Soaring Crane said:
In the Fundamental Theorem of Calculus, it is stated that
lim max deltax_k -> 0 sigma f(x_k*)*deltax_k = L,
where L is the definite integral of f(x) on [a,b].
How do you apply this limit definition to a normal function to find the definite integral? (Could someone give me math examples of this application?)
Is it like the method used with left and right endpoints in the limit definition where n -> oo for area under a curve? How is it different?
Thanks.
Quite frankly, I have never seen anything like that in the Fundamental Theorem of Calclus!
There are two parts to the Fundamental Theorem:
1) If F(x)= \int_a^x f(x) dx, then F'(x)= f(x).
2) If F'(x)= f(x) then \int_a^b f(x)dx= F(b)- F(a)
Neither of those include the
definition of the definite integral which is how you titled this thread. I'm not impressed by the formula you give- it leaves too much unsaid. What exactly do you mean by "deltax_k"? How is x_k* defined?
I THINK what you mean is that you partion the interval from, say, x= a to x= b, into n pieces such that the length of kth "piece" is deltax_k. Choose x_k* to be a value of x in that kth "piece" and sum the products
f(x_k*)deltax_k over k. But for that to be used, you have to give a rule for finding each delta_k not only for each k but for all values of n, as n goes to infinity. I also am not clear what "lim max" means. Do you mean "lim sup"?
The hard part of showing that a function
is integrable is showing that none of those choices affect the final result!
An example that is given in just about every calculus book is this:
Let f(x)= 2x+ 3, a= 0, b= 1. For each n, take all deltax_k= 1/n so that each interval has the same length. Of course, the points separating each interval are x_k= k(deltax_k)= k/n. That is, x_0= 0, the left endpoint, x_1= 1/n, x_2= 2/n,..., up to x_n= n/n= 1, the right end point.
For each interval, take x_k* to be the right endpoint of that interval:
x_k= k/n for k= 1 to n. Then f(x_k*)= 2k/n+ 3. The "Riemann sum" for that n is
\sum_{k=1}^n \left(\frac{2k}{n}+ 3\right)\frac{1}{n}
We can separate that into two sums:
\frac{2}{n^2}\sum_{k=1}^n k+ \frac{3}{n}\sum_{k=1}^n 1
It is well known that
\sum_{k= 1}^n k= \frac{n(n+1)}{2}
and, of course,
\sum_{k=1}^n 1= n.
That means that, for each n, the sum is
\frac{n(n+1)}{n^2}+ \frac{3n}{n}
The limit of that, as n goes to infinity, is 1+ 3= 4.
Of course, the anti-derivative of 2x+ 3 is x
2+ 3x which, evaluated at x=1 and x= 0 gives 4 and 0, which, subracting gives 4 as the value of the integral. That's a lot easier!
By the way, DesCartes, Fermat, and others knew how to find tangent lines (derivatives) and limits of sums (integrals). It was the fact that Newton and Leibniz recognized the identity of "integrals" and "
anti-derivatives" that made them the "founders" of calculus.