Left/right endpoint question

B

bcddd214

Junior Member
Joined
May 16, 2011
Messages
102
I am getting my but handed to me on the simple algebra portion of this one. I am supposed to have lost the radical early I am sure, just can't seem to shake it.... :(

use the upper and lower sums to approximate the area of the given region using the indicated number of (equal) subintervals

y=sqrt(2)

s(n)=?_(i=1)^n ?f(m_i )?x=? ?_(i=1)^n f[2(i-1)/n] (2/n)
=?_(i=1)^n [?(2(i-1)/n)] (2/n)=(?2i/?n-?2/?n)(2/n)
2/n [?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n]
?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n
 
bcddd214 said:
I am getting my but handed to me on the simple algebra portion of this one. I am supposed to have lost the radical early I am sure, just can't seem to shake it.... :(

use the upper and lower sums to approximate the area of the given region using the indicated number of (equal) subintervals

y=sqrt(2)

s(n)=?_(i=1)^n ?f(m_i )?x=? ?_(i=1)^n f[2(i-1)/n] (2/n)
=?_(i=1)^n [?(2(i-1)/n)] (2/n)=(?2i/?n-?2/?n)(2/n)
2/n [?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n]
?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n
Click to expand...

PS, there is a diagram that shows an interval from 0-1 with 4 shaded rectangles in the interval from 0-1
that means n=4, correct?
 
No one has taken a shot at this one?

I get a like farther but I am still seeing what I feel I am supposed to be seeing.
:(

s(n)=?_(i=1)^n?f(m_i )?x=? ?_(i=1)^n f[2(i-1)/n] (2/n)
=?_(i=1)^n[?(2(i-1)/n)] (2/n)=(?2i/?n-?2/?n)(2/n)
2/n [?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n]
2/n ?_(i=1)^n ??2i/?n n^(1/2)/?-2/n ?_(i=1)^n ?2/?n n^(1/2)/
2/n ?_(i=1)^n ??2i n^(1/2) ?-2/n ?_(i=1)^n ?2 n^(1/2)
?_(i=1)^n ?(2?2i)/n ?2n?^(1/2)/n?-2/n ?_(i=1)^n (-2?2)/n n^(1/2)/n
 
bcddd214 said:
I am getting my but handed to me on the simple algebra portion of this one. I am supposed to have lost the radical early I am sure, just can't seem to shake it.... :(

use the upper and lower sums to approximate the area of the given region using the indicated number of (equal) subintervals

y=sqrt(2)

s(n)=?_(i=1)^n ?f(m_i )?x=? ?_(i=1)^n f[2(i-1)/n] (2/n)
=?_(i=1)^n [?(2(i-1)/n)] (2/n)
=(?2i/?n-?2/?n)(2/n) <<<< Cannot understand your work - where did the ? go
2/n [?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n]
?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n
Click to expand...
 
s(n)=?_(i=1)^n ?f(m_i )?x=? ?_(i=1)^n f[2(i-1)/n] (2/n)
=?_(i=1)^n [?(2(i-1)/n)] (2/n)=?_(i=1)^n [?2i/?n-?2/?n] (2/n)
2/n [?_(i=1)^n ?2i/?n-?_(i=1)^n ?2/?n]
2/n ?_(i=1)^n ??2i/?n n^(1/2)/?-2/n ?_(i=1)^n ?2/?n n^(1/2)/
2/n ?_(i=1)^n ??2i n^(1/2) ?-2/n ?_(i=1)^n ?2 n^(1/2)
?_(i=1)^n ?(2?2i)/n ?2n?^(1/2)/n?-2/n ?_(i=1)^n (-2?2)/n n^(1/2)/n
 
I guess you don't want my contribution.
You may know math, but you are apparently cluely on how the internet works.
My time spent here posting these questions is invaluable to you.
Click to expand...
I'd help but I am cluely.
 
bcddd214 said:
approximate the area of the given region using the indicated number of (equal) subintervals

y = sqrt(2)
Click to expand...

I'm guessing that the "given region" is somewhere between the graph of y = sqrt(2) and the x-axis.

Yet, I do not see that you've posted the bounds, and I find no indication of the number of subintervals.

Can you post the entire exercise? :?

 
You must log in or register to reply here.
Top