## Monday, May 7, 2007

### Precalculus 12.5b Area of a Plane Region

12.5b Area of a Plane Region

Area of a Region bounded by:

x = a, x = b, y = 0 and y = f (x) where f(x) is less than or equal to 0.

The boundary of the area is from x0 = a to xn = b.
What is the area under the curve?
George Riemanns - German who came up with technique to find the orange area.

1. Subdivide the interval from a to b into smaller intervals. This is called a partition of [a,b] specifically:

a = x0 ≤ x1 ≤ x2 ≤ x3 ... ≤ xn - 1 ≤ xn = b

2. Choose a point wi in interval [xi - 1, xi] , i = 1 ... n

so...

when i = 1, w1 is contained in [x0, x1]

when i = 2, w2 is contained in [x1, x2]
3. Let Δ xi = xi - xi-1 for i = 1 ... n

i= 1; Δ x1 = x1 - x0

i = 2; Δ x2 = x2 - x1
Δ xi = length of the ith subinterval.

4. Form f(wi)(Δxi) for i = 1 ... n

5. The area of the region:

A = f (w1)(Δ x1) + f(w2)(Δ x2) + ... + f (wn)(Δxn)

This is called a Riemann's Sum.

So by increasing the number of rectangles that you make, you can obtain a closer and closer approximation

Therefore to find the Area of a Plane Region:

Let "f" be continuous function and nonnegative on the interval [a, b]. The Area A of the region bounded by the graph of "f", the x-axis (y = 0), and the vertical lines x = a and x = b is

Area of a rectangle = height times width

Example 1:
Find the Area of the Region: f(x) = 2 - x2, -1 £ x £ 1

Let's let n = 4

so the width = (1 - -1)/4 = 1/2

height =
[-1, -1/2] = 1.75
[-1/2, 0] = 2
[0, 1/2] = 1.75
[1/2, 1] = 1

so the Area when f(x) = 2 - x2 and is divided up into 4 rectangles =
(1/2)(1.75) + (1/2)(2) + (1/2)(1.75) + (1/2)(1) = 3.25

but to find the area with more rectangles, let's use the above formulas:

Therefore to find the area, use the summation formulas:

To check this on the graphing calculator:

put in y1 = 2 - x2

Now on your homescreen, math arrow down to #9 fnInt (Y1, x, -1, 1) enter

should give you 10/3.

In the calculator (function , variable, lower bound, upper bound) = area under the curve.