Definite integrals

Now we can integrate polynomial functions. Other familiar functions are discussed in later chapters. In the meantime, let us look at what benefits the integral function has for us.

The derivative of a function tells us about the change in the original function. If the value of the derivative at some point is a large positive number, the function increases rapidly. Or if the derivative is a small negative number, the function decreases slowly, etc.

In this and the next chapter, we learn that an integral function can be used to calculate the area under the graph of a function at some point. The integral thus describes some kind of quantity or “accumulation” - how many “functions” there are, for example, between [0,1].

In physics, we learn that velocity means the rate of change of a place function - that is, a velocity function is a derivative of a place function. The inverse relationship holds for the integral, if we know the velocity function, we get the distance travelled by integrating the velocity.

The area bounded by the function and the x-axis

The idea is this: look at the area bounded by the function f(x) and the x-axis and try to estimate how large it is in some interval. To facilitate the evaluation, it is divided into slices evenly spaced. The figure shows some example function, the bounded area of which we want to determine between [0,1].

One way to make a rough estimate is to think that the area of one slice is the area of the rectangle where the base is the width of the gap and the height is the value of the function f (x) at the midpoint of the gap [or at the start or end point, this is irrelevant to the end result] , i.e. roughly

In the figure, for example, the area of the fifth slice would be

and the width of the gap is

So the area is

The same goes for the surface area of the other slices. The total area is, of course, obtained as the sum of these, which can be formally written as:

The sum passes over each slice. There are nine slices in the figure above, so i gets the values i = 1,2,3, ..., 9. The smaller the slices we divide the area into, the more accurate an estimate of the area we get. The exact value is obtained as the limit value when the gap shrinks to zero and there are an infinite number of slices. Just as the derivative was defined as the limit value of the separation quotient when the length of the gap approaches zero, the so-called “definite integral” as a limit for the above sum.

Take the positive continuous function f (x) and the general interval [a, b], the definite integral of the function is

Even the integral sign originally came from that sum symbol. It is therefore always useful to think that when integrating, something concrete is summed up.

The area bounded by a positive function would be obtained, but then how is that integral calculated? For this we use the following result, the so-called “Fundamental theorem of calculus”, the proof of which can be found below:

In the middle stage, the so-called “substitute line” is used to help with calculation process because the calculation of a given integral proceeds in practice as follows:

1. We find the integral interval and the function f(x) from the question.

2. We integrate the function f(x), i.e. find all its integral functions.

3. Place the end point b of the interval and then the start point a of the interval in the obtained integral function. Calculate the difference between the two.

Example 1

Notice how the integration constant C cancels out of the calculation. For definite integrals, we don't need to write the constant C.

Proof of Fundamental theorem of calculus

Before we start calculating definite integrals, two things are done. Justify the result that the integral function F(x) of the function f(x) is one whose derivative returns the original function, i.e.

In addition, the connection between a definite integral and an integral function, i.e. the “Fundamental theorem of calculus”, is derived.

Let us examine again the area bounded by the function f(x) by means of the sum. Start from zero and add the slices to some point x. This allows us to define an “area function”.

SUM OF THE AREAS

The variable denoting the slice was changed to t so that it would not be confused with the x denoting the endpoint of the interval. So now the further x is taken, the area grows larger.

INTEGRAL

An integral is obtained for a small interval x -> 0

Which depends on the end point x of the interval.

Let’s go back to the slices and add a small piece to the end of the sum function above and see how much the area increases.

Since we are in the end of the interval, an estimate of the increase in area is obtained

From this, the ratio of the area to the change in the length of the gap is solved

The left side of the expression is the difference quotient of the surface area A(x). By a small gap limit value the derivative is obtained (with the same initial estimates will be accurate results).

We were able to show that the derivative of the area function returns the original function, so the area function is its integral function.

Let's continue a little more. Once the area function is an integral function of the function f (x), then

Adding the integration constant C to the first row gives the result

Let's swap the symbols in the formula so that x → b and t → x, this will change nothing but the names of the symbols. We could also start the area accumulation instead of zero at point a.. As a whole, we have proved the “Fundamental theorem of calculus” [The proof does not need to be remembered, but it helps to understand definite integrals and is part of the general education of mathematics]

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