# What is the anti-derivative of the position

**content**

»Preliminary remark

“The first problem

»Upper and lower sums as approximations

»Upper and lower sums in the limit value

“The definite integral

»The antiderivative and its application

“The indefinite integral

" Remarks

##### Preliminary remark

Using a constructive example, we are trying to get the various terms of integral calculus under one roof and explain their meaning (and their sense). We then deal with the most important integration formulas.

##### The first problem

A property owner would like to calculate the area of his property on a coast, because he no longer knows the exact area of that. Due to the course of the coast, this project is not trivial. With the help of a GPS receiver, he measures some points along the coast

and can thereby define the coast function \ (f \) and describe his property mathematically (in meters) (how to get the coast function \ (f \) is another, very interesting topic in mathematics):

\ begin {align *}

f (x) = & -4.0 \ times 10 ^ {- 9} x ^ 7 + 7.8 \ times 10 ^ {- 7} x ^ 6-5.5 \ times 10 ^ {- 5} x ^ 5 + 1.6 \ cdot 10 ^ {- 3} x ^ 4 - \

& -1.3 \ times 10 ^ {- 2} x ^ 3-1.5 \ times 10 ^ {- 1} x ^ 2 + 2.2 \ times 10 ^ {- 1} x ^ 2 + 60.

\ end {align *}

The (rounded, see notes) function does not seem very handy at first glance, but we will see that the function equation will not be very important to us when discussing the terms. In addition, for the sake of simplicity, we will primarily work with rounded values, the exact readers please forgive us at this point. So our coast becomes a coast function in a coordinate system

The owner now wants to calculate the area between the function \ (f \) and the \ (x \) axis in the interval \ ([0; 35] \).

##### Upper and lower sums as approximations

Since the integral calculus is still unknown to him, he estimates. He knows the longest and shortest part of his property:

This allows him to estimate the area upwards and downwards using two rectangles.

In this case, the width of the rectangle is \ (\ Delta x = 35-0 = 35 \). The heights are given by the highest or lowest value of the interval \ ([0; 35] \), (rounded) 60 and 25. For the actual area \ (F_A \) of the property \ (A \) applies

\ begin {align *}

25 \ cdot 35 \ leq & F_A \ leq 60 \ cdot 35 \

875 \ leq & F_A \ leq 2100.

\ end {align *}

This is now a very large window and we want to improve the approximation. One possibility would be to use the arithmetic mean \ (\ frac {875 + 2100} {2} \). However, we go a different way and want to generalize the idea of rectangles. Instead of approaching the function with a rectangle from above and one from below, we now do this with two each. To do this, we now select the width of the rectangle \ (\ Delta x = \ frac {35-0} {2} = 17.5 \) and the highest or lowest value of the intervals \ ([0; 17.5) as the height ] \) and \ ([17.5; 35] \).

If we read off the rounded values, the area \ (F_A \) now lies between

\ begin {align *}

28.25 \ times 17.5 + 25 \ times 17.5 \ leq & F_A \ times 60 \ times 17.5 + 35 \ times 17.5 \

931.88 \ leq & F_A \ leq 1662.5.

\ end {align *}

We call the sum of the upper rectangles the upper and the lower lower sum, \ (O_2 \) and \ (U_2 \). It is intuitively clear to us that we can continue this procedure and we want to formalize this procedure a little in the following.

##### Upper and lower sums in the limit value

The next step would be to define the upper and lower sums for \ (n = 3 \) rectangles. The widths of the rectangles are then each \ (\ Delta x = \ frac {35-0} {3} \) and the respective heights are again determined by maximum and minimum in these intervals. (Finding these maximas and minimas is not trivial. However, we only need their existence here, and this is what Weierstrass' theorem gives us, see Notes.)

The two sums for \ (O_3 \) and \ (U_3 \) approximate our area even better. We can continue this game and the actual area is "wedged" closer and closer between the upper and lower sums.

Try it yourself in the interactive GeoGebra file.

We can therefore approximate the surface from above and below using two sequences \ (O_n \) and \ (U_n \) and "squeeze in" the actual surface. The first question that arises is, "Can this always be done"? No, generally not! However, one can show that it works for all continuous (and even more) functions and that the two sequences converge towards the same value, the actual area.

##### The definite integral

The area of a continuous function \ (f \), with \ (f> 0 \), in the interval \ ([a; b] \) is then given by the so-called definite integral

\ begin {align *}

\ int_a ^ b f (x) dx.

\ end {align *}

Where does the spelling come from? Define the upper and lower sums \ (O_n \) and \ (U_n \) with the help of the summation sign \ (\ sum \) (where \ (x_i \) are the minimas and \ (x_j \) are the maximums of the intervals)

so we get

\ begin {align *}

U_n \ leq & \ int_a ^ b f (x) dx \ leq O_n \

\ sum_ {i = 1} ^ {n} f (x_i) \ cdot \ Delta x \ leq & \ int_a ^ bf (x) dx \ leq \ sum_ {j = 1} ^ {n} f (x_j) \ cdot \ Delta x \

\ vdots & \ lim_ {n \ to \ infty} \ vdots \

\ sum_ {i = 1} ^ {\ infty} f (x_i) \ cdot \ Delta x \ leq & \ int_a ^ bf (x) dx \ leq \ sum_ {j = 1} ^ {\ infty} f (x_j) \ cdot \ Delta x \

\ end {align *}

The integral can thus be understood as "the infinite sum \ (\ int \) of infinitely narrow strips with width \ (dx \) and height \ (f (x) \)". The integral sign \ (\ int \) arose from the sum sign \ (\ sum \) and the \ (dx \) are the infinitely narrow strips resulting from \ (\ Delta x = \ frac {ba} {n} \) for \ (n \ to \ infty \). Of course, this section measures a lot of mathematical accuracy, you will find the complete description under the notes, only the spelling should be explained.

Why do we assume \ (f> 0 \)? We will see later (in another example) that a certain integral can also be negative and of course this contradicts our intuitive concept of area. The definite integral can also be defined separately from the area calculation. How do you calculate certain integrals and, in our context, the area? In a different context in the previous section we looked for a function \ (s \) to \ (v \) to calculate the distance of a train. Motivated by physics, we demanded with regard to \ (s \) that \ (s' = v \) applies. Later we noticed that the formula for calculating the area under the graph of \ (f \) has the same shape as \ (s \) and that the area under \ (f \) therefore calculates the value of the distance from \ (v \) using \ (s \), corresponds to. The function \ (s \) with \ (s' = v \) is called an antiderivative of \ (v \) and the connection between certain integral and antiderivative is explained by the main theorem of differential and integral calculus.

##### The antiderivative and its application

We call a differentiable function \ (F \) with \ (F '= f \) an antiderivative \ (F \) of \ (f \), a little more precisely, an antiderivative \ (F \) satisfies

\ begin {align *}

F '(x) = f (x) \ quad (\ text {for all} x \ in D_f).

\ end {align *}

Why do we write "a"? If we have found an antiderivative, there automatically exist an infinite number. As a short digression, \ (F_1 (x) = x ^ 2 \) and \ (F_2 (x) = x ^ 2 + 1 \) both \ (F_1 '(x) = F_2' (x) = 2x \) )! So they are both antiderivatives to \ (f (x) = 2x \). It is then intuitively clear to us that two antiderivatives \ (F_1 \) and \ (F_2 \) of a function \ (f \) only differ by one constant \ (C \):

\ begin {align *}

F_1 (x) -F_2 (x) = C.

\ end {align *}

If a function \ (f \) (\ (f> 0 \)) has an antiderivative \ (F \), we can calculate the area with \ (F (b) -F (a) \). This connection explains an application of the main theorem of differential and integral calculus, the so-called Newton-Leibniz formula:

Let \ (f \) be a continuous function on an interval \ ([a; b] \) and \ (F \) an antiderivative, then

\ begin {align *}

\ int_a ^ b f (x) dx = F (b) -F (a).

\ end {align *}

Does an antiderivative always exist? As mentioned before, one can show that every function continuous on an interval \ ([a; b] \) has an antiderivative and we get it via

\ begin {align *}

F (x) = \ int_a ^ x f (t) dt.

\ end {align *}

We come back to our example. The neighbor of the property owner wants to sell part of his property, our property owner can thereby widen his property. Depending on the purchased width \ (x \) and \ (f \), he receives the added area. He now wants to create a function \ (F \) that calculates this area. Starting from the definite integral, the area calculation there and our new theory of antiderivatives, we get

\ begin {align *}

F (x) = \ int_ {35} ^ x f (t) dt.

\ end {align *}

Does this function work the way we want? Yes, \ (F \) calculates the area in the interval \ ([35; x] \) depending on \ (x \). The left border, in our case 35, is "held". Since our independent variable \ (x \) is in the integrant, \ (f (x) \) and \ (dx \) are called \ (f (t) \) and \ (dt \) um, instead of \ (t \) is also often chosen \ (u \) or \ (s \),

\ begin {align *}

F (x) = \ int_ {35} ^ x f (t) dt = \ int_ {35} ^ x f (u) du = \ int_ {35} ^ x f (s) ds.

\ end {align *}

With this we now calculate our additional area.

To the interactive GeoGebra file

##### The indefinite integral

If there is a definite integral, what is the so-called indefinite integral? Here the literature and the teaching are not entirely in agreement. We try to illustrate an often common possibility in the following example: Suppose our landowner wants to pave a path to the sea. He knows the breadth, but is not yet sure of the exact position.

The respective areas to be paved are through

\ begin {align *}

\ int_8 ^ {10} f (x) dx, \ quad \ int_ {20} ^ {22} f (x) dx, \ quad \ int_ {29} ^ {31} f (x) dx

\ end {align *}

given. Should he now use the three antiderivatives

\ begin {align *}

\ int_8 ^ {x} f (t) dt, \ quad \ int_ {20} ^ {x} f (t) dt, \ quad \ int_ {29} ^ {x} f (t) dt

\ end {align *}

form to calculate the areas? If we only have a single antiderivative \ (F \) for \ (f \), we can quickly calculate these areas using

\ begin {align *}

F (10) -F (8), F (22) -F (20), F (31) -F (29).

\ end {align *}

Why again? If we read the term \ (F (31) -F (29) \) again, then this is "the area from \ (a \) to 31 minus the area from \ (a \) to 29, there remains the area between 31 and 29 left ". When looking for an antiderivative (or the set of all antiderivatives) one often simply writes the indefinite integral

\ begin {align *}

\ int f (x) dx = F (x) + C, \ qquad C \ in \ mathbb {R}

\ end {align *}

and chooses "the simplest" antiderivative for arithmetic problems. In the following chapter we will consider what this is in general with a few calculation examples.

##### Remarks

Our function \ (f \) in this chapter is an approximation polynomial through the points drawn. Geogebra's functional term is rounded to ten decimal places.

\ begin {align *}

f (x) = & - 0.000000004x ^ 7 + 0.0000007817x ^ 6 - 0.0000553662x ^ 5 + 0.0016378182x ^ 4 - \

& -0.0133522286x ^ 3-0.1540381437x ^ 2 + 0.2172940595x + 60,

\ end {align *}

As we have seen, however, the complexity and execution of the terms was independent of the functional term.

Are differentiating and integrating a function and an inverse function? Strictly speaking, no, because integrating itself over the indefinite integral is not a function, it maps it to a number of (parent) functions. The main theorem of differential and integral calculus explains the exact relationship (often also called the fundamental theorem of analysis), which says the following:

\ begin {align *}

F (x) = \ int_ {x_0} ^ x f (t) dt

\ end {align *}

is an antiderivative from \ (f \) to \ (I \), so \ (F '(x) = f (x) \) holds for all \ (x \ in I \). Further applies

\ begin {align *}

\ int_ {a} ^ b f (t) dt = F (b) -F (a).

\ end {align *}

Weierstrass' theorem guarantees that when forming the lower and upper sums on the (decreasing) closed intervals, the respective maximum and minimum exist:

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