How to find the area of \u200b\u200ba shape?

Knowing and being able to calculate the areas of various shapes is necessary not only for solving simple geometric problems. You cannot do without this knowledge when drawing up or checking estimates for the repair of premises, calculating the amount of necessary consumables. So let's figure out how to find the areas of different shapes.

The part of a plane enclosed within a closed contour is called the area of \u200b\u200bthis plane. The area is expressed by the number of square units enclosed in it.

To calculate the area of \u200b\u200bbasic geometric shapes, you must use the correct formula.

Area of \u200b\u200ba triangle

Legend:

1. If h, a are known, then the area of \u200b\u200bthe desired triangle is determined as the product of the lengths of the side and the height of the triangle dropped to this side, divided in half: S \u003d (a h) / 2
2. If a, b, c are known, then the required area is calculated by Heron's formula: the square root taken from the product of half of the perimeter of the triangle and three differences of half of the perimeter and each side of the triangle: S \u003d √ (p (p - a) (p - b) (p - c)).
3. If a, b, γ are known, then the area of \u200b\u200ba triangle is determined as half of the product of 2 sides, multiplied by the value of the sine of the angle between these sides: S \u003d (a b sin γ) / 2
4. If a, b, c, R are known, then the required area is determined as the division of the product of the lengths of all sides of the triangle by the four radii of the circumscribed circle: S \u003d (a b c) / 4R
5. If p, r are known, then the required area of \u200b\u200bthe triangle is determined by multiplying half of the perimeter by the radius of the inscribed circle: S \u003d p r

Square area

Legend:

1. If the side is known, then the area of \u200b\u200bthis figure is determined as the square of the length of its side: S \u003d a 2
2. If d is known, then the area of \u200b\u200ba square is defined as half the square of the length of its diagonal: S \u003d d 2/2

Rectangle area

Legend:

• S - determined area,
• a, b - the lengths of the sides of the rectangle.

1. If a, b are known, then the area of \u200b\u200bthis rectangle is determined by the product of the lengths of its two sides: S \u003d a b
2. If the lengths of the sides are unknown, then the area of \u200b\u200bthe rectangle must be divided into triangles. In this case, the area of \u200b\u200ba rectangle is defined as the sum of the areas of its constituent triangles.

Parallelogram area

Legend:

• S is the required area,
• a, b - side lengths,
• h is the length of the height of this parallelogram,
• d1, d2 - lengths of two diagonals,
• α is the angle between the sides,
• γ is the angle between the diagonals.

1. If a, h are known, then the required area is determined by multiplying the lengths of the side and the height lowered to this side: S \u003d a h
2. If a, b, α are known, then the area of \u200b\u200bthe parallelogram is determined by multiplying the lengths of the sides of the parallelogram and the value of the sine of the angle between these sides: S \u003d a b sin α
3. If d 1, d 2, γ are known, then the area of \u200b\u200bthe parallelogram is determined as half of the product of the lengths of the diagonals and the value of the sine of the angle between these diagonals: S \u003d (d 1 d 2 sinγ) / 2

Rhombus area

Legend:

• S is the required area,
• a - side length,
• h - height length,
• α - smaller angle between two sides,
• d1, d2 - lengths of two diagonals.

1. If a, h are known, then the area of \u200b\u200ba rhombus is determined by multiplying the length of the side by the length of the height that is lowered to this side: S \u003d a h
2. If a, α are known, then the area of \u200b\u200bthe rhombus is determined by multiplying the square of the side length by the sine of the angle between the sides: S \u003d a 2 sin α
3. If d 1 and d 2 are known, then the required area is determined as half of the product of the lengths of the diagonals of the rhombus: S \u003d (d 1 d 2) / 2

Trapezium area

Legend:

1. If a, b, c, d are known, then the required area is determined by the formula: S \u003d (a + b) / 2 * √.
2. With known a, b, h, the required area is determined as the product of half the sum of the bases and the height of the trapezoid: S \u003d (a + b) / 2 h

Area of \u200b\u200ba convex quadrilateral

Legend:

1. If d 1, d 2, α are known, then the area of \u200b\u200ba convex quadrilateral is determined as half of the product of the diagonals of the quadrilateral multiplied by the sine of the angle between these diagonals: S \u003d (d 1 d 2 sin α) / 2
2. For known p, r, the area of \u200b\u200ba convex quadrilateral is defined as the product of the half-perimeter of the quadrangle by the radius of a circle inscribed in this quadrilateral: S \u003d p r
3. If a, b, c, d, θ are known, then the area of \u200b\u200ba convex quadrangle is determined as the square root of the products of the difference of the half-perimeter and the length of each side minus the product of the lengths of all sides and the square of the cosine of half the sum of two opposite angles: S 2 \u003d (p - a ) (p - b) (p - c) (p - d) - abcd cos 2 ((α + β) / 2)

Area of \u200b\u200ba circle

Legend:

If r is known, then the required area is determined as the product of the number π by the radius squared: S \u003d π r 2

If d is known, then the area of \u200b\u200ba circle is defined as the product of π by the square of the diameter, divided by four: S \u003d (π d 2) / 4

Complex figure area

A complex one can be broken down into simple geometric shapes. The area of \u200b\u200ba complex figure is defined as the sum or difference of the component areas. Consider, for example, a ring.

Designation:

• S is the area of \u200b\u200bthe ring,
• R, r - radii of the outer and inner circles, respectively,
• D, d - diameters of the outer and inner circles, respectively.

In order to find the area of \u200b\u200bthe ring, it is necessary to subtract the area from the area of \u200b\u200bthe larger circle smaller circle. S \u003d S1-S2 \u003d πR 2 -πr 2 \u003d π (R 2 -r 2).

Thus, if R and r are known, then the area of \u200b\u200bthe ring is determined as the difference between the squares of the radii of the outer and inner circles, multiplied by the number pi: S \u003d π (R 2 -r 2).

If D and d are known, then the area of \u200b\u200bthe ring is determined as a quarter of the difference between the squares of the diameters of the outer and inner circles, multiplied by the number pi: S \u003d (1/4) (D 2 -d 2) π.

Area of \u200b\u200bthe filled figure

Suppose that inside one square (A) is another (B) (smaller), and we need to find the filled cavity between shapes "A" and "B". Let's just say the "frame" of a small square. For this:

1. We find the area of \u200b\u200bfigure "A" (calculated by the formula for finding the area of \u200b\u200ba square).
2. Similarly, we find the area of \u200b\u200bfigure "B".
3. Subtract area "B" from area "A". And thus we get the area of \u200b\u200bthe filled figure.

Now you know how to find the areas of different shapes.

To solve problems in geometry, you need to know formulas - such as the area of \u200b\u200ba triangle or the area of \u200b\u200ba parallelogram - as well as the simple tricks that we will talk about.

First, let's learn the formulas for the areas of the figures. We have specially collected them in a convenient table. Print, Learn & Apply!

Of course, not all geometry formulas are in our table. For example, to solve problems in geometry and stereometry in the second part of the profile USE in mathematics, other formulas for the area of \u200b\u200ba triangle are also used. We will definitely tell you about them.

But what if you need to find not the area of \u200b\u200ba trapezoid or triangle, but the area of \u200b\u200bsome complex figure? There are universal ways! Let's show them with examples from the FIPI job bank.

1. How to find the area of \u200b\u200ba non-standard shape? For example, an arbitrary quadrilateral? A simple trick is to split this figure into those that we all know about, and find its area - as the sum of the areas of these figures.

Divide this quadrilateral with a horizontal line into two triangles with a common base equal to. The heights of these triangles are and. Then the area of \u200b\u200bthe quadrilateral is equal to the sum of the areas of two triangles:.

Answer:.

2. In some cases, the area of \u200b\u200ba figure can be represented as the difference between some areas.

It is not easy to calculate what the base and height are equal to in this triangle! But we can say that its area is equal to the difference between the areas of a square with a side and three right-angled triangles. Do you see them in the picture? We get:.

Answer:.

3. Sometimes in the task it is necessary to find the area of \u200b\u200bnot the whole figure, but its part. Usually, we are talking about the area of \u200b\u200ba sector - a part of a circle. Find the area of \u200b\u200ba sector of a circle of radius, the length of an arc of which is .

In this picture, we see part of a circle. The area of \u200b\u200bthe whole circle is equal since. It remains to find out which part of the circle is depicted. Since the length of the entire circumference is (since), and the length of the arc of this sector is , therefore, the length of the arc is one-fold less than the length of the entire circle. The angle at which this arc rests is also one-fold less than a full circle (that is, degrees). This means that the area of \u200b\u200bthe sector will be one times less than the area of \u200b\u200bthe entire circle.

There are an infinite number of flat figures of the most different shapes, both correct and irregular. A common property of all shapes is that any of them has an area. The areas of shapes are the dimensions of the portion of the plane occupied by those shapes, expressed in specific units. This value is always expressed as a positive number. The unit of measurement is the area of \u200b\u200ba square, whose side is equal to a unit of length (for example, one meter or one centimeter). The approximate value of the area of \u200b\u200bany shape can be calculated by multiplying the number of unit squares into which it is divided by the area of \u200b\u200bone square.

Other definitions of this concept are as follows:

1. The areas of simple figures are scalar positive quantities that satisfy the conditions:

Equal figures have equal areas;

If a figure is divided into parts (simple figures), then its area is the sum of the areas of these figures;

A square with a side of a unit of measure serves as a unit of area.

2. Areas of figures of complex shape (polygons) are positive quantities with the following properties:

Equal polygons have the same area;

If the polygon is made up of several other polygons, its area is equal to the sum of the areas of the latter. This rule is true for non-overlapping polygons.

As an axiom, it is accepted that the areas of figures (polygons) are positive values.

The definition of the area of \u200b\u200ba circle is given separately as the value to which the area of \u200b\u200ba given circle inscribed in a circle tends - despite the fact that the number of its sides tends to infinity.

The areas of irregular shapes (arbitrary shapes) are not defined, only the methods for calculating them are determined.

The calculation of areas was already in ancient times an important practical task in determining the size of land plots. The rules for calculating areas for several hundred years were formulated by Greek scientists and set forth in Euclid's Elements as theorems. It is interesting that the rules for determining the areas of simple figures in them are the same as at the present time. The areas with a curved contour were calculated using the passage to the limit.

The calculation of the areas of a simple rectangle, square), familiar to everyone from school, is quite simple. It is not even necessary to memorize formulas for the areas of figures containing letter designations. It is enough to remember a few simple rules:

2. The area of \u200b\u200ba rectangle is calculated by multiplying its length by its width. In this case, it is necessary that the length and width were expressed in the same units of measurement.

3. The area of \u200b\u200ba complex figure is calculated by dividing it into several simple ones and adding the resulting areas.

4. The diagonal of a rectangle divides it into two triangles, whose areas are equal and equal to half of its area.

5. The area of \u200b\u200ba triangle is calculated as half the product of its height and base.

6. The area of \u200b\u200bthe circle is equal to the product of the square of the radius and the well-known number "π".

7. The area of \u200b\u200bthe parallelogram is calculated as the product of the adjacent sides and the sine of the angle lying between them.

8. The area of \u200b\u200bthe rhombus is ½ of the result of multiplying the diagonals by the sine of the inner angle.

9. The area of \u200b\u200bthe trapezoid is found by multiplying its height by the length of the midline, which is equal to the arithmetic mean of the bases. Another option for determining the area of \u200b\u200ba trapezoid is to multiply its diagonals and the sine of the angle lying between them.

For clarity, children in elementary school are often given tasks: to find the area of \u200b\u200ba figure drawn on paper using a palette or a sheet of transparent paper, cut into cells. Such a sheet of paper is superimposed on the measured figure, the number of full cells (units of area) that fit in its contour is counted, then the number of incomplete cells, which is divided in half.

The knowledge of how to measure the Earth dates back to ancient times and gradually evolved into the science of geometry. This word is translated from the Greek language - "surveying".

The measure of the length and width of a flat area of \u200b\u200bthe Earth is area. In mathematics, it is usually denoted by the Latin letter S (from the English "square" - "area", "square") or the Greek letter σ (sigma). S denotes the area of \u200b\u200ba figure on a plane or the surface area of \u200b\u200ba body, and σ is the cross-sectional area of \u200b\u200ba wire in physics. These are the main symbols, although there may be others, for example, in the field of strength of materials, A is the sectional area of \u200b\u200bthe profile.

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Calculation formulas

Knowing the areas of simple shapes, you can find the parameters of more complex... Ancient mathematicians developed formulas by which they can be easily calculated. These shapes are a triangle, quadrilateral, polygon, circle.

To find the area of \u200b\u200ba complex flat figure, it is broken down into many simple figures such as triangles, trapezoids, or rectangles. Then, by mathematical methods, a formula is derived for the area of \u200b\u200bthis figure. This method is used not only in geometry, but also in mathematical analysis to calculate the areas of figures bounded by curves.

Triangle

Let's start with the simplest shape - a triangle. They are rectangular, isosceles and equilateral. Take any triangle ABC with sides AB \u003d a, BC \u003d b and AC \u003d c (∆ ABC). To find its area, let us recall the theorems of sines and cosines known from the school mathematics course. Releasing all calculations, we come to the following formulas:

• S \u003d √ is the well-known Heron formula, where p \u003d (a + b + c) / 2 is the half-perimeter of the triangle;
• S \u003d a h / 2, where h is the height lowered to side a;
• S \u003d a b (sin γ) / 2, where γ is the angle between sides a and b;
• S \u003d a b / 2, if ∆ ABC - rectangular (here a and b are legs);
• S \u003d b² (sin (2 β)) / 2, if ∆ ABC is isosceles (here b is one of the “hips”, β is the angle between the “hips” of the triangle);
• S \u003d a² √¾ if ∆ ABC is equilateral (here a is the side of the triangle).

Quadrilateral

Let there be a quadrilateral ABCD with AB \u003d a, BC \u003d b, CD \u003d c, AD \u003d d. To find the area S of an arbitrary 4-gon, you need to divide it by the diagonal into two triangles, the areas of which S1 and S2 are generally not equal.

Then, using the formulas, calculate them and add them, that is, S \u003d S1 + S2. However, if a 4-gon belongs to a certain class, then its area can be found using the previously known formulas:

• S \u003d (a + c) h / 2 \u003d e h, if the 4-gon is a trapezoid (here a and c are the bases, e is the middle line of the trapezoid, h is the height lowered to one of the bases of the trapezoid;
• S \u003d a h \u003d a b sin φ \u003d d1 d2 (sin φ) / 2, if ABCD is a parallelogram (here φ is the angle between sides a and b, h is the height dropped to side a, d1 and d2 are diagonals);
• S \u003d a b \u003d d² / 2, if ABCD is a rectangle (d is a diagonal);
• S \u003d a² sin φ \u003d P² (sin φ) / 16 \u003d d1 d2 / 2, if ABCD is a rhombus (a is the side of the rhombus, φ is one of its corners, P is the perimeter);
• S \u003d a² \u003d P² / 16 \u003d d² / 2 if ABCD is a square.

Polygon

To find the area of \u200b\u200ban n-gon, mathematicians break it down into the simplest equal figures - triangles, find the area of \u200b\u200beach of them, and then add them. But if the polygon belongs to the class of regular ones, then use the formula:

S \u003d anh / 2 \u003d a² n / \u003d P² /, where n is the number of vertices (or sides) of the polygon, a is the side of the n-gon, P is its perimeter, h is the apothem, that is, a segment drawn from the center of the polygon to one of its sides at an angle of 90 °.

A circle

A circle is a perfect polygon with an infinite number of sides.... We need to calculate the limit of the expression on the right in the formula for the area of \u200b\u200ba polygon with the number of sides n tending to infinity. In this case, the perimeter of the polygon will turn into the circumference of a circle of radius R, which will be the boundary of our circle, and will become equal to P \u003d 2 π R. Substitute this expression into the above formula. We will get:

S \u003d (π² R² cos (180 ° / n)) / (n sin (180 ° / n)).

Let us find the limit of this expression as n → ∞. To do this, take into account that lim (cos (180 ° / n)) as n → ∞ is equal to cos 0 ° \u003d 1 (lim is the limit sign), and lim \u003d lim as n → ∞ is equal to 1 / π (we translated the degree measure to the radian using the ratio π rad \u003d 180 °, and applied the first remarkable limit lim (sin x) / x \u003d 1 as x → ∞). Substituting the obtained values \u200b\u200binto the last expression for S, we arrive at the well-known formula:

S \u003d π² R² 1 (1 / π) \u003d π R².

Units

System and non-system units are used... System units refer to SI (International System). It is a square meter (square meter, m²) and units derived from it: mm², cm², km².

In square millimeters (mm²), for example, they measure the cross-sectional area of \u200b\u200bwires in electrical engineering, in square centimeters (cm²) - the cross-sections of a beam in structural mechanics, in square meters (m²) - apartments or houses, in square kilometers (km²) - territories in geography ...

However, sometimes non-systemic units of measurement are also used, such as: weaving, ar (a), hectare (ha) and acre (ac). Here are the following relationships:

• 1 weave \u003d 1 a \u003d 100 m² \u003d 0.01 ha;
• 1 hectare \u003d 100 a \u003d 100 ares \u003d 10000 m² \u003d 0.01 km² \u003d 2.471 ac;
• 1 ac \u003d 4046.856 m2 \u003d 40.47 a \u003d 40.47 ares \u003d 0.405 hectares.

Definite integral. How to calculate the area of \u200b\u200ba shape

We pass on to considering applications of integral calculus. In this lesson we will analyze a typical and most common task. - how to calculate the area of \u200b\u200ba flat figure using a definite integral... Finally, those who seek meaning in higher mathematics - may they find it. You never know. We'll have to bring the suburban area closer in life with elementary functions and find its area using a definite integral.

To successfully master the material, you must:

1) Understand the indefinite integral at least at the middle level. Thus, dummies should first read the lesson Not.

2) Be able to apply the Newton-Leibniz formula and calculate a definite integral. You can build warm friendships with definite integrals on the page Definite integral. Solution examples.

In fact, in order to find the area of \u200b\u200ba figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to be able to build a straight line, a parabola and a hyperbola. This can be done (many people need it) with the help of methodological material and an article on geometric transformations of graphs.

Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will not go far ahead of the school curriculum. This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from the hated tower with enthusiasm mastering the course of higher mathematics.

The materials of this workshop are presented simply, in detail and with a minimum of theory.

Let's start with a curved trapezoid.

Curved trapezoid is called a plane figure bounded by an axis, straight lines, and a graph of a continuous function on a segment, which does not change sign on this interval. Let this figure be located not less abscissa axis:

Then the area of \u200b\u200ba curvilinear trapezoid is numerically equal to the definite integral... Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Solution examples I said that a definite integral is a number. And now it's time to state another useful fact. From the point of view of geometry, the definite integral is the AREA.

I.e, a definite integral (if it exists) geometrically corresponds to the area of \u200b\u200bsome figure... For example, consider a definite integral. The integrand defines a curve on the plane that is located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to the area of \u200b\u200bthe corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built CORRECTLY.

When building a drawing, I recommend the following order: first it is better to build all lines (if any) and only later - parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions pointwise, the technique of point-by-point construction can be found in the reference material Graphs and properties of elementary functions... There you can also find very useful material in relation to our lesson - how to quickly build a parabola.

In this problem, the solution might look like this.
Let's draw a drawing (note that the equation defines the axis):

I will not hatch a curved trapezoid, here it is obvious what area we are talking about. The solution continues like this:

On the segment, the graph of the function is located over axis, so:

Answer:

Who has difficulty calculating a definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Solution examples.

After the task is completed, it is always useful to look at the blueprint and estimate if the answer is real. In this case, "by eye" we count the number of cells in the drawing - well, about 9 will be typed, it looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.

Example 2

Calculate the area of \u200b\u200ba shape bounded by lines, and an axis

This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.

What to do if the curved trapezoid is located under the axis?

Example 3

Calculate the area of \u200b\u200ba shape bounded by lines and coordinate axes.

Decision: Let's execute the drawing:

If the curved trapezoid is located under the axis (or at least not higher given axis), then its area can be found by the formula:
In this case:

Attention! The two types of tasks should not be confused:

1) If you are asked to solve just a definite integral without any geometric meaning, then it may be negative.

2) If you are asked to find the area of \u200b\u200ba figure using a definite integral, then the area is always positive! That is why minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of \u200b\u200ba flat figure bounded by lines,.

Decision: First you need to complete the drawing. Generally speaking, when constructing a drawing in problems on an area, we are most interested in the points of intersection of lines. Find the intersection points of the parabola and the line. This can be done in two ways. The first way is analytical. We solve the equation:

Hence, the lower limit of integration, the upper limit of integration.
It is better not to use this method, if possible..

It is much more profitable and faster to construct the lines point by point, while the limits of integration become clear as if “by themselves”. The technique of point-by-point plotting for various charts is discussed in detail in the help. Graphs and properties of elementary functions ... Nevertheless, the analytical method of finding the limits still has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

We return to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing:

I repeat that in the case of a point-by-point construction, the limits of integration are most often found out “automatically”.

And now the working formula: If on a segment some continuous function greater than or equal of some continuous function, then the area of \u200b\u200bthe figure, bounded by the graphs of these functions and straight lines, can be found by the formula:

Here you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which schedule is ABOVE(relative to another graph), and which one 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

Completing the solution might look like this:

The required figure is bounded by a parabola at the top and a straight line at the bottom.
On the segment, according to the corresponding formula:

Answer:

In fact, the school formula for the area of \u200b\u200ba curved trapezoid in the lower half-plane (see simple example No. 3) is a special case of the formula ... Since the axis is given by the equation, and the graph of the function is located not higher axis, then

And now a couple of examples for an independent solution

Example 5

Example 6

Find the area of \u200b\u200bthe figure bounded by lines,.

In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but through inattention ... the area of \u200b\u200bthe wrong figure is found, this is how your humble servant screwed up several times. Here's a real life case:

Example 7

Calculate the area of \u200b\u200bthe figure bounded by lines,,,.

Decision: First we execute the drawing:

... Eh, a lousy drawing came out, but everything seems to be legible.

The figure whose area we need to find is shaded in blue (carefully look at the condition - what the figure is limited to!). But in practice, due to inattention, a "glitch" often arises, that you need to find the area of \u200b\u200bthe figure, which is shaded in green!

This example is also useful in that it calculates the area of \u200b\u200ba figure using two definite integrals. Really:

1) On the segment above the axis there is a line graph;

2) The hyperbola graph is located on the segment above the axis.

It is quite obvious that the areas can (and should) be added, therefore:

Answer:

Let's move on to one more meaningful task.

Example 8

Calculate the area of \u200b\u200ba shape bounded by lines
Let's represent the equations in the "school" form, and execute a point-by-point drawing:

It can be seen from the drawing that our upper limit is "good":.
But what is the lower limit ?! It is clear that this is not an integer, but which one? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well be that. Or root. What if we plotted the graph incorrectly?

In such cases, you have to spend additional time and refine the limits of integration analytically.

Find the intersection points of the line and the parabola.
To do this, we solve the equation:


,

Indeed,.

The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations here are not the easiest ones.

On the segment , according to the corresponding formula:

Answer:

Well, in conclusion of the lesson, we will consider two more difficult tasks.

Example 9

Calculate the area of \u200b\u200ba figure bounded by lines,

Decision: Draw this figure in the drawing.

Damn, I forgot to sign the schedule, but to redo the picture, sorry, not hotts. Not a drawing, in short, today is the day \u003d)

For point-by-point construction, you need to know the appearance of the sinusoid (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table... In a number of cases (as in this one), it is allowed to construct a schematic drawing, on which the graphs and integration limits should be displayed correctly in principle.

There are no problems with the limits of integration, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:

On the segment, the graph of the function is located above the axis, therefore: