In the series: Note 8.

Version: 0.2

By: Albert van der Sel

Doc. Number: Note 8.

For who: for beginners.

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Status: Ready.

Maybe you need to pick up "some" basic "mathematics" rather

So really..., my emphasis is on "rather

So, I am really not sure of it, but I hope that this note can be of use.

Ofcourse, I hope you like my "style" and try the note anyway.

Preceding notes:

Note 1: Basic Arithmetic.

Note 2: Linear Equations.

Note 3: Quadratic Equations and polynomials.

Note 4: The sine/cosine functions.

Note 5: How to differentiate and obtain the derivative function .

Note 6: Analyzing functions.

Note 7: The e

This note: Note 8: Primitive functions or Integral.

Each note in this series, is build "on top" of the preceding ones.

Please be sure that you are on a "level" at least equivalent to the contents up to, and including, note 7.

This note will be a very short note on Primitive fuctions, and "integrals".

The "core" idea is really easy, as you will see in a minute.

Although it might seem a bit awkward at first sight. But it really is only a representation

of a "summation" of very small "elements", as you will see below.

In the expression above, you see the function f(x). Suppose that this function is a continuous

and "well behaved" function. For example, there is no asymptotic behaviour for the "x interval"

for which we will study f(x).

Let's study f(x) in the interval x ∈ [a,b], that is, x "runs" in the interval from "x=a" to "x=b".

Now, let's see what f(x) does, when x "runs" in that interval. Especially, we like to estimate

Take a look at figure 1 below. We see the function f(x) as the red curve. Also, we see the boundaries

defined from "x=a" to "x=b" (green vertical lines). Those boundaries, and the curve, determine the "area"

between the curve and the x axis.

Fig 1. The area defined by the boundaries x=a, x=b, and f(x).

For a rectangle with sides "b" and "h", we now that the area ("A") can be calculated as A=bh.

For a function as shown above, it cannot be that easy.

Well, actually, it will be quite the same procedure.

So suppose we divide the interval [a,b] into smal steps. Say that [a,b] is the range [1,9].

Then we could, for example, divide it in [1,2], [2,3], [3,4], [4,5], [5,6], [6,7], [7,8], [8,9].

Let's call such a smal interval like [2,3], Δx

is often used to denote a small interval, a sort of "delta" from a range of values.

For the example above, that is the set of [1,2], [2,3], [3,4],...,[8,9], each time

If you look at figure 1 again, you might agree that even that set of intervals, correspond to a set

of rectangles. The width of such rectangle is Δx

The height of such rectangle, then would be "f(Δx

So, the area defined by such rectangle would be A = Δx

If you would add all those up, that is add all those area's together, we could write it as:

Ofcourse, if you would make Δx

would get larger and larger.

Note: The Δ symbol, used to compactly write down a large summation. In this example: Σ

it would mean the summation k

The area of f(x) to the x-axis in the interval a ≤ x ≤ b then can be approximated with:

expression can be written as shown below. It's just a matter on how you write things down. Thus:

as in the equation above.

Only, Δx

Note: if Δx

In such case, the summation would have an extremely large number of elements. (going to infinite).

What we have established by now, is that the

where x runs from x=a to x=b, can be written as:

f(x)= |
dF(x) ---- dx |

Or, F '(x)=f(x).

Or, f(x)dx=F '(x)

Remember from note 5 on derivatives, that a derivate function is also denoted by an accent " ' ".

If indeed a function F would exist, in such a way that

of f(x). Also, an expression using ∫ f(x) dx, is called

Now, I will show you how the area actually can be calculated, as is expressed in equation 3.

The area is:

A mathematician would probably look very disturbed when he/she looks at my argument below.

However,

to be "generally" true.

Suppose we have the line f(x)=2x. Figure 2 below, illustrates the graph of y=2x.

Fig 2. The area defined by the boundaries x=a, x=b, and f(x)=2x.

This line goes through the Origin of the XY plane (0,0), and also goes through the point (2,4).

You can probably see that a right-angled triangle exist, defined by the three points (0,0), (2,0) and (2.4).

This triangle is just exactly half of the rectangle with sidelengths 2 and 4.

So, the total surface area of that rectangle is 8, and thus for the right-angled triangle the Surface area is 4.

Thus: the area between f(x) and the x-axis where x runs from x=0 up to x=4, is "4".

Now, let calculate the integral. It should return the value 4 as well.

It's important to see that if f(x)=2x, then F(x) must be x

is f(x)=2x.

Indeed, the area is 4, just as we could see from the fact that the area of the right-angled triangle is exactly half

of the area of the rectangle defined by the points (0,0), (2,0), (0,4) and (2,4).

This reasoning will always work with linear equations of the form y=ax+b

Although ofcourse equation 4 is not proven, it is (hopefully) made likely and acceptable, by providing

just one simple example.

A key point is thus, to find F(x), if you only happen to know what f(x) is.

Remember the "derivative" function, we saw in note 5? Here we found for example, that suppose we have f(x)=x

Then the

In note 5, we also discussed some methods to calculate the "derivative", for several types of functions.

The

and therefore it determines the "slope", or gradient, of the tangent line to f(x), at "x".

This time, it's a bit the otherway around. Suppose we have f(x). Then what is F(x), in such a way that F

Suppose f(x)=4x

Then, what is F(x)?

Answer:

F(x)=x

Since the derivative of F(x) is F '(x)= f(x) =4x

Here, you knew how to obtain the derivative of x

actually, "F(x)=x

We introduce a new symbol, used in the equation of finding the primitive function, namely: ∫

Suppose we have funcion f(x). Then what is F(x), in such a way that F

Suppose f(x)=x

Then:

Thus F(x) = 1/(k+1) x

Again: Just think of for example the derivative of x

Thus, the primitive of 3x

You see how to interpret the "1/(k+1) x

Thus the derivative of 1/(k+1) x

Suppose h(x)=x

Then what is the primitive function H(x)?

You can view h(x) as the sum of x

It's true since the derivative of "1/6 x

You see? The "core" idea is not hard to understand. But I agree that specific exercises can be quite spicy.

Remember the "chain" rule from note 5?

It said that the derivative of the (compound) function f(x)=u(v(x)),

is f '(x) = u '(v(x)) . v '(x)

So..., it means that whenever you "reckognize" a function that "looks" like (or might be viewed as) u '(v(x)) . v '(x)

then it's primitive is u(v(x))

Doing many excercises will make you experienced in calculating primitives. For my purpose, understanding

the 'core' idea will suffice for the moment.

calculating the "area" below the function f(x) in two dimensions, like the XY plane.

Would we do a similar action in three dimensions, then we would calculate the "volume" bounded by some surface.

Later we will see more on the latter (not this note).

My statement now is, that in many cases:

Suppse we have the line f(x)=2x.

This line goes through the Origin of the XY plane (0,0), and also goes through the point (2,4).

You can probably see that a right-angled triangle exist, defined by the three points (0,0), (2,0) and (2.4).

This triangle is just exactly half of the rectangle with sidelengths 2 and 4.

So, the total surface area of that rectangle is 8, and thus for the right-angled triangle the Surface area is 4.

Now, let's calculate the integral. It should return the value 4 as well:

Therefore, I hope that I made it "plausible" to be true in general as well.

If you believe that the example above is a representative example for a general case, then

we arrive as what is often seen as a "Fundamental Theorem of Integral Calculus":

Please remember that the indefinite integral (without the boundaries "a" and "b"),

And, the definite integral (over an interval)

Determine the definite integral of f(x)=2x

Answer:

We need to calculate:

Thus:

In note 5 we found that:

If f(x)=sin(x) then f '(x)=cos(x)

If f(x)=cos(x) yhen f '(x)= -sin(x)

Thus:

(We left out the constant "c", but you should include it).

Suppose we have f(x)=x

In general, try to rewrite the expression, so to get rid of the "(" and ")".

This way, often it is easier to apply one of the rules listed above.

x

Now we can apply our rules:

F(x)=1/4x

This is the solution, but we need to rewrite it a bit more neatly:

1/4x

F(x)=(1/4) x

Suppose we have f(x)=sin(2x). What is F(x)?

You always need a book of "tricks".

If you look at the derivative of cos(2x), then using the chain rule (see note 5), we have

-sin(2x).2 = -2sin(2x).

Thus:

F(x)=

Suppose we have f(x)=sin

First, note that sin

Now, in a former note we have seen that sin

Thus: sin

So: sin

Now, d/dx (cos(x))=-sin(x). Thus d(cos(x))=-sin(x)dx.

Thus:

sin

=(cos

Now, we use the "substitution trick": u=cos(x).

Then we get:

(cos

Thus:

The next note (note 9) is a super quick intro into "complex numbers",

since I need those too for note 10 (which is about differential equations).