Multiple choice def integral question

[nadineM]

Homework Statement

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)

Homework 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

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.

 

[siddharth]

You've already written the answer here

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

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

 

[nadineM]

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

 

[HallsofIvy]

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.

 

[nadineM]

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.

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?

 

[HallsofIvy]

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.

 

[nadineM]

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.

Thank you so much you are always so helpful and patient! and are able to explain things clearly! THANKS