How Do Limit and Integral Methods Relate for Finding Area Under a Curve?

[vanmaiden]

Homework Statement

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?

Homework Equations

\intf(x) dx = F(x)|^{b}_{a} = F(b) - F(a) = Area

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

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.

 

[LCKurtz]

That is the fundamental theorem of calculus. You might start by reading here:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

 

[vanmaiden]

That is the fundamental theorem of calculus. You might start by reading here:

http://en.wikipedia.org/wiki/Fundamental_theorem_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?

 

[LCKurtz]

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.

Does that help answer your question?

 

[vanmaiden]

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]

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 isf(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.

 

[vanmaiden]

If f(0) = 4, then shouldn't f(x) - f(0) = \frac{x^{3}}{3} - 4?

 

[LCKurtz]

If f(0) = 4, then shouldn't f(x) - f(0) = \frac{x^{3}}{3} - 4?

f(x) - 4 = \frac{x^3} 3