Instruction

1

When graphing two given functions in the area of their intersection is formed a closed figure bounded by these curves and two straight lines x=a and x=b, where a and b are the ends of the considered interval. This figure visually displays the touch. Its area can be calculated proindeksirovat difference functions.

2

Feature located above on the chart, is the larger value, hence the formula of its expression will be the first: S = ∫f1 ∫f2, where f1 > f2 on the interval [a, b]. However, taking into account that the quantitative characteristic of any geometric object is a positive value, we can calculate the area of the figure bounded the graphs of the functions in the module:

S = |∫f1 ∫f2|.

S = |∫f1 ∫f2|.

3

This option is especially useful if there is no opportunity or time to build the graph. When calculating a definite integral you use the rule of the Newton-Leibniz, which involves the substitution in the final result the limit values of the interval. Then the area of the shape equal to the difference between two values of the integral found at the integration phase, of greater F(b) and the smaller F(a).

4

Sometimes a closed figure on a given interval is formed by the complete intersection of the graphs of functions, i.e., the endpoints of the interval are points that belong to both curves. Example: find the points of intersection of the lines y = x/2 + 5 and y = 3•x – x2/4 + 3 and calculate the area.

5

Solution.

To find the intersection points, write down the equation:

x/2 + 5 = 3•x – x2/4 + 3 → x2 – 10•x + 8 = 0

D = 100 - 64 = 36 → x1,2 = (10 ± 6)/2.

To find the intersection points, write down the equation:

x/2 + 5 = 3•x – x2/4 + 3 → x2 – 10•x + 8 = 0

D = 100 - 64 = 36 → x1,2 = (10 ± 6)/2.

6

So you've found the ends of the interval of integration is [2; 8]:

S = |∫ (3•x – x2/4 + 3 – x/2 - 5)DX| = |(5•x2/4 – X3/12 - 2•x)| ≈ 59.

S = |∫ (3•x – x2/4 + 3 – x/2 - 5)DX| = |(5•x2/4 – X3/12 - 2•x)| ≈ 59.

7

Consider another example: U1 = √(4•x + 5); Y2 = x and the equation of the line x = 3.

In this task, only one end of the interval x=3. This means that the second value you want to find the schedule. Construct a line defined by the functions U1 and U2. It is obvious that the value x=3 is the upper limit, therefore, need to determine the lower limit. For this Paranaita expression:

√(4•x + 5) = x ↑2

4•x + 5 = x2 → x2 – 4•x – 5 = 0

In this task, only one end of the interval x=3. This means that the second value you want to find the schedule. Construct a line defined by the functions U1 and U2. It is obvious that the value x=3 is the upper limit, therefore, need to determine the lower limit. For this Paranaita expression:

√(4•x + 5) = x ↑2

4•x + 5 = x2 → x2 – 4•x – 5 = 0

8

Find the roots of the equation:

D = 16 + 20 = 36 → x1 = 5; x2 = -1.

Look at the chart, the lower interval value is -1. Since U1 is above U2, then:

S = ∫(√(4•x + 5) - x)DX on the interval [-1; 3].

S = (1/3•√((4•x + 5)3) – x2/2) = 19.

D = 16 + 20 = 36 → x1 = 5; x2 = -1.

Look at the chart, the lower interval value is -1. Since U1 is above U2, then:

S = ∫(√(4•x + 5) - x)DX on the interval [-1; 3].

S = (1/3•√((4•x + 5)3) – x2/2) = 19.

# Advice 2 : How to find the area of the figure bounded by the lines

The geometric meaning of the definite integral – the area of the curvilinear trapezoid. To find the area of the figurebounded by lines, is used one of the properties of the integral, which is the additivity of the space of integrable on the same interval functions.

Instruction

1

By definition of the integral, it is equal to the area of the curvilinear trapezoid bounded by the graph of the given function. When you want to find the area of the figure bounded by lines, it is about curves given on the graph the two functions f1(x) and f2(x).

2

Suppose that on some interval [a, b] given two functions defined and continuous. And one of the functions of the chart is higher than the other. Thus, there is a visual figure, bounded by lines of functions and lines x = a, x = b.

3

Then the area of the shape can be expressed by the formula integrating the difference function on the interval [a, b]. Evaluation of the integral is made by the law of Newton-Leibniz, according to which the result is equal to the difference between the integral of the function from the boundary values of the interval.

4

Example1.

Find the area of the figure bounded by straight lines y = -1/3·x – ½, x = 1, x = 4 and the parabola y = -x2 + 6·x – 5.

Find the area of the figure bounded by straight lines y = -1/3·x – ½, x = 1, x = 4 and the parabola y = -x2 + 6·x – 5.

5

Solution.

Plot all lines. You can see that the line of the parabola is above the line y = -1/3·x – ½. Therefore, under the integral sign in this case there must be a difference between the equation of a parabola and the given straight line. The interval of integration, respectively, is located between the points x = 1 and x = 4:

S = ∫(-x2 + 6·x – 5 – (-1/3·x – 1/2))dx = (-x2 +19/3·x – 9/2)dx on the interval [1, 4].

Plot all lines. You can see that the line of the parabola is above the line y = -1/3·x – ½. Therefore, under the integral sign in this case there must be a difference between the equation of a parabola and the given straight line. The interval of integration, respectively, is located between the points x = 1 and x = 4:

S = ∫(-x2 + 6·x – 5 – (-1/3·x – 1/2))dx = (-x2 +19/3·x – 9/2)dx on the interval [1, 4].

6

Find the integral for the resulting expanded expression:

F(-x2 + 19/3x – 9/2) = -1/3x3 + 19/6x2 – 9/2x.

F(-x2 + 19/3x – 9/2) = -1/3x3 + 19/6x2 – 9/2x.

7

Substitute the values of the cut ends:

S = (-1/3·43 + 19/6·42 – 9/2·4) – (-1/3·13 + 19/6·12 – 9/2·1) = 13.

S = (-1/3·43 + 19/6·42 – 9/2·4) – (-1/3·13 + 19/6·12 – 9/2·1) = 13.

8

Example 2.

Calculate the area of the figure bounded by the lines y = √(x + 2), y = x and the line x = 7.

Calculate the area of the figure bounded by the lines y = √(x + 2), y = x and the line x = 7.

9

Solution.

This task is more complex than the previous one, because there is a second straight line, parallel to the x-axis. This means that the second boundary value of the integral is indeterminate. Therefore, it must be found from the graph. Build the specified line.

This task is more complex than the previous one, because there is a second straight line, parallel to the x-axis. This means that the second boundary value of the integral is indeterminate. Therefore, it must be found from the graph. Build the specified line.

10

You will see that the straight line y = x goes diagonally along the coordinate axes. And the graph of the function root is the positive half of a parabola. It is obvious that the lines on the graph intersect, so the intersection point will be the lower limit of integration.

11

Find the point of intersection, solving the equation:

x = √(x + 2) → x2 = x + 2 [x ≥ -2] → x2 – x – 2 = 0.

x = √(x + 2) → x2 = x + 2 [x ≥ -2] → x2 – x – 2 = 0.

12

Determine the roots of the quadratic equation using the discriminant:

D = 9 → x1 = 2; x2 = -1.

D = 9 → x1 = 2; x2 = -1.

13

It is obvious that -1 is not a solution because the abscissa of the crossing currents – positive magnitude. Consequently, the second limit of integration x = 2. The function y = x on the graph above of the function y = √(x + 2) so the integral it will be the first.

Integrate the resulting expression on the interval [2, 7] and find the area of the shape:

S = ∫(x - √(x + 2))dx = (x2/2 – 2/3·(x + 2)^(3/2)).

Integrate the resulting expression on the interval [2, 7] and find the area of the shape:

S = ∫(x - √(x + 2))dx = (x2/2 – 2/3·(x + 2)^(3/2)).

14

Substitute the range values:

S = (72/2 – 2/3·9^(3/2)) – (22/2 – 2/3·4^(3/2)) = 59/6.

S = (72/2 – 2/3·9^(3/2)) – (22/2 – 2/3·4^(3/2)) = 59/6.

# Advice 3 : How to calculate the area of the figure bounded by the parabola

Even from a school course it is known that for finding areas of figures on the coordinate plane requires knowledge of such concepts as integral. For its application to determine the areas of curvilinear trapezoids - exactly what are these figures enough to know certain algorithms.

Instruction

1

To calculate the area of the figure bounded by the parabola, draw it in the Cartesian coordinate system. For the image of the parabola should know a minimum of three points, one needs to be a top. To find the coordinate of the vertices along the X-axis, substitute the known data into the formula x=-b/2a, Y-axis, substitute the obtained value of the argument to the function. After that, analyze the graphics data included in the problem statement. If the vertex is below the x axis, the branches are directed upwards, if the above — down. The remaining 2 points are coordinates of intersection with the axis OX. Stroke resulting figure. This will greatly facilitate the solution of this problem.

2

Then, determine the limits of integration. They are usually specified in the problem statement using variables a and b. These values are put in the upper and lower parts of the symbol of the integral, respectively. After the integral, enter the value of the function and multiply it by dx (for example, (x2)dx in the case of the parabola). Then calculate the integral value of a function in General form using a special table at the link given in the section "Additional sources", and then substitute back the limits of integration and find the difference. The difference will be the area.

3

There is also the possibility of calculating the integral and using software. To do this, follow the link in the section "Additional sources", a special mathematical website. In the open text box, enter the integral of f(x), where f(x) — recording function, the schedule limits the area of the figure on the coordinate plane. After that press the button with the symbol equal to. The opening page portray the resulting figure, and show the calculations of its area.