Exploring Relationship of Limit & Integral Methods for Finding Area Under Curve

Yes, you are correct. I mistyped it. f(x) - 4 = \frac{x^3} 3 is the same as f(x) = \frac{x^3} 3 + 4 Therefore, the direct integration method for finding the area under a curve is related to the limit method through the fundamental theorem of calculus, where the antiderivative of a function represents the area under the curve. In summary, the two methods of finding the area under a curve, the limit method and the direct integral method, are related through the fundamental theorem of calculus, where the antiderivative of a function represents the area under
  • #1
vanmaiden
102
1

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


[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

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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  • #3


LCKurtz said:
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) = [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?
 
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  • #4


vanmaiden said:
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?
 
  • #5


LCKurtz said:
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?
 
  • #6


vanmaiden said:
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.
 
  • #7


If f(0) = 4, then shouldn't f(x) - f(0) = [itex]\frac{x^{3}}{3}[/itex] - 4?
 
  • #8


vanmaiden said:
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]
 

1. What is the relationship between limits and integrals?

The relationship between limits and integrals is that the integral is essentially the inverse operation of the derivative, and limits are used to evaluate integrals. The limit of a function as the interval size approaches zero is used to find the area under a curve, which is the basis of integral calculus.

2. How are limits used to find the area under a curve?

Limits are used in the process of taking the integral of a function to find the area under a curve. The limit of a function as the interval size approaches zero is used to calculate the sum of infinitely many infinitesimal rectangles, which approximates the area under the curve.

3. What is the significance of finding the area under a curve using integral methods?

Finding the area under a curve using integral methods is significant because it allows us to calculate the total accumulation of a quantity over an interval. This can be applied to real-world problems such as calculating volume, work, or displacement.

4. What are the limitations of using limits and integrals to find the area under a curve?

One limitation is that it can be a time-consuming and tedious process, especially for complex functions. Additionally, it may not be possible to find an exact solution for certain functions, requiring the use of approximations or numerical methods.

5. How does the relationship between limits and integrals extend beyond finding the area under a curve?

The relationship between limits and integrals extends beyond finding the area under a curve to other applications in mathematics, physics, and engineering. For example, limits and integrals are used in optimization problems, finding volumes of solids of revolution, and calculating the work done by a force.

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