator and I don't see exactly how this works.

[itex]lim_{n→∞}[/itex] [itex]\sum^{n}_{i = 1}[/itex] [itex]f(x_{i})[/itex]Δx = Δy of F(x) = Area

I'm trying to seek an explanation as to why the limit method yields the same result as the direct integral method.
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LCKurtz
#2
Nov30-11, 10:13 PM
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That is the fundamental theorem of calculus. You might start by reading here:

http://en.wikipedia.org/wiki/Fundame...em_of_calculus
vanmaiden
#3
Dec3-11, 03:08 AM
P: 101
Quote Quote by LCKurtz View Post
That is the fundamental theorem of calculus. You might start by reading here:

http://en.wikipedia.org/wiki/Fundame...em_of_calculus
Ah yes, I read through a bit of it. I'm rather confused on the proof for the First Fundamental Theorem of Calculus where it is
F(x) = [itex]\int^{x}_{a}f(t) dt[/itex]
I've just never seen an antiderivative represented in this way before. Could you interpret this for me? Why does an antiderivative have an upper and lower bound?

LCKurtz
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Dec3-11, 12:07 PM
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Finding the Relationship between the Limit Method and Direct Integral Method for Area

Quote Quote by vanmaiden View Post
Ah yes, I read through a bit of it. I'm rather confused on the proof for the First Fundamental Theorem of Calculus where it is
F(x) = [itex]\int^{x}_{a}f(t) dt[/itex]
I've just never seen an antiderivative represented in this way before. Could you interpret this for me? Why does an antiderivative have an upper and lower bound?
Let's say you have a function f(x) and its antiderivative F(x) so you might have written [tex]F(x)=\int f(x)\, dx + C[/tex] where F'(x) = f(x). If you were going a definite integral you would write [tex]\int_a^b f(x)\,dx = (F(x)+C)|_a^b = F(b) - F(a)[/tex] and the C is usually omitted since it cancels out anyway.

Now the x in that definite integral is a dummy variable, not affecting the answer, so that line could as well have been written[tex]\int_a^b f(t)\,dt = F(t)|_a^b = F(b) - F(a)[/tex] Since this is true for any a and b, let's choose to let b be a variable x:[tex]\int_a^x f(t)\,dt = F(t)|_a^x = F(x) - F(a)[/tex] Since these are equal you still have F'(x) = f(x) so the left side is an antiderivative of f(x). Since a can be anything, the F(a) is like the constant of integration in our first equation.

Does that help answer your question?
vanmaiden
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Dec3-11, 12:16 PM
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Quote Quote by LCKurtz View Post
Since a can be anything, the F(a) is like the constant of integration in our first equation.

Does that help answer your question?
Yes!!! So, to make sure I have this correct, F(a) = C in this case, correct?
LCKurtz
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Dec3-11, 12:32 PM
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Quote Quote by vanmaiden View Post
Yes!!! So, to make sure I have this correct, F(a) = C in this case, correct?
I would leave it as F(a). Here's an example. Suppose you are trying to find the function whose derivative is x2 and whose value at x = 0 is 4. You might do it this way:[tex]f(x) = \int x^2\, dx =\frac {x^3}{3}+C[/tex] Then you plug in x = 0 to require that f(0) = 4 and that tells you that C = 4 so your answer is[tex]f(x) = \frac{x^3} 3+4[/tex] Alternatively you could have solved the problem this way:[tex]f(x)-f(0)=\int_0^x t^2\, dt = \frac{t^3} 3 |_0^x =\frac {x^3} 3[/tex] where f(0) = 4, which you could have put in in the first place. Same answer, slightly different methods.
vanmaiden
#7
Dec3-11, 01:00 PM
P: 101
If f(0) = 4, then shouldn't f(x) - f(0) = [itex]\frac{x^{3}}{3}[/itex] - 4?
LCKurtz
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Dec3-11, 01:44 PM
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Quote Quote by vanmaiden View Post
If f(0) = 4, then shouldn't f(x) - f(0) = [itex]\frac{x^{3}}{3}[/itex] - 4?
f(x) - 4 = [itex]\frac{x^3} 3[/itex]


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