This lesson is the first in a series of videos on integration. In it we will analyze what an antiderivative of a function is, and also study the elementary methods of calculating these very antiderivatives.
In fact, there is nothing complicated here: essentially it all comes down to the concept of derivative, which you should already be familiar with. :)
I’ll note right away that since this is the very first lesson in our new topic, there won't be any today complex calculations and formulas, but what we will study today will form the basis for much more complex calculations and constructions when calculating complex integrals and squares.
In addition, when starting to study integration and integrals in particular, we implicitly assume that the student is already at least familiar with the concepts of derivatives and has at least basic skills in calculating them. Without a clear understanding of this, there is absolutely nothing to do in integration.
However, here lies one of the most common and insidious problems. The fact is that, when starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, in exams and independent work stupid and offensive mistakes are made.
Therefore, now I will not give a clear definition of an antiderivative. In return, I suggest you see how it is calculated using a simple specific example.
What is an antiderivative and how is it calculated?
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is calculated simply:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look carefully at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can write it this way, according to the definition of a derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of an antiderivative. But to write it correctly, you need to write the following:
Let us write the following expression in the same way:
If we generalize this rule, we can derive the following formula:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now we can formulate a clear definition.
An antiderivative of a function is a function whose derivative is equal to the original function.
Questions about the antiderivative function
It would seem a fairly simple and understandable definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, okay, this formula is correct. However, in this case, with $n=1$, we have problems: “zero” appears in the denominator, and we cannot divide by “zero”.
- The formula is limited to degrees only. How to calculate the antiderivative, for example, of sine, cosine and any other trigonometry, as well as constants.
- Existential question: is it always possible to find an antiderivative? If yes, then what about the antiderivative of the sum, difference, product, etc.?
On last question I'll answer right away. Unfortunately, the antiderivative, unlike the derivative, is not always considered. No such thing universal formula, by which from any initial construction we will obtain a function that will be equal to this similar construction. As for powers and constants, we’ll talk about that now.
Solving problems with power functions
\[((x)^(-1))\to \frac(((x)^(-1+1)))(-1+1)=\frac(1)(0)\]
As we see, this formula for $((x)^(-1))$ does not work. The question arises: what works then? Can't we count $((x)^(-1))$? Of course we can. Let's just remember this first:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has studied this topic at least a little will remember that this expression is equal to the derivative of the natural logarithm:
\[((\left(\ln x \right))^(\prime ))=\frac(1)(x)\]
Therefore, we can confidently write the following:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
You need to know this formula, just like the derivative of a power function.
So what we know so far:
- For a power function - $((x)^(n))\to \frac(((x)^(n+1)))(n+1)$
- For a constant - $=const\to \cdot x$
- A special case of a power function is $\frac(1)(x)\to \ln x$
And if we start multiplying and dividing the simplest functions, how then can we calculate the antiderivative of a product or quotient. Unfortunately, analogies with the derivative of a product or quotient do not work here. Any standard formula does not exist. For some cases, there are tricky special formulas - we will get acquainted with them in future video lessons.
However, remember: general formula, a similar formula for calculating the derivative of a quotient and a product does not exist.
Solving real problems
Task No. 1
Let's each power functions Let's calculate separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Problem No. 2
As I already said, prototypes of works and private “right through” are not considered. However, here you can do in the following way:
We broke down the fraction into the sum of two fractions.
Let's do the math:
The good news is that knowing the formulas for calculating antiderivatives, you are already able to calculate more complex designs. However, let's go further and expand our knowledge a little more. The fact is that many constructions and expressions, which, at first glance, have nothing to do with $((x)^(n))$, can be represented as a power with rational indicator, namely:
\[\sqrt(x)=((x)^(\frac(1)(2)))\]
\[\sqrt[n](x)=((x)^(\frac(1)(n)))\]
\[\frac(1)(((x)^(n)))=((x)^(-n))\]
All these techniques can and should be combined. Power expressions Can
- multiply (degrees add);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving power expressions with rational exponent
Example #1
Let's calculate each root separately:
\[\sqrt(x)=((x)^(\frac(1)(2)))\to \frac(((x)^(\frac(1)(2)+1)))(\ frac(1)(2)+1)=\frac(((x)^(\frac(3)(2))))(\frac(3)(2))=\frac(2\cdot (( x)^(\frac(3)(2))))(3)\]
\[\sqrt(x)=((x)^(\frac(1)(4)))\to \frac(((x)^(\frac(1)(4))))(\frac( 1)(4)+1)=\frac(((x)^(\frac(5)(4))))(\frac(5)(4))=\frac(4\cdot ((x) ^(\frac(5)(4))))(5)\]
In total, our entire construction can be written as follows:
Example No. 2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore we get:
\[\frac(1)(((x)^(3)))=((x)^(-3))\to \frac(((x)^(-3+1)))(-3 +1)=\frac(((x)^(-2)))(-2)=-\frac(1)(2((x)^(2)))\]
In total, collecting everything into one expression, we can write:
Example No. 3
To begin with, we note that we have already calculated $\sqrt(x)$:
\[\sqrt(x)\to \frac(4((x)^(\frac(5)(4))))(5)\]
\[((x)^(\frac(3)(2)))\to \frac(((x)^(\frac(3)(2)+1)))(\frac(3)(2 )+1)=\frac(2\cdot ((x)^(\frac(5)(2))))(5)\]
Let's rewrite:
I hope I won't surprise anyone if I say that what we've just studied is just the most simple calculations primitive, the most elementary structures. Let's now look a little more complex examples, in which, in addition to the tabular antiderivatives, you will also need to remember school curriculum, namely, abbreviated multiplication formulas.
Solving more complex examples
Task No. 1
Let us recall the formula for the squared difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the prototype of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
Let's put everything together into a common design:
Problem No. 2
In this case, we need to expand the difference cube. Let's remember:
\[((\left(a-b \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Taking this fact into account, we can write it like this:
Let's transform our function a little:
We count as always - for each term separately:
\[((x)^(-3))\to \frac(((x)^(-2)))(-2)\]
\[((x)^(-2))\to \frac(((x)^(-1)))(-1)\]
\[((x)^(-1))\to \ln x\]
Let us write the resulting construction:
Problem No. 3
At the top we have the square of the sum, let's expand it:
\[\frac(((\left(x+\sqrt(x) \right))^(2)))(x)=\frac(((x)^(2))+2x\cdot \sqrt(x )+((\left(\sqrt(x) \right))^(2)))(x)=\]
\[=\frac(((x)^(2)))(x)+\frac(2x\sqrt(x))(x)+\frac(x)(x)=x+2((x) ^(\frac(1)(2)))+1\]
\[((x)^(\frac(1)(2)))\to \frac(2\cdot ((x)^(\frac(3)(2))))(3)\]
Let's write the final solution:
Now attention! Very important thing, with which it is connected lion's share errors and misunderstandings. The fact is that until now, counting antiderivatives with the help of derivatives and bringing transformations, we did not think about what the derivative of a constant is equal to. But the derivative of a constant is equal to “zero”. This means that you can write the following options:
- $((x)^(2))\to \frac(((x)^(3)))(3)$
- $((x)^(2))\to \frac(((x)^(3)))(3)+1$
- $((x)^(2))\to \frac(((x)^(3)))(3)+C$
This is very important to understand: if the derivative of a function is always the same, then the same function has an infinite number of antiderivatives. We can simply add any constant numbers to our antiderivatives and get new ones.
It is no coincidence that in the explanation of the problems that we just solved it was written “Write down general form primitives." Those. It is already assumed in advance that there is not one of them, but a whole multitude. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks we will correct what we did not complete.
Once again we rewrite our constructions:
In such cases, you should add that $C$ is a constant - $C=const$.
In our second function we get the following construction:
And the last one:
And now we really got what was required of us in the original condition of the problem.
Solving problems of finding antiderivatives with a given point
Now that we know about constants and the peculiarities of writing antiderivatives, it is quite logical that next type problems when, from the set of all antiderivatives, it is required to find one single one that would pass through given point. What is this task?
The fact is that all antiderivatives of a given function differ only in that they are shifted vertically by a certain number. And this means that no matter what point on coordinate plane we didn’t take it, one antiderivative will definitely pass, and, moreover, only one.
So, the problems that we will now solve are formulated as follows: not just find the antiderivative, knowing the formula of the original function, but choose exactly the one that passes through the given point, the coordinates of which will be given in the problem statement.
Example #1
First, let’s simply count each term:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((x)^(3))\to \frac(((x)^(4)))(4)\]
Now we substitute these expressions into our construction:
This function must pass through the point $M\left(-1;4 \right)$. What does it mean that it passes through a point? This means that if instead of $x$ we put $-1$ everywhere, and instead of $F\left(x \right)$ we put $-4$, then we should get the correct numerical equality. Let's do this:
We see that we have an equation for $C$, so let's try to solve it:
Let's write down the very solution we were looking for:
Example No. 2
First of all, it is necessary to reveal the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original construction will be written as follows:
Now let's find $C$: substitute the coordinates of point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As final chord In addition to what we have just discussed, I propose to consider two more complex tasks, which contain trigonometry. In them, in the same way, you will need to find antiderivatives for all functions, then select from this set the only one that passes through the point $M$ on the coordinate plane.
Looking ahead, I would like to note that the technique that we will now use to find antiderivatives of trigonometric functions, in fact, is a universal technique for self-testing.
Task No. 1
Let's remember the following formula:
\[((\left(\text(tg)x \right))^(\prime ))=\frac(1)(((\cos )^(2))x)\]
Based on this, we can write:
Let's substitute the coordinates of point $M$ into our expression:
\[-1=\text(tg)\frac(\text( )\!\!\pi\!\!\text( ))(\text(4))+C\]
Let's rewrite the expression taking this fact into account:
Problem No. 2
This will be a little more difficult. Now you'll see why.
Let's remember this formula:
\[((\left(\text(ctg)x \right))^(\prime ))=-\frac(1)(((\sin )^(2))x)\]
To get rid of the “minus”, you need to do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Let's substitute the coordinates of point $M$:
In total, we write down the final construction:
That's all I wanted to tell you about today. We studied the term antiderivatives itself, how to count them from elementary functions, as well as how to find an antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you at least a little to understand this complex topic. In any case, it is on antiderivatives that indefinite and indefinite integrals are constructed, so it is absolutely necessary to calculate them. That's all for me. See you again!
The emergence of the concept of integral was due to the need to find the antiderivative of a function from its derivative, as well as to determine the amount of work, area complex figures, the distance traveled, with parameters outlined by curves described by nonlinear formulas.
and that work is equal to the product of force and distance. If all movement occurs with constant speed or the distance is overcome with the application of the same force, then everything is clear, you just need to multiply them. What is the integral of a constant? of the form y=kx+c.
But the strength can change throughout the work, and in some kind of natural dependence. The same situation arises with the calculation of the distance traveled if the speed is not constant.
So, it’s clear why the integral is needed. Defining it as the sum of products of the values of a function by an infinitesimal increment of the argument completely describes main meaning this concept as the area of a figure, bounded at the top by the line of the function, and at the edges by the boundaries of the definition.
Jean Gaston Darboux, a French mathematician, in the second half of the 19th century, very clearly explained what an integral is. He made it so clear that in general it would not be difficult for even a schoolchild to understand this issue. junior classes high school.
Let's say there is a function of any complex shape. The ordinate axis on which the values of the argument are plotted is divided into small intervals, ideally they are infinitesimal, but since the concept of infinity is quite abstract, it is enough to simply imagine small segments, the value of which is usually denoted Greek letterΔ (delta).
The function turned out to be “chopped” into small bricks.
Each argument value corresponds to a point on the ordinate axis, on which the corresponding function values are plotted. But since the selected area has two boundaries, there will also be two function values, greater and lesser.
The sum of products of larger values by the increment Δ is called a large amount Darboux, and is denoted as S. Accordingly, smaller values in a limited area, multiplied by Δ, all together form a small Darboux sum s. The site itself resembles rectangular trapezoid, since the curvature of the function line with an infinitesimal increment can be neglected. The easiest way to find the area is like this geometric figure- this is to add the products of the larger and smaller function values by the Δ-increment and divide by two, that is, define it as the arithmetic mean.
This is what the Darboux integral is:
s=Σf(x) Δ - small amount;
S= Σf(x+Δ)Δ is a large amount.
So what is an integral? The area limited by the line of the function and the boundaries of the definition will be equal to:
∫f(x)dx = ((S+s)/2) +c
That is, the arithmetic mean of large and small Darboux sums.s is a constant value that is reset during differentiation.
Based on the geometric expression of this concept, it becomes clear physical meaning integral. outlined by the speed function, and limited by the time interval along the x-axis, will be the length of the distance traveled.
L = ∫f(x)dx on the interval from t1 to t2,
f(x) is a function of speed, that is, the formula by which it changes over time;
L - path length;
t1 - start time of the journey;
t2 is the end time of the journey.
Exactly the same principle is used to determine the amount of work, only the distance will be plotted along the abscissa, and the amount of force applied at each specific point will be plotted along the ordinate.
Let's consider the movement of a point along a straight line. Let it take time t from the beginning of the movement the point has traveled a distance s(t). Then instantaneous speed v(t) equal to the derivative of the function s(t), that is v(t) = s"(t).
In practice it occurs inverse problem: at a given speed of point movement v(t) find the path she took s(t), that is, find such a function s(t), whose derivative is equal to v(t). Function s(t), such that s"(t) = v(t), is called the antiderivative of the function v(t).
For example, if v(t) = аt, Where A– given number, then the function
s(t) = (аt 2) / 2v(t), because
s"(t) = ((аt 2) / 2) " = аt = v(t).
Function F(x) called the antiderivative of the function f(x) on some interval, if for all X from this gap F"(x) = f(x).
For example, the function F(x) = sin x is the antiderivative of the function f(x) = cos x, because (sin x)" = cos x; function F(x) = x 4 /4 is the antiderivative of the function f(x) = x 3, because (x 4 /4)" = x 3.
Let's consider the problem.
Task.
Prove that the functions x 3 /3, x 3 /3 + 1, x 3 /3 – 4 are antiderivatives of the same function f(x) = x 2.
Solution.
1) Let us denote F 1 (x) = x 3 /3, then F" 1 (x) = 3 ∙ (x 2 /3) = x 2 = f(x).
2) F 2 (x) = x 3 /3 + 1, F" 2 (x) = (x 3 /3 + 1)" = (x 3 /3)" + (1)" = x 2 = f( x).
3) F 3 (x) = x 3 /3 – 4, F" 3 (x) = (x 3 /3 – 4)" = x 2 = f (x).
In general, any function x 3 /3 + C, where C is a constant, is an antiderivative of the function x 2. This follows from the fact that the derivative of the constant is zero. This example shows that for given function its antiderivative is determined ambiguously.
Let F 1 (x) and F 2 (x) be two antiderivatives of the same function f(x).
Then F 1 "(x) = f(x) and F" 2 (x) = f(x).
The derivative of their difference g(x) = F 1 (x) – F 2 (x) is equal to zero, since g"(x) = F" 1 (x) – F" 2 (x) = f(x) – f (x) = 0.
If g"(x) = 0 on a certain interval, then the tangent to the graph of the function y = g(x) at each point of this interval is parallel to the Ox axis. Therefore, the graph of the function y = g(x) is a straight line parallel to the Ox axis, i.e. e. g(x) = C, where C is some constant. From the equalities g(x) = C, g(x) = F 1 (x) – F 2 (x) it follows that F 1 (x) = F 2 (x) + S.
So, if the function F(x) is an antiderivative of the function f(x) on a certain interval, then all antiderivatives of the function f(x) are written in the form F(x) + C, where C is an arbitrary constant.
Let's consider the graphs of all antiderivatives of a given function f(x). If F(x) is one of the antiderivatives of the function f(x), then any antiderivative of this function is obtained by adding to F(x) some constant: F(x) + C. Graphs of functions y = F(x) + C are obtained from the graph y = F(x) by shift along the Oy axis. By choosing C, you can ensure that the graph of the antiderivative passes through a given point.
Let us pay attention to the rules for finding antiderivatives.
Recall that the operation of finding the derivative for a given function is called differentiation. The inverse operation of finding the antiderivative for a given function is called integration(from Latin word "restore").
Table of antiderivatives for some functions it can be compiled using a table of derivatives. For example, knowing that (cos x)" = -sin x, we get (-cos x)" = sin x, from which it follows that all antiderivative functions sin x are written in the form -cos x + C, Where WITH– constant.
Let's look at some of the meanings of antiderivatives.
1) Function: x p, p ≠ -1. Antiderivative: (x p+1) / (p+1) + C.
2) Function: 1/x, x > 0. Antiderivative: ln x + C.
3) Function: x p, p ≠ -1. Antiderivative: (x p+1) / (p+1) + C.
4) Function: e x. Antiderivative: e x + C.
5) Function: sin x. Antiderivative: -cos x + C.
6) Function: (kx + b) p, р ≠ -1, k ≠ 0. Antiderivative: (((kx + b) p+1) / k(p+1)) + C.
7) Function: 1/(kx + b), k ≠ 0. Antiderivative: (1/k) ln (kx + b)+ C.
8) Function: e kx + b, k ≠ 0. Antiderivative: (1/k) e kx + b + C.
9) Function: sin (kx + b), k ≠ 0. Antiderivative: (-1/k) cos (kx + b).
10)
Function: cos (kx + b), k ≠ 0. Antiderivative: (1/k) sin (kx + b). 
Integration rules can be obtained using differentiation rules. Let's look at some rules.
Let F(x) And G(x)– antiderivatives of functions respectively f(x) And g(x) at some interval. Then:
1) function F(x) ± G(x) is the antiderivative of the function f(x) ± g(x);
2) function аF(x) is the antiderivative of the function аf(x).
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The calculation of area is fundamental to area theory. The question arises about its location when the figure has irregular shape or it is necessary to resort to calculating it through an integral.
This article talks about calculating area curved trapezoid in a geometric sense. This makes it possible to identify the relationship between the integral and the area of a curvilinear trapezoid. If a function f (x) is given, and it is continuous on the interval [ a ; b ] , the sign in front of the expression does not change.
Yandex.RTB R-A-339285-1 Definition 1
A figure designated as G, bounded by lines of the form y = f(x), y = 0, x = a and x = b, is called curved trapezoid. It takes the designation S(G).
Let's look at the figure below.
To calculate a curved trapezoid, you need to split the segment [a; b ] for the number n of parts x i - 1 ; x i, i = 1, 2, . . . , n with points defined at a = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b , причем дать обозначение λ = m a x i = 1 , 2 , . . . , n x i - x i - 1 с точками x i , i = 1 , 2 , . . . , n - 1 . Необходимо выбрать так, чтобы λ → 0 при n → + ∞ , тогда фигуры, которые соответствуют нижней и upper parts Darboux are considered to be incoming P and outgoing Q polygonal shapes for G. Consider the figure below.
From here we have that P ⊂ G ⊂ Q, and with an increase in the number of partition points n, we obtain an inequality of the form S - s< ε , где ε является малым positive number, s and S are upper and lower Dabroux sums from the interval [a; b ] . Otherwise it will be written as lim λ → 0 S - s = 0 . This means that when referring to the concept definite integral Darboux, we obtain that lim λ → 0 S = lim λ → 0 s = S G = ∫ a b f (x) d x .
From the last equality we obtain that a definite integral of the form ∫ a b f (x) d x is the area of a curvilinear trapezoid for a given continuous function of the form y = f (x) . That's what it is geometric meaning definite integral.
When calculating ∫ a b f (x) d x, we obtain the area of the desired figure, which is limited by the lines y = f (x), y = 0, x = a and x = b.
Comment: When the function y = f (x) is non-positive from the interval [ a ; b ], then we find that the area of a curvilinear trapezoid is calculated based on the formula S (G) = - ∫ a b f (x) d x.
Example 1
Calculate the area of the figure, which is limited by given lines of the form y = 2 · e x 3, y = 0, x = - 2, x = 3.
Solution
In order to solve, it is first necessary to construct a figure on a plane where there is a line y = 0 coinciding with O x, with lines of the form x = - 2 and x = 3, parallel to the axis o y, where the curve y = 2 e x 3 is constructed using geometric transformations graph of the function y = e x. Let's build a graph.
This shows that it is necessary to find the area of a curved trapezoid. Recalling the geometric meaning of the integral, we find that the desired area will be expressed by a certain integral, which must be resolved. This means that it is necessary to apply the formula S (G) = ∫ - 2 3 2 · e x 3 d x . This indefinite integral is calculated based on the Newton-Leibniz formula
S (G) = ∫ - 2 3 2 e x 3 d x = 6 e x 3 - 2 3 = 6 e 3 3 - 6 e - 2 3 = 6 e - e - 2 3
Answer: S (G) = 6 e - e - 2 3
Comment: To find the area of a curved trapezoid, it is not always possible to construct a figure. Then the solution is carried out as follows. Given a known function f (x) that is non-negative or non-positive on the interval [ a ; b ] , a formula of the form S G = ∫ a b f (x) d x or S G = - ∫ a b f (x) d x is used.
Example 2
Calculate the area bounded by lines of the form y = 1 3 (x 2 + 2 x - 8), y = 0, x = - 2, x = 4.
Solution
To construct this figure, we find that y = 0 coincides with O x, and x = - 2 and x = 4 are parallel to O y. The graph of the function y = 1 3 (x 2 + 2 x - 8) = 1 3 (x + 1) 2 - 3 is a parabola with the coordinates of the point (- 1 ; 3) being its vertex with branches pointing upward. To find the intersection points of the parabola with O x, you need to calculate:
1 3 (x 2 + 2 x - 8) = 0 ⇔ x 2 + 2 x - 8 = 0 D = 2 2 - 4 1 (- 8) = 36 x 1 = - 2 + 36 2 = 2 , x 2 = - 2 - 36 2 = - 4
This means that the parabola intersects oh at points (4; 0) and (2; 0). From this we get that the figure designated as G will take the form shown in the figure below.
This figure is not a curvilinear trapezoid, because a function of the form y = 1 3 (x 2 + 2 x - 8) changes sign on the interval [ - 2 ; 4 ] . The figure G can be represented as a union of two curvilinear trapezoids G = G 1 ∪ G 2, based on the property of area additivity, we have that S (G) = S (G 1) + S (G 2). Consider the graph below.
Segment [ - 2 ; 4 ] is considered a non-negative area of the parabola, then from this we obtain that the area will have the form S G 2 = ∫ 2 4 1 3 (x 2 + 2 x - 8) d x . Segment [ - 2 ; 2 ] is non-positive for a function of the form y = 1 3 (x 2 + 2 x - 8), which means, based on the geometric meaning of the definite integral, we obtain that S (G 1) = - ∫ - 2 2 1 3 (x 2 + 2 x - 8) d x . It is necessary to make calculations using the Newton-Leibniz formula. Then the definite integral will take the form:
S (G) = S (G 1) + S (G 2) = - ∫ - 2 2 1 3 (x 2 + 2 x - 8) d x + ∫ 2 4 1 3 (x 2 + 2 x - 8) d x = = - 1 3 x 3 3 + x 2 - 8 x - 2 2 + 1 3 x 3 3 + x 2 - 8 x 2 4 = = - 1 3 2 3 3 + 2 2 - 8 2 - - 2 3 3 + (- 2) 2 - 8 · (- 2) + + 1 3 4 3 3 + 4 3 - 8 · 4 - 2 3 3 + 2 2 - 8 · 2 = = - 1 3 8 3 - 12 + 8 3 - 20 + 1 3 64 3 - 16 - 8 3 + 12 = 124 9
It is worth noting that finding the area is not correct according to the principle S (G) = ∫ - 2 4 1 3 (x 2 + 2 x - 8) d x = 1 3 x 3 3 + x 2 - 8 x - 2 4 = = 1 3 4 3 3 + 4 3 - 8 4 - - 2 3 3 + - 2 2 - 8 - 2 = 1 3 64 3 - 16 + 8 3 - 20 = - 4
Since the resulting number is negative and represents the difference S (G 2) - S (G 1).
Answer: S (G) = S (G 1) + S (G 2) = 124 9
If the figures are limited by lines of the form y = c, y = d, x = 0 and x = g (y), and the function is equal to x = g (y), and is continuous and has a constant sign on the interval [ c; d ], then they are called curvilinear tarpeziums. Consider in the figure below.
∫ c d g (y) d y is that its value is the area of a curvilinear trapezoid for a continuous and non-negative function of the form x = g (y) located on the interval [ c ; d] .
Example 3
Calculate the figure, which is limited by the ordinate axis and the lines x = 4 ln y y + 3, y = 1, y = 4.
Solution
Plotting a graph of x = 4 ln y y + 3 is not easy. Therefore, it is necessary to solve without a drawing. Recall that the function is defined for all positive values y. Let's consider the function values available on the interval [ 1 ; 4 ] . From the properties of elementary functions we know that logarithmic function increases throughout the entire domain of definition. Then not the segment [ 1 ; 4 ] is non-negative. This means that ln y ≥ 0. The existing expression ln y y , defined on the same segment, is non-negative. We can conclude that the function x = 4 ln y y + 3 is positive on the interval equal to [ 1 ; 4 ] . We find that the figure on this interval is positive. Then its area should be calculated using the formula S (G) = ∫ 1 4 4 ln y y + 3 d y .
A calculation needs to be made indefinite integral. To do this you need to find antiderivative of function x = 4 ln y y + 3 and apply the Newton-Leibniz formula. We get that
∫ 4 ln y y + 3 d y = 4 ∫ ln y y d y + 3 ∫ d y = 4 ∫ ln y d (ln y) + 3 y = = 4 ln 2 y 2 + 3 y + C = 2 ln 2 y + 3 y + C ⇒ S (G) = ∫ 1 4 4 ln y y + 3 d y = 2 ln 2 + y + 3 y 1 4 = = 2 ln 2 4 + 3 4 - (2 ln 2 1 + 3 1) = 8 ln 2 2 + 9
Consider the drawing below.
Answer: S (G) = 8 ln 2 2 + 9
Results
In this article, we identified the geometric meaning of a definite integral and studied the relationship with the area of a curvilinear trapezoid. It follows that we have the opportunity to calculate the area of complex figures by calculating the integral for a curved trapezoid. In the section on finding areas and figures that limited lines y = f (x), x = g (y), these examples are discussed in detail.
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