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### Calculus (Area of a Plane Region)

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Date: 5/20/96 at 23:21:56
From: Anonymous
Subject: Calculus (Area of a Plane Region)

The problem is like this:
y = 4-x2 ; x axis

a) draw a figure showing the region and a rectangular element of area;
b) express the area of the region as the limit of a Riemann sum;
c) find the limit in part b by evaluating a definite integral by the
second fundamental theorem of the calculus.
```

```
Date: 6/1/96 at 17:54:40
From: Doctor Anthony
Subject: Re: Calculus (Area of a Plane Region)

The graph of y = 4-x^2 will be an inverted parabola, passing through
the points (-2,0), (0,4) and (2,0).  We must find the area between
this curve and the x axis.  By symmetry we can find the area between
x=0 (y axis) and x=2, and then double the result.

Divide the region by n vertical strips, each of width 2/n.  For thin
strips we can consider the area of a strip as height times width, with
height given by the y value at that position (=4-x^2) and width =2/n.
The x value for the rth strip will be r(2/n) where r takes all values
from 1 to n. By summing all these areas, we obtain the area under the
curve, provided n is taken sufficiently large.

Area under curve  = SIGMA[4-4r^2/n^2](2/n)  for r = 1 to n

= (8/n)SIGMA[1 - r^2/n^2]  for r = 1 to n

= (8/n)[n - (1/n^2){n(n+1)(2n+1)/6}]

= (8/n)[n - (1/6n^2){2n^3+3n^2+n}]
= (8/n)[n - (1/6){2n+ 3 + 1/n}]
= (8/n)[n - (1/3)n + 1/2 + 1/6n]
= (8/n)[(2/3)n + 1/2 + 1/6n]
= (16/3) + 4/n + 4/3n^2

Now let n -> infinity and the last two terms tend to zero, so the area
is 16/3.  The total area required is twice this value, i.e. 32/3

If we do normal integration of 4-x^2 between -2 an +2 we have

INT[(4-x^2)dx] from -2 to +2

= [4x-x^3/3] from -2 to +2
= {8-8/3}-{-8+8/3}
= 8 - 8/3 + 8 - 8/3
= 32/3

And so this illustrates the theory that the integral gives the area
under the curve.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus

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