CHAPTER 10

### Chapter 10 – Theory of Integration

The goal of this chapter is to undo

all that has been done in the previous chapters on differentiation. We saw

how

the derivative can be used to approximate the anti-derivative, but how do

we

use it to give us the exact behavior of its anti-derivative?

Differentiation involves two main

operations. The first is subtraction or letting a change in x,

get smaller. The

second operation is division to calculate the relative change of the

dependent

variable, f with respect to the independent variable, x. Therefore, the

focus

of this chapter will be on summation and multiplication.

To begin our study, let us look at

the definition of the derivative of a function,

What this definition does is analyze

a changing function over an infinitely small interval to calculate the

rate of

change of the function at that instant. By letting

go to zero, the terms

interval and point become synonymous such that the derivative gives the

rate

of change at any point x. If we multiply both sides of the equation by dx

,where

we get:

We can calculate the approximate

change in the function,

by replacing the

infinitesimal dx with a discrete

The answer is only an approximation

because the equation assumes that the rate of change or derivative is

constant

over the interval

. For example consider the following linear function:

or is a constant. At

any point on the graph of

, the rate of change is reflected by the constant steepness

of the graph. Now let me pose a simple question. How much does the

function,

, change when x goes from 0 to 4? By evaluating

at 0 and 4 we get:

The new symbol [ means to evaluate the function at the two values

specified on the symbol and subtract the difference.

When x changes from 0 to 4,� the dependent variable, f, changes by

8. How

else could we arrive at the same answer? From the definition of the

derivative

we know that:

Over an interval

Δx,

the change in f(x), can be approximated by evaluating

for the interval. Let

us see how this applies to the function we were studying. We have shown

that as

x changes from 0 to 4, f changed from 0 to 8. Now let us find the same

answer

using the above equation:

The answer obtained from the

derivative is exact because the

rate

of change of the function is constant over the discrete interval,

But what if the derivative is

not a constant but varies with x? As we shall soon prove, the total change

of a

function f(x) from x = a to x = b is found by taking the infinite sum of

from a to b where

each

gives an

infinitely� small change in the

function, df,� over the infinitely

small

interval, dx.

Let us look at a function whose

derivative varies with x:

As x changes the rate of change of

f(x) increases. We can begin by asking ourselves what is the change in

f(x),� when x changes from 0 to 8?

In

the graph this translates to:

For the function,

, when

Δx

goes from 0 to 8,

Δf

is

given by the equation:

How else could we

calculate the change in f? From the derivative we know that:

Since the derivative of

the function changes with x, then we need to evaluate the above equation

over

small intervals of

Δx

to

get an accurate answer. Remember the rate of change of f(x) is changing

and is

not constant over a discrete interval,

Δx.

By breaking up the

derivative,

into 8 intervals of

Δx = 1, we

can approximate the change in� of

from x = 0 to

x = 8� by using the derivative only. A small

Δf of the graph is found by evaluating

over each interval,

The net change in f(x)

is found by summing up the individual approximations of

, for each interval.

Substituting known values in:

Adding up the

, we get 28. This corresponds to an error of 4 since we

calculated the actual

Δf

of f(x) to be 32 as:

To approximate this

we evaluated the

derivative,

over small

sub-intervals to calculate the net change in the function,

. Relating the instantaneous rate of change to the slope of the tangent

is:

By evaluating the

derivative over each interval

Δx

we got the approximate

Δf

of

f(x) over that interval only.

How

could we get a more accurate result? The solution is to further divide the

interval into a smaller

, such that the derivative is assumed to be constant over the infinitely small

interval. If we let

equal 0.5, we will

have sixteen intervals over which we can calculate the

You can clearly see that

by further reducing the interval, the error is difference between the

approximate and actual

is very small. By using sixteen intervals, we reduced the error from 4 to 2. We can

go on to conclude that as the interval,

, goes to zero, then the net change in f(x), will correspond

exactly to the infinite sum of the derivatives evaluated over each

interval,

Thus the net change in a

function from a to b is exactly equal to the infinite sum of its

derivative

evaluated over an infinitely small interval,

One argument

against such a conclusion is that as

decreases, the error

gets smaller for each individual interval however the total error remains

the

same. This is because as

, we are adding up an infinite sum of small errors as opposed

to a discrete sum of 8 or 10 for example. We have shown intuitively that

this

does not happen,� but how can we

prove

it?

Returning to the

definition of rate of change between two points on a graph, separated by a

distance

Notice that we are not

taking any limit as

. The above equation holds true for any two points on the

graph of

. Multiplying both sides by

gives us:

If we let

, then x is the derivative function such that:

Therefore

/2 represents the error associated with calculating

when we ignore it.

The total sum of errors from x = a to x = b, where

is:

Thus the total error for

calculating the net change in

is:

Remember that

, therefore:

This important result

confirms the fact that the sum of the derivatives evaluated over infinite

times

over infinitely small intervals of dx from x = a to x = b corresponds

exactly

to the net change in the anti-derivative, f(x). We have proven this for

the

case of

, but how do we prove it for the general case for any f(x)

and its derivative? More specifically we want to prove the following

fundamental theorem of Calculus.

Here, the integral sign,

, replaces the summation sign,

, and represents the infinite sum of the derivative evaluated

over an infinitely small interval, dx. To understand this let us take a

more

conceptual look at how the rate of change of a function is defined. The

following formula give us the average rate of change of f(x), through any

two

points on the graph of f(x).

As we let

, the two points,

come closer together.

The average rate of change converges to the instantaneous rate of change of the function at that point, x.

������

For any function,

, as

the rate of change

converges to a constant value over an infinitely small interval,

. Since the instantaneous rate of change, given by

is constant over the

interval, dx, then an infinitely

small change in f(x), df, is found by

multiplying the definition of the derivative by dx, where

Thus the net change in

is the infinite sum

of the

from x = a to x-b is:

The first term

converges to

, while the second term,

represents the net

change in the function,

from x = a to x = b,

where

. Finally the last term is the infinite sum of the derivative

evaluate over each interval, dx. This

infinite sum is represented by the integral sign,

This the fundamental theorem of Calculus.

For example if f'(x)= 2x and we sum up 2xdx as

delta x goes to zero over an interval, we get the CHANGE in the anti-derivative function

F(x) over that same interval, where F(x) = x^2 .

Next section ->

Section 11.1 – Understanding Integration