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 x_n = 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 ... ≤ x_{n - 1} ≤ x_n = b

2. Choose a point w_i in interval [x_{i - 1}, x_i] , 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 x_i = x_i - x_i-1 for i = 1 ... n

i= 1; Δ x₁ = x₁ - x₀

i = 2; Δ x₂ = x₂ - x₁

Δ x_i = length of the i^th subinterval.

4. Form f(w_i)(Δx_i) for i = 1 ... n

5. The area of the region:

A = f (w₁)(Δ x₁) + f(w₂)(Δ x₂) + ... + f (w_n)(Δx_n)

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

check out this website: http://tutorial.math.lamar.edu/classes/calcI/areaproblem.aspx

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.

Precalculus 12.5 The Area Problem

12.5 The Area Problem

I. Limits of Summation

S = a₁ + a₁r + a₁r² + ... =

_¥
å (a₁(r^i-1)) =
^{i = 1}

a₁/(1-r) provided that the absolute value of r < 1

Using Limit Notation, this sum can be written as:

S =
¥
lim å (a₁ r^i-1)
^{n®¥ i = 1}

= lim a₁ (1 - rⁿ)/(1-r)
^n®¥

Recall that since r < 1 thus making it a fraction, when you take the limit or rⁿ as x approaches infinity, this would equal zero. So:

= (a₁ (1 - 0))/ (1 - r)

= a₁/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of
i² when i = 1 to i = 30?

=(30)(30 + 1)(2(30) + 1)/6 = (30)(31)(61)/6 = 56730/6 = 9455

Example 2: Evaluating a Summation:

What is the sum of:
(2i + 1) when i = 1 to i = 50?

You have to break this down into parts:

2( å i ) from i = 1 to i = 50 AND
å(1) from i = 1 to i = 50 is equal to

2(50(50+1)/2) + 1(50) = (50)(51) + 50 = 2550 + 50 = 2600

II. Finding the Limit of a Summation

_n
å (2i + 3)/(n²)
^{i = 1}

= (1/n²)(2(n(n+1))/2) + 3n) = (1/n²)(n(n+1) + 3n)
= (1/n)(n + 1 + 3) = (n + 4)/n

which in essence is infinity over infinity so
Therefore the Summation of
lim S_n =
^n®¥

lim (n + 4)/n = 1/1 = 1 so
^n®¥

lim S_n = 1
^n®¥

You can check this by using a table of values:

(n, S(n)) = {(1, 5), (10, 1.4), (100, 1.04), (1000, 1.004), (10000, 1.0004)...}
So you can see the values of S(n) approach 1.

Example 3: Finding the Limit of a Summation:

_n
å (3/n³)(1 + i²)=
^{i = 1}

= (3/n³)(n + (n(n + 1)(2n + 1)/6) =
(3/n² + (3(2n² + 3n + 1)(6n²)
= 3/n² + (6n² + 9n + 3)/(6n²)

lim 3/n² = 0 and
^{n ®¥}

lim (6n² + 9n + 3)/(6n²) = 6/6 = 1
^{n ®¥}

So:

lim S_n = 1
^{n ®¥}

Again, you can check this using a table of values:
(n, S(n)) = {(1, 6), (10, 1.185), (100, 1.0154), (1000, 1.0015), (10000, 1.0002)...}

Precalculus 12.4 Limits at Infinity and Limits of Sequences

12.4 Limits at Infinity and Limits of Sequences

Definition of Limits at Infinity
If "f" is a function and L₁ and L₂ are real numbers, the statements

lim f(x) = L₁
^x®-¥

AND

lim f(x) = L₁
^x®+¥

denote the limits at infinity. The first is read "the limit of f(x) as x approaches negative infinity is L₁," and the second is read "the limit of f(x) as x approaches infinity is L₂"

To help evaluate limits at infinity, you can use the following:

Limits at Infinity
If "r" is a positive real number, then

lim (1/(x^r)) = 0 Limit toward the right.
^x®+¥

Furthermore, if x^r is defined when x is less then 0, then

lim (1/(x^r) = 0 the limit toward the left.
^{x®- ¥}

Thought: -1/10 = -.1, -1/100 = -.01, -1/1000 = -.001, -1/10000 = -.0001
so as you can see, in the denominator as x approaches negative infinity, f(x) approaches zero.

Example 1: Evaluating a Limit at Infinity:

lim (5/(2x)) = 0
^x®¥

As you can see when you graph this function, as x gets larger, f(x) gets closer to zero so that is how we can conclude the the limit equals zero.

Example 2:
lim (1 - 3/(x²)) =
^x®¥

We can separate each part of the function to:

lim (1 )= 1
^x®¥

and

lim (3/(x²)) = 0
^x®¥

so
lim (1 - 3/(x²)) = 1 - 0 = 1
^x®¥

Compare the following limits:
Find the limit as x approaches infinity:

a). f(x) = (-5x + 2)/(4x² - 1)

b). g(x) = (4x² + 1)/(3x² - 1)

c). j(x) = (-3x⁴ + 1)/(2x³-1)


When you take the limit of each of these functions as x approaches infinity, you get:

1st, look in the denominator and find the largest exponent and use this term by dividing all the terms of the function by that variable raised to the largest exponent.

a).
lim (-5x + 2)/(4x² - 1)=
^x®¥

so divide each term by x²

lim [(-5x)/(x²) + 2/(x²)]/[(4x²)/(x²) - 1/(x²)]
^x®¥

= (-0 + 0)/(4 - 0) = 0

b).
lim (4x² + 1)/(3x² - 1)=
^x®¥

so divide each term by x²

lim [4 + 1/(x²)]/[3 - 1/(x²)]=
^x®¥

(4 + 0)/(3 - 0) = 4/3

c).
lim (-3x⁴ + 1)/(2x³-1)
^x®¥

and so divide each of the terms by x³

lim (-3x + 1/x³)/(2 - 1/x³)=
^x®¥

-3x/2 so by putting infinity in for x, the limit does not exist because the numerator decreases without bound as the denominator approaches 2.

Therefore we can conclude:

Limits at Infinity for Rational Functions

for the rational function f(x) = N(x)/D(x) when

N(x) = a_nxⁿ + ... + a₀

And

D(x) = b_mx^m + ... + b₀

the limit as x approaches positive or negative infinity is as follows:

lim f(x) = 0, when n is less than m
^x®¥

lim f(x) = a_n/b_m, when n = m
^x®¥

And if n is greater than m, the limit does not exist.

Limit of a Sequence:
Let "f" be a function of a real variable, such that
lim f(x) = L
^x®¥

If {a_n} is a sequence such that f(n) = a_n
for every positive integer "n", then the
lim a_n = L
^n®¥

Example:
Given: 1/3, 1/9, 1/27, 1/81, ...

As n increases without bound, the terms of the sequence
a_n = 1/3ⁿ is said to

CONVERGE to zero.

lim 1/3ⁿ = 0
^n®¥

___________________________________________

A sequence that does not converge is said to DIVERGE!

Example of this is: the sequence 1, -1, 1, -1, ... this diverges.

Find the limit of the Sequence:

Example 1:
lim (4x - 1)/(x + 3)=
^x®¥
{3/4, 7/5, 11/6, 15/7, ...} = 4

Example 2:
lim (3x + 2)/(2x² - 1)=
^x®¥
{5/1, 8/7, 11/17, 14/31, ...} = 0

Example 3:
lim (7x² - 1)/(8x²)=
^x®¥
{6/8, 27/32, 62/72, 111/128, ...} = 7/8

Use the site: http://www.calculus-help.com/funstuff/phobe.html

Chapter 1, lesson 4 to help with this lesson.