Application of the integral to the solution of applied problems
Area calculation
The definite integral of a continuous non-negative function f(x) is numerically equal to the area of a curvilinear trapezoid bounded by the curve y = f(x), the O x axis and the straight lines x = a and x = b. In accordance with this, the area formula is written as follows:
Let's look at some examples of calculating the areas of plane figures.
Task No. 1. Calculate the area bounded by the lines y = x 2 +1, y = 0, x = 0, x = 2.
Solution. Let's construct a figure whose area we will have to calculate.
y = x 2 + 1 is a parabola whose branches are directed upward, and the parabola is shifted upward by one unit relative to the O y axis (Figure 1).
Figure 1. Graph of the function y = x 2 + 1
Task No. 2. Calculate the area bounded by the lines y = x 2 – 1, y = 0 in the range from 0 to 1.
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Solution. The graph of this function is a parabola of branches that are directed upward, and the parabola is shifted relative to the O y axis down by one unit (Figure 2).
Figure 2. Graph of the function y = x 2 – 1
Task No. 3. Make a drawing and calculate the area of the figure bounded by the lines
y = 8 + 2x – x 2 and y = 2x – 4.
Solution. The first of these two lines is a parabola with its branches directed downward, since the coefficient of x 2 is negative, and the second line is a straight line intersecting both coordinate axes.
To construct a parabola, we find the coordinates of its vertex: y’=2 – 2x; 2 – 2x = 0, x = 1 – abscissa of the vertex; y(1) = 8 + 2∙1 – 1 2 = 9 is its ordinate, N(1;9) is the vertex.
Now let’s find the intersection points of the parabola and the straight line by solving the system of equations:
Equating the right sides of an equation whose left sides are equal.
We get 8 + 2x – x 2 = 2x – 4 or x 2 – 12 = 0, whence 
.
So, the points are the intersection points of a parabola and a straight line (Figure 1).
Figure 3 Graphs of functions y = 8 + 2x – x 2 and y = 2x – 4
Let's construct a straight line y = 2x – 4. It passes through the points (0;-4), (2;0) on the coordinate axes.
To construct a parabola, you can also use its intersection points with the 0x axis, that is, the roots of the equation 8 + 2x – x 2 = 0 or x 2 – 2x – 8 = 0. Using Vieta’s theorem, it is easy to find its roots: x 1 = 2, x 2 = 4.
Figure 3 shows a figure (parabolic segment M 1 N M 2) bounded by these lines.
The second part of the problem is to find the area of this figure. Its area can be found using a definite integral according to the formula 
.
In relation to this condition, we obtain the integral: 
2 Calculation of the volume of a body of rotation
The volume of the body obtained from the rotation of the curve y = f(x) around the O x axis is calculated by the formula:
When rotating around the O y axis, the formula looks like:
Task No. 4. Determine the volume of the body obtained from the rotation of a curved trapezoid bounded by straight lines x = 0 x = 3 and curve y = around the O x axis.
Solution. Let's draw a picture (Figure 4).
Figure 4. Graph of the function y =
The required volume is 
Task No. 5. Calculate the volume of the body obtained from the rotation of a curved trapezoid bounded by the curve y = x 2 and straight lines y = 0 and y = 4 around the O y axis.
Solution. We have:
Review questions
A)
Solution.
The first and most important point in the decision is drawing.
Let's make the drawing:
The equation y=0 sets the “x” axis;
- x=-2 And x=1- straight, parallel to the axis OU;
- y=x 2 +2 - a parabola, the branches of which are directed upward, with the vertex at the point (0;2).
Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x=0 find the intersection with the axis OU and solving the corresponding quadratic equation, find the intersection with the axis Oh .
The vertex of a parabola can be found using the formulas: 
You can also build lines point by point.
On the interval [-2;1] the graph of the function y=x 2 +2 located above the axis Ox, That's why:
Answer: S=9 sq. units
After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, “by eye” we count the number of cells in the drawing - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.
What to do if a curved trapezoid is located under the axis Oh?
b) Calculate the area of the figure bounded by the lines y=-e x , x=1 and coordinate axes.
Solution.
Let's make a drawing.
If a curved trapezoid is completely located under the axis Oh , then its area can be found using the formula:
Answer: S=(e-1) sq. units" 1.72 sq. units
Attention! The two types of tasks should not be confused:
1) If you are asked to solve simply a definite integral without any geometric meaning, then it may be negative.
2) If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.
In practice, most often the figure is located in both the upper and lower half-plane.
c) Find the area of a flat figure bounded by lines y=2x-x 2, y=-x.
Solution.
First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola 
and straight 
This can be done in two ways. The first method is analytical.
We solve the equation:
This means that the lower limit of integration a=0, upper limit of integration b=3 .
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We build the given lines: 1. Parabola - vertex at point (1;1); axis intersection Oh - points (0;0) and (0;2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. And now Attention! If on the segment [ a;b] some continuous function f(x) greater than or equal to some continuous function g(x), then the area of the corresponding figure can be found using the formula: And it doesn’t matter where the figure is located - above the axis or below the axis, but what matters is which graph is HIGHER (relative to another graph), and which is BELOW. In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from |
.You can construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational).
The desired figure is limited by a parabola above and a straight line below.
On the segment 
, according to the corresponding formula:
Answer: S=4.5 sq. units
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Any fractal is constructed according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.
The iterative algorithm for constructing a Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 cubes adjacent to it along the faces are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.
A figure bounded by the graph of a continuous non-negative function $f(x)$ on the segment $$ and the lines $y=0, \ x=a$ and $x=b$ is called a curvilinear trapezoid.
The area of the corresponding curvilinear trapezoid is calculated by the formula:
$S=\int\limits_(a)^(b)(f(x)dx).$ (*)
We will conditionally divide problems to find the area of a curvilinear trapezoid into $4$ types. Let's look at each type in more detail.
Type I: a curved trapezoid is specified explicitly. Then immediately apply the formula (*).
For example, find the area of a curvilinear trapezoid bounded by the graph of the function $y=4-(x-2)^(2)$ and the lines $y=0, \ x=1$ and $x=3$.
Let's draw this curved trapezoid.
Using formula (*), we find the area of this curvilinear trapezoid.
$S=\int\limits_(1)^(3)(\left(4-(x-2)^(2)\right)dx)=\int\limits_(1)^(3)(4dx)- \int\limits_(1)^(3)((x-2)^(2)dx)=4x|_(1)^(3) – \left.\frac((x-2)^(3) )(3)\right|_(1)^(3)=$
$=4(3-1)-\frac(1)(3)\left((3-2)^(3)-(1-2)^(3)\right)=4 \cdot 2 – \frac (1)(3)\left((1)^(3)-(-1)^(3)\right) = 8 – \frac(1)(3)(1+1) =$
$=8-\frac(2)(3)=7\frac(1)(3)$ (units$^(2)$).
Type II: the curved trapezoid is specified implicitly. In this case, the straight lines $x=a, \ x=b$ are usually not specified or partially specified. In this case, you need to find the intersection points of the functions $y=f(x)$ and $y=0$. These points will be points $a$ and $b$.
For example, find the area of a figure bounded by the graphs of the functions $y=1-x^(2)$ and $y=0$.
Let's find the intersection points. To do this, we equate the right-hand sides of the functions.
Thus, $a=-1$ and $b=1$. Let's draw this curved trapezoid.
Let's find the area of this curved trapezoid.
$S=\int\limits_(-1)^(1)(\left(1-x^(2)\right)dx)=\int\limits_(-1)^(1)(1dx)-\int \limits_(-1)^(1)(x^(2)dx)=x|_(-1)^(1) – \left.\frac(x^(3))(3)\right|_ (-1)^(1)=$
$=(1-(-1))-\frac(1)(3)\left(1^(3)-(-1)^(3)\right)=2 – \frac(1)(3) \left(1+1\right) = 2 – \frac(2)(3) = 1\frac(1)(3)$ (units$^(2)$).
Type III: the area of a figure limited by the intersection of two continuous non-negative functions. This figure will not be a curved trapezoid, which means you cannot calculate its area using formula (*). How to be? It turns out that the area of this figure can be found as the difference between the areas of curvilinear trapezoids bounded by the upper function and $y=0$ ($S_(uf)$), and the lower function and $y=0$ ($S_(lf)$), where the role of $x=a, \ x=b$ is played by the $x$ coordinates of the points of intersection of these functions, i.e.
$S=S_(uf)-S_(lf)$. (**)
The most important thing when calculating such areas is not to “miss” with the choice of the upper and lower functions.
For example, find the area of a figure bounded by the functions $y=x^(2)$ and $y=x+6$.
Let's find the intersection points of these graphs:
According to Vieta's theorem,
$x_(1)=-2,\x_(2)=3.$
That is, $a=-2,\b=3$. Let's draw a figure:
Thus, the top function is $y=x+6$, and the bottom function is $y=x^(2)$. Next, we find $S_(uf)$ and $S_(lf)$ using formula (*).
$S_(uf)=\int\limits_(-2)^(3)((x+6)dx)=\int\limits_(-2)^(3)(xdx)+\int\limits_(-2 )^(3)(6dx)=\left.\frac(x^(2))(2)\right|_(-2)^(3) + 6x|_(-2)^(3)= 32 .5$ (units$^(2)$).
$S_(lf)=\int\limits_(-2)^(3)(x^(2)dx)=\left.\frac(x^(3))(3)\right|_(-2) ^(3) = \frac(35)(3)$ (units$^(2)$).
Let's substitute what we found into (**) and get:
$S=32.5-\frac(35)(3)= \frac(125)(6)$ (units$^(2)$).
Type IV: the area of a figure bounded by a function(s) that does not satisfy the non-negativity condition. In order to find the area of such a figure, you need to be symmetrical about the $Ox$ axis ( in other words, put “minuses” in front of the functions) display the area and, using the methods outlined in types I – III, find the area of the displayed area. This area will be the required area. First, you may have to find the intersection points of the function graphs.
For example, find the area of a figure bounded by the graphs of the functions $y=x^(2)-1$ and $y=0$.
Let's find the intersection points of the function graphs:
those. $a=-1$, and $b=1$. Let's draw the area.
Let's display the area symmetrically:
$y=0 \ \Rightarrow \ y=-0=0$
$y=x^(2)-1 \ \Rightarrow \ y= -(x^(2)-1) = 1-x^(2)$.
The result is a curvilinear trapezoid bounded by the graph of the function $y=1-x^(2)$ and $y=0$. This is a problem to find a curved trapezoid of the second type. We have already solved it. The answer was: $S= 1\frac(1)(3)$ (units $^(2)$). This means that the area of the required curvilinear trapezoid is equal to:
$S=1\frac(1)(3)$ (units$^(2)$).
Let's consider a curved trapezoid bounded by the Ox axis, the curve y=f(x) and two straight lines: x=a and x=b (Fig. 85). Let's take an arbitrary value of x (just not a and not b). Let's give it an increment h = dx and consider a strip bounded by straight lines AB and CD, the Ox axis and the arc BD belonging to the curve under consideration. We will call this strip an elementary strip. The area of an elementary strip differs from the area of the rectangle ACQB by the curvilinear triangle BQD, and the area of the latter is less than the area of the rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As side h decreases, side Du also decreases and simultaneously with h tends to zero. Therefore, the area of the BQDM is second-order infinitesimal. The area of an elementary strip is the increment of the area, and the area of the rectangle ACQB, equal to AB-AC ==/(x) dx> is the differential of the area. Consequently, we find the area itself by integrating its differential. Within the figure under consideration, the independent variable l: changes from a to b, so the required area 5 will be equal to 5= \f(x) dx. (I) Example 1. Let us calculate the area bounded by the parabola y - 1 -x*, straight lines X =--Fj-, x = 1 and the O* axis (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration are a = - and £ = 1, therefore J [*-t]\- -fl -- Г -1-±Л_ 1V1 -l-l-Ii-^ 3) |_ 2 3V 2 / J 3 24 24* Example 2. Let's calculate the area limited by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain A 2 S= J sinxdx= [-cos x]Q =0 -(-1) = lf Example 3. Calculate the area limited by the arc of the sinusoid ^у = sin jc, enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of the previous example. However, let's do the calculations: I 5= | s\nxdx= [ - cosх)* - - cos i-(-cos 0)= 1 + 1 = 2. o Indeed, our assumption turned out to be correct. Example 4. Calculate the area bounded by the sinusoid and the Ox axis at one period (Fig. 88). Preliminary calculations suggest that the area will be four times larger than in Example 2. However, after making calculations, we obtain “i Г,*i S - \ sin x dx = [ - cos x]0 = = - cos 2l -(-cos 0) = - 1 + 1 = 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area limited by the same sinusoid y = sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we obtain 2l $2l sin xdx=[ - cosх]l = -cos 2i~)-c05i=- 1-1 =-2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in exercise 3, we find that their absolute values are the same, but the signs are different. If we apply property V (see Chapter XI, § 4), we get 2l I 2l J sin xdx= J sin * dx [ sin x dx = 2 + (- 2) = 0What happened in this example is not an accident. Always the area located below the Ox axis, provided that the independent variable changes from left to right, is obtained when calculated using integrals. In this course we will always consider areas without signs. Therefore, the answer in the example just discussed will be: the required area is 2 + |-2| = 4. Example 5. Let's calculate the area of the BAB shown in Fig. 89. This area is limited by the Ox axis, the parabola y = - xr and the straight line y - = -x+\. Area of a curvilinear trapezoid The required area OAB consists of two parts: OAM and MAV. Since point A is the intersection point of a parabola and a straight line, we will find its coordinates by solving the system of equations 3 2 Y = mx. (we only need to find the abscissa of point A). Solving the system, we find l; = ~. Therefore, the area has to be calculated in parts, first square. OAM and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 U 2. QAM-^x)