# [SOLVED] Lower and Upper Riemann sums?

• Aug 1st 2009, 09:52 AM
PTL
[SOLVED] Lower and Upper Riemann sums?
Let $\displaystyle f(x) = sin x$. Estimate (integral from 0 to pi/2) $\displaystyle \int {f(x)} dx$ by calculating the lower Riemann sums $\displaystyle L(f,P_n)$ and the upper Riemann sums $\displaystyle U(f,P_n)$ for n = 3 with respect to a partition of $\displaystyle P_n$ $\displaystyle [0,\frac{\pi}{2}]$ into n subintervals of equal length. Verify that
$\displaystyle L(f,P_3)<= 1 <= U(f,P_3)$
and explain why $\displaystyle L(f,P_n) <= 1 <= U(f,P_n)$ for all $\displaystyle n > 0$

I don't even know what this means!! (Doh)
• Aug 1st 2009, 10:03 AM
VonNemo19
Quote:

Originally Posted by PTL
Let $\displaystyle f(x) = sin x$. Estimate (integral from 0 to pi/2) $\displaystyle \int {f(x)} dx$ by calculating the lower Riemann sums $\displaystyle L(f,P_n)$ and the upper Riemann sums $\displaystyle U(f,P_n)$ for n = 3 with respect to a partition of $\displaystyle P_n$ $\displaystyle [0,\frac{\pi}{2}]$ into n subintervals of equal length. Verify that
$\displaystyle L(f,P_3)<= 1 <= U(f,P_3)$
and explain why $\displaystyle L(f,P_n) <= 1 <= U(f,P_n)$ for all $\displaystyle n > 0$

I don't even know what this means!! (Doh)

$\displaystyle n=3\Rightarrow\Delta{x}=\frac{\pi}{2n}=\frac{\pi}{ 6}$

$\displaystyle \text{ Let }x_i$ be the right end point of each subinterval, then

$\displaystyle x_i=(0+\frac{\pi{i}}{6})$, therefore

$\displaystyle U=\sum_{i=1}^{3}\sin{\frac{\pi{i}}{6}}*\frac{\pi}{ 6}$

Can you proceed?
• Aug 1st 2009, 10:07 AM
flyingsquirrel
Quote:

Originally Posted by PTL
I don't even know what this means!! (Doh)

I suggest you look for the definitions of an upper/lower Riemann sum in your notes or in your textbook. Once you know the definitions this problem will be easy to solve.
• Aug 1st 2009, 11:01 AM
Random Variable
$\displaystyle P_{3} = \{ 0, \frac{\pi}{6} , \frac{\pi}{3}, \frac{\pi}{2} \}$

let $\displaystyle M_{i}$ be the supremum of f(x) on interval i, and let $\displaystyle m_{i}$ be the infinum of f(x) on interval i

then $\displaystyle M_{1} = \frac{1}{2}, \ M_{2} = \frac{\sqrt{3}}{2}, \ M_{3} = 1$

and $\displaystyle m_{1} = 0, \ m_{2} = \frac{1}{2}, \ m_{3} = \frac{\sqrt{3}}{2}$

and since the length of each interval is $\displaystyle \frac{\pi}{6}$

$\displaystyle U(f, P_{3}) = \frac {\pi}{6} \Big(\frac{1}{2}+\frac{\sqrt{3}}{2} +1\Big) \approx 1.24$

$\displaystyle L(f, P_{3}) = \frac {\pi}{6} \Big(0+\frac{1}{2} +\frac{\sqrt{3}}{2}\Big) \approx 0.715$
• Aug 1st 2009, 11:05 AM
VonNemo19
Here's a picture of the right sum
• Aug 1st 2009, 11:05 AM
Random Variable
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• Aug 5th 2009, 05:47 AM
PTL
Thanks guys.

Question - how does one mark a thread as Solved?

Also - there's a bit at the end of the question, where I'm asked to
"Explain why L(f,P_n) <= 1 <= U(f,P_n) for all n > 0".
Can I just say it's because the function is monotonically increasing and therefore the Upper Sum will be an overestimation, and the Lower Sum will be an underestimation?
Where does the '1' come into that though?
• Aug 5th 2009, 06:11 AM
Jester
Quote:

Originally Posted by PTL
Thanks guys.

Question - how does one mark a thread as Solved?

Also - there's a bit at the end of the question, where I'm asked to
"Explain why L(f,P_n) <= 1 <= U(f,P_n) for all n > 0".
Can I just say it's because the function is monotonically increasing and therefore the Upper Sum will be an overestimation, and the Lower Sum will be an underestimation?
Where does the '1' come into that though?

The 1 is the actual area under the curve, i.e. $\displaystyle \int_0^{\pi/2} \sin x \, dx = 1$
• Aug 5th 2009, 06:37 AM
flyingsquirrel
Quote:

Originally Posted by PTL
Question - how does one mark a thread as Solved?

Go to the top of the page, click on "Thread Tools" and then on "Mark this thread as solved".
• Aug 5th 2009, 06:50 AM
PTL
Thanks.