Basic arithmetic/calculus.
In the series: Note 8.

Subject: Primitive functions or Integral functions.

Date : 5 March, 2016
Version: 0.2
By: Albert van der Sel
Doc. Number: Note 8.
For who: for beginners.
Remark: Please refresh the page to see any updates.
Status: Ready.

This note is especially for beginners.

Maybe you need to pick up "some" basic "mathematics" rather quickly.
So really..., my emphasis is on "rather quickly".

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 ex and ln(x) functions.

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, or sometimes called, the "antiderivatives", of a function.
Although some specific excercises can be spicy, the "core" idea is real easy, as you will see in a minute.

1. Primitive fuction: The indefinite integral f(x) dx.

Below we use the simple assumption that f(x) is a "well-behaved" function, with no gaps, and no asymptotes in the "neighborhood".

Remember the "derivative" function, we saw in note 5? Here we found for example, that suppose we have f(x)=x2 + b.

Then the "derivative" function is f '(x)=2x.

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

The interpretation of the "derivative" function was, that it described "the rate of change of f(x) : rate of change of x",
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 '(x)=f(x)?


Suppose f(x)=4x3

Then, what is F(x)?



Since the derivative of F(x) is F '(x)= f(x) =4x3

Here, you knew how to obtain the derivative of x4, which is 4x3

actually, "F(x)=x4 + c", where c is a constant. The derivative of a constant is zero.

Notation of finding the primitive (integral) function:

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 '(x)=f(x)? Thus f(x) is the "derivative" function of F(x).

f(x) dx = F(x) + c

So, F(x) is called the antiderivative, or primitive function of f(x). It's just a way to write down stuff.

Some rules (among many rules) of finding the primitive (integral) function:

- rule 1:

Suppose f(x)=xk


xk dx = 1/(k+1) xk+1 + c

This is true since the derivative of 1/(k+1) xk+1 + c = xk

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

Again: Just think of for example the derivative of x3. The derivative is 3x2
Thus, the primitive of 3x2, is ofcourse x3.
You see how to interpret the "1/(k+1) xk+1" stuff?

Thus the derivative of 1/(k+1) xk+1 = (k+1) 1/(k+1) xk+1-1 = xk

- rule 2:

a f(x) dx = a f(x) dx

where a is some constant.

- rule 3:

( f(x) + g(x) ) dx = f(x) dx + g(x) dx


Suppose h(x)=x5 + x3

Then what is the primitive function H(x)?

You can view h(x) as the sum of x5 and x3. So we can use rule 3, but rule 1 too:

H(x)=1/6 x6 + 1/4 x4

Indeed, you may say that we need to add a constant "c" too.

It's true since the derivative of "1/6 x6 + 1/4 x4" is indeed "x5 + x3".

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

- rule 4:

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.

2. Integration over an interval "a ≤ x ≤ b": ab f(x) dx.

This time, we will see an interpretation too. As we will see, integration over an interval means
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:

ab f(x) dx = F(b) - F(a) = Area of f(x) in interval a ≤ x ≤ b

Ofcourse, it can be proved in a formal way. However, I like to provide an illustrative example.


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 calculate the integral. It should return the value 4 as well:

2x dx = x2


ab f(x) dx = F(b) - F(a) = F(2) - F(0) = 22 - 02 = 4

So indeed. The integral returns the same value, as the Surface area we calculated before.
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":

ab f(x) dx = F(b) - F(a) = Area of f(x) in interval a ≤ x ≤ b

By the way, F(b) - F(a) is also sometimes denoted as [F(x)]ab

Please remember that the indefinite integral (without the boundaries "a" and "b"), returns primitive functions.
And, the definite integral (over an interval) returns a value, to be intepreted as the Surface area below f(x).


Determine the definite integral of f(x)=2x3 -5, in the interval 1 ≤ x ≤ 3.


We need to calculate:

13 (2x3 -5) dx

The primitive function of f(x)=2x3 -5, is F(x)=1/2x4 -5x

13 (2x3 -5) dx = F(3) - F(1) = (1/2 * 34 - 5 * 3) - (1/2 * 14 - 5 * 1) = 30

That's it ! Hope you liked it.

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).