Approximating the Area of a Plane Region

Nov 2014
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Please help me with this. I've seen two online lectures, attended my class lecture, read what the book says, even checked WolframWorld, to no avail.

"use left and right endpoints and the given
number of rectangles to find two approximations of the area of
the region between the graph of the function and the x -axis
over the given interval"

I can't use anti-derivatives yet. The formula we got is:

A = The limit as x approaches infinity of: The sum of the product of: f(xi) * 'delta x', with n = number of rectangles, and i = 1

I really don't get this problem:

f (x) = 9 - x
The region is [2,4]
the number of rectangles, n = 6

This is my latest attempt, can anybody tell me what I'm doing wrong???

{See below for screenshot of my attempts please}

I only have today and tomorrow morning to figure this out, does anybody have resources where I can look up how to do this? I've read a lot of stuff but I just can't get it.

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Last edited:
Aug 2006
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"use left and right endpoints and the given
number of rectangles to find two approximations of the area of
the region between the graph of the function and the x -axis
over the given interval"

f (x) = 9 - x
The region is [2,4]
the number of rectangles, n = 6
OK, $n=6$ and $\dfrac{4-2}{6}=\dfrac{1}{3}=\Delta x$
So you left hand endpoints are $\ell_k=2+k\cdot\Delta x,~k=0,1,2,3,4,5$ and left sum is $\sum\limits_{\ell = 0}^5 {f({\ell_k})\Delta x}$

And you left right hand endpoints are $r_k=2+k\cdot\Delta x,~k=1,2,3,4,5,6$ and right sum is $\sum\limits_{k = 1}^6 {f({r_k})\Delta x}$

You do the arithmetic.
 
Last edited:
Nov 2013
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see if this picture helps any

Clipboard01.jpg
 
Nov 2014
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Home
OK, $n=6$ and $\dfrac{4-2}{6}=\dfrac{1}{3}=\Delta x$
So you left hand endpoints are $\ell_k=2+k\cdot\Delta x,~k=0,1,2,3,4,5$ and left sum is $\sum\limits_{\ell = 0}^5 {f({\ell_k})\Delta x}$

And you left right hand endpoints are $r_k=2+k\cdot\Delta x,~k=1,2,3,4,5,6$ and right sum is $\sum\limits_{k = 1}^6 {f({r_k})\Delta x}$

You do the arithmetic.
Yeah I got that far, but I don't get what r subscript k is supposed to be. Is it: "2 + (1/3)i"?
 
Aug 2006
22,432
8,612
Yeah I got that far, but I don't get what r subscript k is supposed to be. Is it: "2 + (1/3)i"?
Are you saying that you have no idea how to evaluate $f(\ell_k)=9-(\ell_k)~?$
 
Last edited:
Nov 2014
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Part of your problem is the fact these are finite sums. You never use $\infty$.
Look at $\sum\limits_{k = 0}^5 {f({x_k})\Delta x}$ there are but six terms to add up.
Oh, so I'm not supposed to use this approach when solving this problem? I just manually add the sums then? I'll try that and see if it works
 
Jun 2008
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$f(x) = 9-x$

Left sum ...

$\displaystyle \dfrac{1}{3} \sum_{i=0}^5 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[f\left(\dfrac{6}{3}\right) + f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) \bigg]$

Right sum ...

$\displaystyle \dfrac{1}{3} \sum_{i=1}^6 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[ f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) + f\left(\dfrac{12}{3}\right)\bigg]$
 
Nov 2014
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$f(x) = 9-x$

Left sum ...

$\displaystyle \dfrac{1}{3} \sum_{i=0}^5 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[f\left(\dfrac{6}{3}\right) + f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) \bigg]$

Right sum ...

$\displaystyle \dfrac{1}{3} \sum_{i=1}^6 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[ f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) + f\left(\dfrac{12}{3}\right)\bigg]$
Thanks, I figured out how to solve the "Right sum"

but the "left sum" poses problems. i=0 in that case, I only know how to solve if i = 1. Do you know how to fix this?