# Definition of the integral question (Riemann sums?)

• May 19th 2010, 10:16 PM
s3a
Definition of the integral question (Riemann sums?)
The following sum
http://gauss.vaniercollege.qc.ca/web...d534baa891.png
is a right Riemann sum for the definite integral where http://gauss.vaniercollege.qc.ca/web...dd497c3d61.png = ? = 3
and http://gauss.vaniercollege.qc.ca/web...910e13a9c1.png = ? = sqrt(9-x^2)

The part of the question I'm stuck on is:
The limit of these Riemann sums as http://gauss.vaniercollege.qc.ca/web...63147a7bc1.png is = ?

I know that I need to evaluate:
lim(n->inf) sum from 0 to n of sqrt(9-(3i/n)^2) but after scaring the 3i/n part, I'm stuck.

Any help would be GREATLY appreciated!

P.S.
Does this whole question involve a topic called "Riemann sums?"
• May 19th 2010, 11:00 PM
matheagle
Are you supposed to obtain the limit of the Riemann sum OR integrate?
And n goes to infinity not x.
• May 20th 2010, 04:32 AM
HallsofIvy
Quote:

Originally Posted by s3a
The following sum
http://gauss.vaniercollege.qc.ca/web...d534baa891.png
is a right Riemann sum for the definite integral where http://gauss.vaniercollege.qc.ca/web...dd497c3d61.png = ? = 3
and http://gauss.vaniercollege.qc.ca/web...910e13a9c1.png = ? = sqrt(9-x^2)

The part of the question I'm stuck on is:
The limit of these Riemann sums as http://gauss.vaniercollege.qc.ca/web...63147a7bc1.png is = ?

I know that I need to evaluate:
lim(x->inf) sum from 0 to n of sqrt(9-(3i/n)^2) but after scaring the 3i/n part, I'm stuck.

Any help would be GREATLY appreciated!

P.S.
Does this whole question involve a topic called "Riemann sums?"

Well, the whole point is that since that sum is a Riemann sum for the integral $\int_0^3 \sqrt{9- x^2} dx$ its limit as n goes to infinity must be the integral!

And you don't even have to do an integral. If $y= \sqrt{9- x^2}$, then $x^2+ y^2= 9$ except that $y\ge 0$- the upper half of a circle of radius 3 with center at the origin. And since the integral is from x= 0 to x= 3 rather than -3 to 3, that integral is the area of 1/4 of a circle of radius 3.
• May 20th 2010, 12:48 PM
s3a
Quote:

Originally Posted by matheagle
Are you supposed to obtain the limit of the Riemann sum OR integrate?
And n goes to infinity not x.

Ya n->inf not x (sorry force of habit). And I'm supposed to evaluate the limit of the Riemann sum for this. How do I do this algebraically?
• May 20th 2010, 03:05 PM
Plato
Quote:

Originally Posted by s3a
Ya n->inf not x (sorry force of habit). And I'm supposed to evaluate the limit of the Riemann sum for this. How do I do this algebraically?

Quote:

Originally Posted by HallsofIvy
Well, the whole point is that since that sum is a Riemann sum for the integral $\int_0^3 \sqrt{9- x^2} dx$ its limit as n goes to infinity must be the integral!
And you don't even have to do an integral. If $y= \sqrt{9- x^2}$, then $x^2+ y^2= 9$ except that $y\ge 0$- the upper half of a circle of radius 3 with center at the origin. And since the integral is from x= 0 to x= 3 rather than -3 to 3, that integral is the area of 1/4 of a circle of radius 3.

• May 20th 2010, 04:25 PM
s3a
Oh I misread that thinking he wanted me to do the integral. But still, is there not an algebraic way? (I did the graphing way)
• May 20th 2010, 04:56 PM
matheagle
let $x=3\sin u$
• May 21st 2010, 04:57 AM
HallsofIvy
Quote:

Originally Posted by s3a
Oh I misread that thinking he wanted me to do the integral. But still, is there not an algebraic way? (I did the graphing way)

What do you mean by an "algebraic way" (and, for that matter, what is the "graphing way")?

As I told you, that is 1/4 of a circle of radius 3. Its area is $(1/4)\pi (3^2)$.