Graph of curved trapezium. The area of ​​the crooked trapezium. The stage of reconversion of the home office

Viznachennya. The figure, surrounded by a graph without interruption, constant sign function f (x), with abciss and straight lines x = a, x = b, is called a curved trapezoid.

Ways to know the area of ​​the crooked trapeze

Theorem. As f (x) is without interruption and is a non-negative function in terms of growth, then the area of ​​the same curved trapezium will increase the growth of the first ones.

Given: f (x) - uninterrupted unspecified. function, Xо.

Bring: S = F (b) - F (a), de F (x) - first f (x).

Delivered:

1) I can easily access the function S (x). To the skin Xo it is possible to put it in the form of that part of the curved trapezium, as to lie more straight (Fig. 2), but to pass through the point from the center of the abcss and parallel to the axis of ordinates.

From the same S (a) = 0 і S (b) = Str

So S (a) is the first thing f (x).

D (f) = D (S) =

S "(x0) = lim (S (x0 + Dx) - S (x0) / Dx), for Dx®0 DS is a rectangular

Dx®0 with sides Dx і f (x0)

S "(x0) = lim (Dx f (x0) / Dx) = lim f (x0) = f (x0): if x0 is a point, then S (x) is

Dx®0 Dx®0 is primary f (x).

From the previous theorem about the primordial vigilance S (x) = F (x) + C.

S (a) = 0, then S (a) = F (a) + C

S = S (b) = F (b) + C = F (b) -F (a)

1). Rozib'єmo vіdrizok on n rіvnih parts. Croc rosbittia (fig. 3)

Dx = (b-a) / n. For tsom Str = lim (f (x0) Dx + f (x1) Dx + ... + f (xn)) Dx = n®Ґ = lim Dx (f (x0) + f (x1) + ... + f (xn))

When n®Ґ is negligible, who Str = Dx (f (x0) + f (x1) + ... + f (xn))

I call the value of the integral integral.

The sum stands before the border, is called an integral sum.

Singing integral tse between the integral sumi on the change at n®Ґ. The integral sum goes as between the sums of the creations of the age of the child, which is cut off during the development of the area of ​​the value of the function in any point of the interval.

a - lower boundary of integration;

b - top.

Newton-Leibnitz formula.

The formulas of the curved trapezoidal trapezoidal area are modified:

if F is primary for b on, then

m f (x) dx = F (b) -F (a)

m f (x) dx = F (x) φ = F (b) - F (a)

The power of the designated integral.

t f (x) dx = t f (z) dz

m f (x) dx = F (a) - F (a) = 0

m f (x) dx = - m f (x) dx

t f (x) dx = F (a) - F (b) t f (x) dx = F (b) - F (a) = - (F (a) - F (b))

If a, b і c be-like points between I, in which the function f (x) is infinitely superior, then

t f (x) dx = t f (x) dx + t f (x) dx

F (b) - F (a) = F (c) - F (a) + F (b) - F (c) = F (b) - F (a)

(The power of additivity of the singing integral)

If l and m are of constant magnitude, then

t (lf (x) + m j (x)) dx = l t f (x) dx + m Tj (x)) dx -

The power of the linearity of the singing integral.

t (f (x) + g (x) + ... + h (x)) dx = t f (x) dx + t g (x) dx + ... + t h (x) dx

m (f (x) + g (x) + ... + h (x)) dx = (F (b) + G (b) + ... + H (b)) - (F (a) + G (a) + ... + H (a)) + C = F (b) -F (a) + C1 + G (b) -G (a) + C2 + ... + H (b) - H (a) + Cn = bbb = t f (x) dx + t g (x) dx + ... + t h (x) dx

A set of standard pictures (fig. 4, 5, 6, 7, 8)

Small. 4

Small. 6 Small. 7

Oskіlki f (x)<0, то формулу Ньютона-Лейбница составить нельзя, теорема верна только для f(x)і0.

Require: view the symmetry of the function for the OX axis. ABCD®A "B" CD b

S (ABCD) = S (A "B" CD) = m -f (x) dx

S = t f (x) dx = t g (x) dx

S = t (f (x) -g (x)) dx + t (g (x) -f (x)) dx

S = m (f (x) + m-g (x) -m) dx =

m (f (x) - g (x)) dx

m ((f (x) -g (x)) dx

S = m (f (x) + m-g (x) -m) dx =

T (f (x) - g (x)) dx

If on top of f (x) Іg (x), then the area between the graphs of the road

m ((f (x) -g (x)) dx

Functions f (x) and g (x) good and bad

S = t f (x) dx - t g (x) dx = t (f (x) -g (x)) dx

Any singing integral (like іsnu) є has even a good geometric sense. At the level, I said that the integral values ​​are the same number. And at once the time has come to state one corny fact. From the point of view of the geometrical values ​​of the integral - Tse AREA.

Tobto, singing integral (yaksho vin isnu) geometrically resembles the area of ​​the deyakoi figuri... For example, an intelligible integral. The integral function is set on an area like a curve (it can be done when you are well-fed), and the singing integral itself is numerically larger than the area of ​​a curved trapezium.

butt 1

Tse typical formuluvannya zavdannya. The first and most important moment of the decision is the motivation for the chair... Moreover, the chair must be RIGHT.

When prompted for a chair, I recommend the offensive order: a collection of more beautifully stay all straight (like the smell є) and only potim- parabols, hyperbolas, graphs of other functions. Graphs of functions in the image of the buduvati pointwise The technique of pointwise inducement can be learned in pre-materiel.

In the same place you can find another hundred percent of our lesson material - like a quick parabola.

At the end of the day, you can see it this way.
Viconmo armchair (beastful respect, vnyannya is asked to hang):

I will not hatch a curvilinear trapezium, here it is obvious, about yak squares. The decision is simple as follows:

On the top of the graph of the function of rosetting over vissu, Tom:

as follows:

Who has found it difficult to count the singing integral and the Newton-Leibnitz formula , Beat up to the lecture Integral values. put on the solution.

In addition, as a viconano, be sure to look at the armchair and figure out what the real Wiishov has seen. In this vipad "on the eye", there is a number of clitches in the armchair - well, about 9, it looks like the truth. Quite zealously, as if we have Wiishov, say, they say: 20 square ones, then, obviously, there is a pity here - there is clearly no room in the figurine of 20 cells, as there are only a dozen. If you see the views as negative, then the default is probably incorrect.

butt 2

Count the area of ​​figurines, surrounded by lines, and

Tse butt for an independent solution. Outside the decision and see in the end of the lesson.

Scho robiti, yakscho curvilinear trapezoid roztashovana go to see?

butt 3

Count the area of ​​figurines, surrounded by lines, and coordinate axes.

Solution: Viconaєmo chair:

Yaksho curved trapeze on the other hand That її area can be known for the formula:
In this vipadku:

Uwaga! Do not swindle two types of people:

1) If you are proponated with just a singing integral without any geometric sense, then it can be negative.

2) If you are encouraged to know the area of ​​the figuri behind the help of the singing integral, then the area is positive! The very fact is, in only the form of figurine minus.

In practice, most often the figure is roasted in the upper and in the lower area, and that, from the simplest school tasks, it is possible to move to larger applications.

butt 4

Know the area of ​​flat figurines, surrounded by lines.

Solution: You need a seat for a visitor. Apparently, when prompted by the chair in the tasks on the area we are most likely to point out the lines. We know the point of overflow of the parabola and straight. The price can be changed in two ways. The first method is analytical. Virishuєmo rivnyannya:

This means, the lower boundary of the integration, the upper boundary of the integration.
In this way, more beautifully, if possible, do not krystuvatisya.

Nagato vigіdnіshe and shvidvaty lіnії point-by-point, with a wide range of integration, yak bi "by itself". The technique of point-by-point inducement for young graphs was presented in the report Graphs and power of elementary functions... Protest, the analytical way of knowing between all the same is brought in one stasis, as, for example, the graph to finish the great, for the flow of prompting did not appear between the integration (the stench can be shot, or irrational). This stock can be easily seen.

We turn it over to our factory: more rational way to get a straight line and only through a parabola. Viconaєmo armchair:

I repeat, when the pointwise inducement of the integration is most common, it will be automatic.

And now the working formula is: As soon as the function goes without interruption more or more Since there is a continuous function, then the area of ​​the similar figuri can be known by the formula:

Here, there is no need to think, the figure is de-roasted - over the top or the bottom of the head, і, roughly, importantly, like graph VISCHE(Schedule of the іnshy graphyka), and yaky - LOWER.

When the butt is open, it is obvious that the parabola will grow straight on the side of the parabola, and that it is necessary to

The completed solution can be seen as follows:

Shukana figur is surrounded by a parabola at the top and straight at the bottom.
On the basis of the following formula:

as follows:

For a school formula for the area of ​​the curved trapezium in the lower pivot (div. Simple butt No. 3) - an award for the formulas ... Oskilki won’t be asked, and the graph of the function of the expansion below the axis, then

And at the same time, a sprinkle of butts for an independent solution

butt 5

butt 6

Know the area of ​​figuri, surrounded by lines.

In the course of the resolution of tasks on the calculation of the area behind the additional designation of the integral, there is a great incident. The viconano's chair is correct, the rosrahunks are correct, or through disrespect ... there is no area of ​​\ u200b \ u200btієї figuri, It’s just the same as your dear servant. The axis of the real life vision:

butt 7

Count the area of ​​figurines, surrounded by lines ,,,.

A collection of viconєmo armchair:

The figure, the area that we need to know, is shaded with blue color(It is important to wonder at the mind - what the figure is surrounded by!). Ale on practice, through the lack of importance, it is not easy to win, so it is necessary to know the area of ​​the figuri, the yak is shaded with green color!

Tsey butt cinnamon tim, which in the new area of ​​the figurie to vvazhaya for the help of two-valued integrals. action:

1) On the line over the top of the wrap-around graph;

2) On the way over the top of the swash over the graph of the hyperbole.

All in all, it is obvious that the area is possible (and necessary), if only:

as follows:

butt 8

Count the area with figurines, surrounded by lines,
Uyavimo іvnyannya in the "school" viglyadі, і viconaєmo point-to-point chair:

You can see from the armchair that the upper boundary between us is "good" :.
Why don't you get the lower border ?! Zrozumіlo, is it not a whole number, ale yak? Maybe bootie? Ale de garant_ya, where the armchair of the viconies is with ideal accuracy, at all times can come. Abo root. And how did you get the graph wrong?

In such cases, it takes a long time to clarify the integer integration in an analytical way.

We know the point to cross the straight line and parabola.
For ts'go virishuєmo rivnyannya:

Otzhe ,.

Further, the solution is trivial, smut, do not get lost in the settings and signs, there is no simple calculation here.

on vidrizka , For a general formula:

as follows:

Well, and at the end of the lesson, two folds are visible.

butt 9

Count the area with figurines, surrounded by lines,

Solution: Imagine a qiu figure on a chair.

For a pointwise inducement of an armchair, it is necessary for the nobility of the callous viglyad of sinusoid (and of the nobility of the nobility graphs of all elementary functions), And also the meaning of the sinus, you can know in trigonometric tables... In a number of vipadks (as in the whole), it is allowed to create a schematic chair, on which it is fundamentally correct to be guilty of displaying graphics and between integration.

There are no problems with the integration between them, the stench goes straight from the mind: - "x" changes from zero to "pi". Let's make out the solution:

On the top of the graph of the function of rosetting over the top, to that:

(1) How to integrate sine and cosine in unpaired steps, you can wonder at the level Integrals of trigonometric functions... Tse typical priyom, one sinus appears.

(2) Vikorist's most basic trigonometric sameness in viglyad

(3) We will change the change, todi:

New changes in integration:

Whoever has the name of the villains, ask for the replacements, please go to the lesson The method of substitution in the unassigned integral... To whom is not even less intelligent the algorithm of replacing in the singing integration, see the side Integral values. put on the solution.

Butt 1 . Count the area of ​​the figurines, surrounded by lines: x + 2y - 4 = 0, y = 0, x = -3, і x = 2

Vicona will induce figuri (div. Fig.) I will be straight x + 2y - 4 = 0 at two points A (4; 0) and B (0; 2). After hanging y through x, we can deduce mo y = -0.5x + 2. By formula (1), de f (x) = -0.5x + 2, a = -3, b = 2, we know

S = = [-0.25 = 11.25 sq. od

Butt 2. Count the area of ​​figurines, surrounded by lines: x - 2y + 4 = 0, x + y - 5 = 0 and y = 0.

Decision. Viconaєmo will induce the figuri.

Let's stay on the straight line x - 2y + 4 = 0: y = 0, x = - 4, A (-4; 0); x = 0, y = 2, B (0; 2).

Let's stay on the straight line x + y - 5 = 0: y = 0, x = 5, C (5; 0), x = 0, y = 5, D (0; 5).

We know the point of overflowing straight, having violated the system of rivnyans:

x = 2, y = 3; M (2; 3).

For the calculation of the shukanoy area of ​​the rosib'єmo AMC tricycle for two AMN and NMC tricycles, so when changing from A to N, the area is surrounded by straight lines, and when changing from N to C - straight

For tricycle AMN maєmo :; y = 0.5x + 2, i.e. f (x) = 0.5x + 2, a = - 4, b = 2.

For tricycle NMC maєmo: y = - x + 5, i.e. F (x) = - x + 5, a = 2, b = 5.

Having calculated the area of ​​the skin with tricycles and the results, it is known:

sq. od.

sq. od.

9 + 4.5 = 13.5 sq. od. Pereirka: = 0.5АС = 0.5 sq. od.

Butt 3. Count the area of ​​figurines, surrounded by lines: y = x 2 , Y = 0, x = 2, x = 3.

In this view, it is necessary to calculate the area of ​​the curved trapezium, surrounded by a parabola y = x 2 , Straight lines x = 2 і x = 3і vіssu Oh (div. Fig.) For the formula (1) we know the area of ​​the curved trapezoid

= = 6kV. od.

Butt 4. Count the area of ​​figurines, surrounded by lines: y = - x 2 + 4 i y = 0

Viconaєmo will induce the figuri. Shukana of the area is laid with a parabola y = - x 2 + 4 і віссю Oh.

We know the point of the parabolic crossover from the vissu Oh. Vvazhayuchi y = 0, we know x = So if the figure is symmetrical as to the axis Oy, then the area of ​​the figure is calculated, the right-handed person is removed from the axis Oy, and the result is subtracted: = + 4x] sq. od. 2 = 2 sq. od.

Butt 5. Count the area of ​​figurines, surrounded by lines: y 2 = X, yx = 1, x = 4

Here it is necessary to calculate the area of ​​the curved trapezium, surrounded by the upper parabolic head 2 = X, віссю Ох і straight lines x = 1іx = 4 (div. Fig.)

For formula (1), de f (x) = a = 1 і b = 4 maєmo = (= square.

butt 6 . Count the area of ​​the figurines, surrounded by lines: y = sinx, y = 0, x = 0, x =.

Shukana of the area is surrounded by a sinusoidal sinusoidal і vissyu Oh (div. Fig.).

Mahmo - cosx = - cos = 1 + 1 = 2 sq. od.

Butt 7. Count the area of ​​figurines, surrounded by lines: y = - 6x, y = 0 and x = 4.

Figure roztashovana pid vissyu Oh (div. Fig.).

Otzhe, її area is known for the formula (3)

= =

Butt 8. Count the area of ​​the figurines, surrounded by lines: y = і х = 2. The curve y = will move along the points (div. Fig.). In this rank, the area of ​​the figuri is known by the formula (4)

butt 9 .

NS 2 + y 2 = r 2 .

Here it is necessary to count the area, surrounded by a stake x 2 + y 2 = r 2 , TE The area of ​​the radius r with the center on the cob of coordinates. We know the fourth part of the whole area, taking the boundaries of integration from 0

dor; maєmo: 1 = = [

already, 1 =

Butt 10. Count the area with figuri, surrounded by lines: y = x 2 i y = 2x

A given figure is surrounded by a parabola y = x 2 і straight y = 2x (div. fig.) 2 - 2x = 0 x = 0 і x = 2

Vikoristovuchi for the known area of ​​the formula (5), we take

= І nehai F (x)- deyaka її Pervisna. todі number F (b) -F (a) be called an integral of a before b functions f (x) i know

.

parity
be called the Newton-Leibnitz formula.

The tsia formula pov'yazu zavdannya znakhozhennya areas of flat figurines with integral.

At the zagalny vipadku, as the figure is surrounded by graphs of functions y = f (x);y = g (x) (f (x)> g (x)) I straight x = a;x = b, That її її dorіvnyu area:

.

Butt 2. At the same point, the graph of the function y = x 2 + 1 request to conduct a dotted line y = 0, X = 0, X = 1 trapezium of the best area?

Decision. hey M 0 (x 0 , y 0 ) - point of graph function y = x 2 + 1, a shukana was carried out in yak.

We know the dots y = y 0 + f (x 0 ) (X-x 0 ) .

maєmo:

Tom

.

I know the area of ​​the trapeze OABS.

.

B- the point of cross-flow is dotted with a straight line x = 1 

The zavdannya rang up to the most significant function

S(x)= -x 2 + X + 1 for vidrizka. we know S (x)=– 2x + 1. I know that I’m critical point of mind S (x)= 0  x =.

Bachimo, the function of the reach is of the greatest value when x =... we know
.

as follows: I will do the exact same thing.
.

Apparently, it is often the case that the knowledge of the integral, which is outgoing geometrical sense, is being developed. It is shown on the butt, how to show the same zavdannya.

Butt 4. Vikoristovuchi geometric sense integral count

a )
; b)
.

Decision.

a)
- road areas of curved trapezium, surrounded by lines.

NS transform

- the upper half of the circle with the center R(1; 0) i radius R = 1.

Tom
.

as follows:
.

b) Similarly, the area will be surrounded by graphs. – 2x + 2, similar to it in points A
, B(4;2)

y =–9x- 59, parabolic y = 3x 2 + Ax + 1, which is visible, which is similar to a parabola at points x = - 2 get started Ox kut size arctg 6.

know a, Yaksho vidomo, where the area is curved trapezium, surrounded by lines y = 3x 3 + 2x, x = a, y = 0, door unit.

Know the least value of the area of ​​the figuri, surrounded by a parabola y = x 2 + 2x- 3 i straight y = kx + 1.

6. Stage of information about home management.

Manager: Protect intelligence by studying meti, wisdom and ways of making homework assignments. # 18, 19, 20, 21 unpaired

7. Summarizing the lesson.

Supervisor: I will give an assessment of the robot class and okremikh scientists.

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