# Multiple choice def integral question

1. Apr 27, 2007

1. The problem statement, all variables and given/known data

If f is a continuous real valued function, then the def integral from 0 to 1 of f(x)dx =
a. limit as n-> + infinity of 1/n * the sumation from k=0 to n-1 of f(k/n)
b. limit as n-> + infinity of the sumation from k=1 to n of (1/k)*f(k/n)
c. limit as n-> + infinity of 1/n * the sumation from k=0 to n-1 of kf(k/n)
d. limit as n-> + infinity of the sumation from k=0 to n+1 of kf(k+1/n)
e. the summation from k=0 to +infinity of kf(k/n)

2. Relevant equations
the definite intergral from a to b of f(x)dx = limit as n -> infinity of the summation i=1 to n of f(xi) delta x
where delta x = (b-a)/n

3. The attempt at a solution

Sorry that this is all in words I have no idea how to do the symbols, if someone could help me with that so it is easier for you to read and me to write, that would be great.....
Any ways..... This was a homework question I got last week and have been assigned to present the answer to the class.
Only I have no idea which is right or why.
I believe that the answer is either a or c, leaning more towards a, but I have no idea. I do not even know if the revalant equation that I put down is even what I should be using...or something else...any pushes in the right direction will be greatly appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 27, 2007

### siddharth

In your given problem, what's a and b? Can you find the answer now?

3. Apr 28, 2007

a=0 b=1. that is why I said choices a or c because then delta x is 1/n. but then the summation is different in the definition, the summation goes from i=1 to n. the choices I am given are summations from k=0 to n-1...so it is confusing me. I also see now that b is close but the delta x is 1/k instead of 1/n....still stumped

4. Apr 28, 2007

### HallsofIvy

Staff Emeritus
There are, of course, an infinite number of possible "Riemann sums" that will give a specific integral. One method, in particular is to divide the interval (from 0 to 1 in this case) into "n" pieces, so that delta x is 1/n, and then choose the "point at which to evaluate f(x) in each interval" to be the left endpoint of the interval. Since all of the endpoints are k/n for k from 0 to n, the left endpoints will be from k= 0 to n-1 (k= n is a right endpoint only while every other k is a left endpoint of one interval). The "Riemann sum" is the sum of the "areas of the rectangles" f(k/n)(1/n) summed for k going from 0 to n-1.

Another way is to do it exactly as above except that you choose the right endpoint of each interval as the point at which to evaluate f(x). In that case, k= 0 is a left endpoint only while every other endpoint, from k= 1 to n, is a right endpoint of some interval. The Riemann sum is the "area of the rectangle" f(k/n)(1/n) summed for k going from 1 to n.

5. Apr 28, 2007

ok thanks I think I get it now. The answer is a( the limit as n -> infinity of 1/n * the summation from k=0 to n-1 of (k/n)) as I suspected correct?
but one more question. Why can you bring the delta x that is 1/n out in front of the summation notation?

Last edited: Apr 28, 2007
6. Apr 28, 2007

### HallsofIvy

Staff Emeritus
Yes, that is correct. Had they had a choice of (1/n) sum from 1 to n, f(x/n) that would also b correct. But the only sum from 1 to n did NOT have the factor of 1/n.

You can take (1/n) out of the sum because it does not depend on k!
(1/n)f(1/n)+ (1/n)f(2/n)+ (1/n)f(3/n)= (1/n)(f(1/n)+ f(2/n)+ f(3/n)) by the distributive law.

7. Apr 28, 2007