# estimate integral

• Aug 25th 2007, 12:36 PM
runner07
estimate integral
Estimate

ln 1.5 = (integral from 1 to 1.5) dt/t

using the approximation (1/2)[Lf(P) + Uf(P)] with
P = {1 = (4/4) , (5/4), (6/4) , (7/4) , (8/4), (9/4), (10/4) = (5/2)} .

sorry everyone! the book is online so i had copied and pasted this question but for some reason the numbers changed and i didnt notice. im sorry for the confusion. thank you all for letting me know.
[note- the f's in Lf and Uf are like L(sub)f and U(sub)f]
• Aug 25th 2007, 03:58 PM
TKHunny
Quote:

Originally Posted by runner07
P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.

1) I cannot imagine what that might mean.
2) If you invented this notation, please don't invent new notation. There is already way too much.
3) Can you provide the EXACT wording of the question and some indication of your efforts? If your textbook's author invented the notation, the explanation has to be in there, somewhere.
• Aug 25th 2007, 04:23 PM
Jhevon
Quote:

Originally Posted by runner07
Estimate

ln 1.5 = (integral from 1 to 1.5) dt/t

using the approximation (1/2)[Lf(P) + Uf(P)] with
P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.

I believe he is describing a Darboux integral

P describes the partitions and L(f,P) is the lower Darboux sum and U(f,P) is the upper Darboux sum
• Aug 25th 2007, 04:37 PM
TKHunny
That may be, but I'm still wondering how 1 = 8 and 12 = 1.5!

Perhaps this means the number of subintervals?
• Aug 25th 2007, 04:38 PM
Jhevon
Quote:

Originally Posted by TKHunny
That may be, but I'm still wondering how 1 = 8 and 12 = 1.5!

Perhaps this means the number of subintervals?

yeah, he lost me there too
• Aug 25th 2007, 05:03 PM
ThePerfectHacker
Quote:

Originally Posted by runner07
Estimate

ln 1.5 = (integral from 1 to 1.5) dt/t

using the approximation (1/2)[Lf(P) + Uf(P)] with
P = {1 = 8 , 9 , 10 , 11 , 12 = 1. 5}.

Say the partition is $P=\{1,1.1,1.2,1.3,1.4,1.5\}$.

Then, $U(f,P) = \sup_{[1,1.1]}\{ f\}\cdot (1.1-1) + \sup_{[1.1,1.2]}\cdot (1.2-1.1) + ... + \sup_{[1.4,1.5]}\cdot (1.5-1.4)$
Since $f$ is non-increasing the supremum is the left endpoint.
Thus,
$U(f,P) = f(1)(.1)+f(1.1)(.1)+...+f(1.4)(.1) = .1 \left( 1+\frac{1}{1.1}+...+\frac{1}{1.4} \right)$
Similarly,
$L(f,P) = f(1.1)(.1)+f(1.2)(.1)+...+f(1.5)(.1) = .1\left( \frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{1.5} \right)$
• Aug 26th 2007, 09:47 AM
runner07
sorry everyone! the book is online so i had copied and pasted this question but for some reason the numbers changed and i didnt notice. im sorry for the confusion. thank you all for letting me know.
• Aug 26th 2007, 07:33 PM
TKHunny
No worries. Use TPH's very clear demonstration and let's see what you get.