Thread: How do I use Riemman sums to integrate (4-x^2)^(1/2) from 0 to 2?

1. How do I use Riemman sums to integrate (4-x^2)^(1/2) from 0 to 2?

My calc. AB teacher decided against teaching Riemann sums with sigma notation and as I have been studying BC independently I sort of came up with this problem. The answer is pi of course, but if I use the sigma notation and add "infinite rectangles" I need to have some way of coming up with interval widths in a different manner than the normal (b-a)/n. the largest problem seems to be that I don't know how to figure out how to evaluate this type of integral where the width changes. On top of this I know pi is irrational so I figure my answer should come out to an irreducible infinite series but I haven't a clue how to come to that point using these riemman sums. Definite integration is not much problem, I can use arcsin and tables I think, I just want to better understand this method and my teacher hasn't taught anything above math heuristics for a long time. Thanks!

2. Originally Posted by eulcer
My calc. AB teacher decided against teaching Riemann sums with sigma notation and as I have been studying BC independently I sort of came up with this problem. The answer is pi of course, but if I use the sigma notation and add "infinite rectangles" I need to have some way of coming up with interval widths in a different manner than the normal (b-a)/n. the largest problem seems to be that I don't know how to figure out how to evaluate this type of integral where the width changes. On top of this I know pi is irrational so I figure my answer should come out to an irreducible infinite series but I haven't a clue how to come to that point using these riemman sums. Definite integration is not much problem, I can use arcsin and tables I think, I just want to better understand this method and my teacher hasn't taught anything above math heuristics for a long time. Thanks!
If you start with a trig substitution and change the bounds, you will be able to obtain the pi term.

Let $x=2\sin{\theta} \ \ \ dx=2\cos{\theta}d\theta$

$\displaystyle\int_0^{\frac{\pi}{2}}\sqrt{4(1-\sin^2{\theta})}\cos{\theta}d\theta=4\int_0^{\frac {\pi}{2}}\cos^2{\theta}d\theta$

3. Originally Posted by eulcer
My calc. AB teacher decided against teaching Riemann sums with sigma notation and as I have been studying BC independently I sort of came up with this problem. The answer is pi of course, but if I use the sigma notation and add "infinite rectangles" I need to have some way of coming up with interval widths in a different manner than the normal (b-a)/n. the largest problem seems to be that I don't know how to figure out how to evaluate this type of integral where the width changes. On top of this I know pi is irrational so I figure my answer should come out to an irreducible infinite series but I haven't a clue how to come to that point using these riemman sums. Definite integration is not much problem, I can use arcsin and tables I think, I just want to better understand this method and my teacher hasn't taught anything above math heuristics for a long time. Thanks!
The problem is that practically calculating Riemann Sums even in the more tractable sense taught in calculus is damn near impossible for most functions. This is why one has the Fundamental Theorem of Calculus, which does this in general for continuous functions. One may hope to follow the proof the FTC for a clue how to do this, but unfortunately it uses existence tricks etc. to circumvent any real ugliness. Thus, in general something like $\sqrt{4-x^2}$ is not a function conducive to using Riemann sums. Now, if you wanted to do $4-x^2$ that'd be a different story

4. Originally Posted by dwsmith
If you start with a trig substitution and change the bounds, you will be able to obtain the pi term.

Let $x=2\sin{\theta} \ \ \ dx=2\cos{\theta}d\theta$

$\displaystyle\int_0^{\frac{\pi}{2}}\sqrt{4(1-\sin^2{\theta})}\cos{\theta}d\theta=4\int_0^{\frac {\pi}{2}}\cos^2{\theta}d\theta$
Once again, you have completely misread the question! The question was about using Riemann sums to approximate an integral, not about doing the integral by finding an anti-derivative. While you are usually very helpful, occaisionally, you go off on strange tangents! Please read the problems more carefully.

5. Originally Posted by HallsofIvy
Once again, you have completely misread the question! The question was about using Riemann sums to approximate an integral, not about doing the integral by finding an anti-derivative. While you are usually very helpful, occaisionally, you go off on strange tangents! Please read the problems more carefully.

Edit:

As you can see (http://mathworld.wolfram.com/RiemannSum.html), it will work.

$\displaystyle \Delta x=\frac{b-a}{n}=\frac{\frac{\pi}{2}-0}{n}=\frac{\pi}{2n}$

$\displaystyle f(c_i)=a+i\Delta x=4\cos^2{\left(0+\frac{i\pi}{2n}\right)}$

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x$

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n \left(4\cos^2{\left(\frac{i\pi}{2n}\right)}\left(\ frac{\pi}{2n}\right)\right)=\lim_{n\to\infty}\left[\frac{2\pi}{n}\sum_{i=1}^n \cos^2{\left(\frac{i\pi}{2n}\right)}\right]=\pi$

7. Thank's dwsmith. I wouldn't have known your intentions without that next post, but that's my fault. It makes sense all up to the last step. How do you determine the limit as n-> infinity? Trig identities?

8. Originally Posted by eulcer
Thank's dwsmith. I wouldn't have known your intentions without that next post, but that's my fault. It makes sense all up to the last step. How do you determine the limit as n-> infinity? Trig identities?
First, solve the summation.

$\displaystyle\sum_{i=1}^n \cos^2{\left(\frac{i\pi}{2n}\right)}$

9. yeah I think I got that.
(cos(pi(n)/4+pi/4))^2, right?

10. Originally Posted by eulcer
yeah I think I got that.
(cos(pi(n)/4+pi/4))^2, right?
$\displaystyle\frac{1}{4} \left(-1+2 n-\text{Csc}\left[\frac{\pi }{2 n}\right] \text{Sin}\left[\frac{-\pi -2 n \pi }{2 n}\right]\right)$

11. Originally Posted by dwsmith
$\displaystyle\frac{1}{4} \left(-1+2 n-\text{Csc}\left[\frac{\pi }{2 n}\right] \text{Sin}\left[\frac{-\pi -2 n \pi }{2 n}\right]\right)$
Alright, so I understand why what I did doesn't work, but where does this come from?

12. Originally Posted by eulcer
Alright, so I understand why what I did doesn't work, but where does this come from?
$\displaystyle\sum_{i=1}^n \cos^2{\left(\frac{i\pi}{2n}\right)}=\frac{1}{4} \left(-1+2 n-\text{Csc}\left[\frac{\pi }{2 n}\right] \text{Sin}\left[\frac{-\pi -2 n \pi }{2 n}\right]\right)$

13. No, I understand what you're saying. I just don't have a clue how it came to be.

How do I solve summations when the "i" is in a trig fxn?

14. Originally Posted by eulcer
No, I understand what you're saying. I just don't have a clue how it came to be.

How do I solve summations when the "i" is in a trig fxn?
Note that $\displaystyle \sum_{k=1}^{n}\cos^2\left(\frac{\pi k}{2n}\right)=\frac{1}{2}\sum_{k=1}^{n}\left(1+\co s\left(\frac{\pi k}{n}\right)\right)=\frac{n}{2}+\frac{1}{2}\text{R e}\sum_{k=1}^{n}e^{\frac{i\pi k}{n}}$. Is any of that familiar looking to you?

15. Originally Posted by eulcer
No, I understand what you're saying. I just don't have a clue how it came to be.

How do I solve summations when the "i" is in a trig fxn?
Drexel used the identity

$\displaystyle \cos^2=\frac{1+\cos{2u}}{2}$

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