# Thread: Approximating the Area of a Plane Region

1. ## Approximating the Area of a Plane Region

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.

2. ## Re: Approximating the Area of a Plane Region

Originally Posted by Awesome31312
"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.

3. ## Re: Approximating the Area of a Plane Region

see if this picture helps any

4. ## Re: Approximating the Area of a Plane Region

Originally Posted by Plato
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"?

5. ## Re: Approximating the Area of a Plane Region

Originally Posted by Awesome31312
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 $f(\ell_k)=9-(\ell_k)~?$

6. ## Re: Approximating the Area of a Plane Region

Originally Posted by Plato
Are you saying that you have no idea how to evaluate $f(2+\ell_k)=9-(2+\ell_k)~?$
I tried that, got the wrong answer, I'm so angry at myself

7. ## Re: Approximating the Area of a Plane Region

Originally Posted by Awesome31312
I tried that, got the wrong answer, I'm so angry at myself
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.

8. ## Re: Approximating the Area of a Plane Region

Originally Posted by Plato
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

9. ## Re: Approximating the Area of a Plane Region

$f(x) = 9-x$

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 ...

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

10. ## Re: Approximating the Area of a Plane Region

Originally Posted by skeeter
$f(x) = 9-x$

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?

11. ## Re: Approximating the Area of a Plane Region

Originally Posted by Awesome31312
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?
You my Awesome friend need to rethink taking this level of calculus.
Take a look at Skeeter's sum:
Left sum ...
$\displaystyle \dfrac{1}{3} \sum_{i=0}^5 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[f\left(\dfrac{6}{3}\right) + f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) \bigg]$
Can't you just do those calculations?
The sum index starts with $i=0\to i=5$
The Riemann subdivision is $\left[2,\frac{7}{3}\right],~\left[\frac{7}{3},\frac{8}{3}\right],~\left[\frac{8}{3},\frac{9}{3}\right],~\left[\frac{9}{3},\frac{10}{3}\right],~\left[\frac{10}{3},\frac{11}{3}\right]$

I think you simply do not understand the underlying concepts of summation. I hope I am wrong.

12. ## Re: Approximating the Area of a Plane Region

Originally Posted by Plato
You my Awesome friend need to rethink taking this level of calculus.
Take a look at Skeeter's sum:
Left sum ...
$\displaystyle \dfrac{1}{3} \sum_{i=0}^5 f\left(2+\dfrac{i}{3}\right) = \dfrac{1}{3} \bigg[f\left(\dfrac{6}{3}\right) + f\left(\dfrac{7}{3}\right) + f\left(\dfrac{8}{3}\right) + f\left(\dfrac{9}{3}\right) + f\left(\dfrac{10}{3}\right) + f\left(\dfrac{11}{3}\right) \bigg]$
Can't you just do those calculations?
The sum index starts with $i=0\to i=5$
The Riemann subdivision is $\left[2,\frac{7}{3}\right],~\left[\frac{7}{3},\frac{8}{3}\right],~\left[\frac{8}{3},\frac{9}{3}\right],~\left[\frac{9}{3},\frac{10}{3}\right],~\left[\frac{10}{3},\frac{11}{3}\right]$

I think you simply do not understand the underlying concepts of summation. I hope I am wrong.

I have an A in the class so far. We 'just' got introduced to the concept of sums. I'm in an elementary calculus class. If there is a way to do it by rewriting the sums, that's what I was asking about. I can do the calculations manually, but our professor would rather have us rewrite it and solve it through simplification and summation rules.

Thanks a lot for your help by the way!!!

13. ## Re: Approximating the Area of a Plane Region

Originally Posted by Awesome31312
I have an A in the class so far. We 'just' got introduced to the concept of sums. I'm in an elementary calculus class.!
A grade in early in a basic course is not always a guide to success.
There are many many ways to write a sum.
$\large{\displaystyle \dfrac{1}{3} \sum_{i=0}^5 f\left(2+\dfrac{i}{3}\right) = \displaystyle \dfrac{1}{3} \sum_{i=1}^6 f\left(2+\dfrac{i-1}{3}\right)=\displaystyle \dfrac{1}{3} \sum_{i=19}^{24} f\left(2+\dfrac{i-19}{3}\right)}$