# Thread: Definition of the integral question (Riemann sums?)

1. ## Definition of the integral question (Riemann sums?)

The following sum

is a right Riemann sum for the definite integral
where = ? = 3
and = ? = sqrt(9-x^2)

The part of the question I'm stuck on is:
The limit of these Riemann sums as 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?"

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

3. Originally Posted by s3a
The following sum

is a right Riemann sum for the definite integral
where = ? = 3
and = ? = sqrt(9-x^2)

The part of the question I'm stuck on is:
The limit of these Riemann sums as 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.

4. 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?

5. 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?
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.

6. 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)

7. let $x=3\sin u$

8. 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 $(1/4)\pi (3^2)$.