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

by vanmaiden
Tags: area, curve, direct integral, limit
 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 $\int$f(x) dx = F(x)|$^{b}_{a}$ = F(b) - F(a) = Area $lim_{n→∞}$ $\sum^{n}_{i = 1}$ $f(x_{i})$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. $lim_{n→∞}$ $\sum^{n}_{i = 1}$ $f(x_{i})$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.
 HW Helper Thanks PF Gold P: 7,659 That is the fundamental theorem of calculus. You might start by reading here: http://en.wikipedia.org/wiki/Fundame...em_of_calculus
P: 101
 Quote by LCKurtz 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) = $\int^{x}_{a}f(t) dt$
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?

HW Helper
Thanks
PF Gold
P: 7,659
Finding the Relationship between the Limit Method and Direct Integral Method for Area

 Quote by vanmaiden 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) = $\int^{x}_{a}f(t) dt$ 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 $$F(x)=\int f(x)\, dx + C$$ where F'(x) = f(x). If you were going a definite integral you would write $$\int_a^b f(x)\,dx = (F(x)+C)|_a^b = F(b) - F(a)$$ 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$$\int_a^b f(t)\,dt = F(t)|_a^b = F(b) - F(a)$$ Since this is true for any a and b, let's choose to let b be a variable x:$$\int_a^x f(t)\,dt = F(t)|_a^x = F(x) - F(a)$$ 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.

P: 101
 Quote by LCKurtz 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?
HW Helper
Thanks
PF Gold
P: 7,659
 Quote by vanmaiden 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:$$f(x) = \int x^2\, dx =\frac {x^3}{3}+C$$ Then you plug in x = 0 to require that f(0) = 4 and that tells you that C = 4 so your answer is$$f(x) = \frac{x^3} 3+4$$ Alternatively you could have solved the problem this way:$$f(x)-f(0)=\int_0^x t^2\, dt = \frac{t^3} 3 |_0^x =\frac {x^3} 3$$ where f(0) = 4, which you could have put in in the first place. Same answer, slightly different methods.
 P: 101 If f(0) = 4, then shouldn't f(x) - f(0) = $\frac{x^{3}}{3}$ - 4?
HW Helper
Thanks
PF Gold
P: 7,659
 Quote by vanmaiden If f(0) = 4, then shouldn't f(x) - f(0) = $\frac{x^{3}}{3}$ - 4?
f(x) - 4 = $\frac{x^3} 3$

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