The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=\frac(2)(3)x^3-20x^2+201x-\frac(5)(9)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323383. Prototype No:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(4)(9)x^3-\frac(34)(3)x^2-\frac(280)(3)x-\frac(18)(5 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323385. Prototype No:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(1)(6)x^3-\frac(17)(4)x^2-35x-\frac(5)(11)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323387. Prototype No:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(1)(5)x^3-\frac(9)(2)x^2-30x-\frac(11)(8)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323389. Prototype No:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(11)(30)x^3-\frac(33)(4)x^2-\frac(297)(5)x-\frac(1)(2 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323391. Prototype No:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(7)(27)x^3-\frac(35)(6)x^2-42x-\frac(7)(4)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323393. Prototype No:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(1)(4)x^3-\frac(21)(4)x^2-\frac(135)(4)x-\frac(13)(2 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323395. Prototype No:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-x^3-21x^2-144x-\frac(11)(4)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323397. Prototype No:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(5)(8)x^3-\frac(105)(8)x^2-90x-\frac(1)(2)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task No.:
323399. Prototype No:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(1)(10)x^3-\frac(21)(10)x^2-\frac(72)(5)x-\frac(4)(3 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
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Job type: 7
Topic: Antiderivative of function
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight segments). Using the figure, calculate F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x).
Show solutionSolution
According to the Newton-Leibniz formula, the difference F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid limited by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5.
From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and height 3. Its area is equal
\frac(4+3)(2)\cdot 3=10.5.
Job type: 7
Topic: Antiderivative of function
Condition
Answer
Solution
The figure shows a graph of the function y=F(x) - one of the antiderivatives of some function f(x) defined on the interval (-5; 5).
\frac(4+3)(2)\cdot 3=10.5.
Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].
Job type: 7
Topic: Antiderivative of function
Condition
According to the definition of an antiderivative, the equality holds: F"(x)=f(x). Therefore, the equation f(x)=0 can be written as F"(x)=0.
Solution
Since the figure shows the graph of the function y=F(x), we need to find those points in the interval [-3; 4], in which the derivative of the function F(x) is equal to zero. It is clear from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph.
From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and height 3. There are exactly 7 of them in the indicated interval (four minimum points and three maximum points).
\frac(4+3)(2)\cdot 3=10.5.
Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].
Job type: 7
Topic: Antiderivative of function
Condition
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Solution
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight segments). Using the figure, calculate F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x).
According to the Newton-Leibniz formula, the difference F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid limited by the graph of the function y=f(x), straight lines y=0 , x=5 and x=0.
\frac(4+3)(2)\cdot 3=10.5.
Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].
Job type: 7
Topic: Antiderivative of function
Condition
The figure shows a graph of some function y=f(x).
The function F(x)=-x^3+4.5x^2-7 is one of the antiderivatives of the function f(x).
Solution
Find the area of the shaded figure. The shaded figure is a curvilinear trapezoid bounded from above by the graph of the function y=f(x), straight lines y=0, x=1 and x=3. According to the Newton-Leibniz formula, its area S is equal to the difference F(3)-F(1), where F(x) is the antiderivative of the function f(x) specified in the condition. That's why 6,5-(-3,5)= 10.
\frac(4+3)(2)\cdot 3=10.5.
Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].
Job type: 7
Topic: Antiderivative of function
Condition
S=
F(3)-F(1)=
-3^3 +(4.5)\cdot 3^2 -7-(-1^3 +(4.5)\cdot 1^2 -7)=
The figure shows a graph of some function y=f(x). The function F(x)=x^3+6x^2+13x-5 is one of the antiderivatives of the function f(x). Find the area of the shaded figure.
Showing the connection between the sign of the derivative and the nature of the monotonicity of the function.
Please be extremely careful about the following. Look, the schedule of WHAT is given to you! Function or its derivative
If given a graph of the derivative
, then we will be interested only in the function signs and zeros.
We are not interested in any “hills” or “hollows” in principle!
Task 1.
The figure shows a graph of a function defined on the interval. Determine the number of integer points at which the derivative of the function is negative.
If given a graph of the derivative
Solution: In the figure, the areas of decreasing function are highlighted in color: These decreasing regions of the function contain 4 integer values.
Task 2.
The figure shows a graph of a function defined on the interval. Find the number of points at which the tangent to the graph of the function is parallel to or coincides with the line.
Once the tangent to the graph of a function is parallel (or coincides) with a straight line (or, which is the same thing), having
slope
, equal to zero, then the tangent has an angular coefficient .
If given a graph of the derivative
This in turn means that the tangent is parallel to the axis, since the slope is the tangent of the angle of inclination of the tangent to the axis.
Therefore, we find extremum points (maximum and minimum points) on the graph - it is at these points that the functions tangent to the graph will be parallel to the axis.
There are 4 such points.
Task 3.
The figure shows a graph of the derivative of a function defined on the interval. Find the number of points at which the tangent to the graph of the function is parallel to or coincides with the line.
The figure shows a graph of a function defined on the interval. Find the number of points at which the derivative of the function is 0.
If given a graph of the derivative
The derivative is equal to zero at extremum points. We have 4 of them:
Task 5.
The figure shows a graph of a function and eleven points on the x-axis:. At how many of these points is the derivative of the function negative?
If given a graph of the derivative
On intervals of decreasing function, its derivative takes negative values. And the function decreases at points. There are 4 such points.
Task 6.
The figure shows a graph of a function defined on the interval. Find the sum of the extremum points of the function.
If given a graph of the derivative
Extremum points– these are the maximum points (-3, -1, 1) and minimum points (-2, 0, 3).
Sum of extremum points: -3-1+1-2+0+3=-2.
Task 7.
The figure shows a graph of the derivative of a function defined on the interval. Find the intervals of increase of the function. In your answer, indicate the sum of integer points included in these intervals.
If given a graph of the derivative
The figure highlights the intervals where the derivative of the function is non-negative.
There are no integer points on the small increasing interval; on the increasing interval there are four integer values: , , and .
Their sum:
Task 8.
The figure shows a graph of the derivative of a function defined on the interval. Find the intervals of increase of the function. In your answer, indicate the length of the largest of them.
If given a graph of the derivative
In the figure, all intervals on which the derivative is positive are highlighted in color, which means the function itself increases on these intervals.
The length of the largest of them is 6.
Task 9.
The figure shows a graph of the derivative of a function defined on the interval. At what point on the segment does it take on the greatest value?
If given a graph of the derivative
Let's see how the graph behaves on the segment, which is what we are interested in only the sign of the derivative .
The sign of the derivative on is minus, since the graph on this segment is below the axis.
Hello, friends! In this article we will look at tasks for antiderivatives. These tasks are included in the Unified State Examination in mathematics. Despite the fact that the sections themselves - differentiation and integration - are quite capacious in the algebra course and require a responsible approach to understanding, the tasks themselves, which are included in the open bank of tasks in mathematics and will be extremely simple on the Unified State Exam and can be solved in one or two steps.
It is important to understand exactly the essence of the antiderivative and, in particular, the geometric meaning of the integral. Let us briefly consider the theoretical foundations.
Geometric meaning of the integral
Briefly about the integral we can say this: the integral is the area.
Definition: Let a graph of a positive function f defined on the segment be given on the coordinate plane. A subgraph (or curvilinear trapezoid) is a figure bounded by the graph of a function f, the lines x = a and x = b and the x-axis.
Definition: Let a positive function f be given, defined on a finite segment. The integral of a function f on a segment is the area of its subgraph.
As already said F′(x) = f (x).What can we conclude?
It's simple. We need to determine how many points there are on this graph at which F′(x) = 0. We know that at those points where the tangent to the graph of the function is parallel to the x axis. Let's show these points on the interval [–2;4]:
These are the extremum points of a given function F (x). There are ten of them.
Answer: 10
323078. The figure shows a graph of a certain function y = f (x) (two rays with a common starting point). Using the figure, calculate F (8) – F (2), where F (x) is one of the antiderivatives of the function f (x).
Let us write down the Newton–Leibniz theorem again:Let f be a given function, F its arbitrary antiderivative. Then
And this, as already said, is the area of the subgraph of the function.
Thus, the problem comes down to finding the area of the trapezoid (interval from 2 to 8):
It is not difficult to calculate it by cells. We get 7. The sign is positive, since the figure is located above the x-axis (or in the positive half-plane of the y-axis).
Even in this case, one could say this: the difference in the values of the antiderivatives at the points is the area of the figure.
Answer: 7
323079. The figure shows a graph of a certain function y = f (x). The function F (x) = x 3 +30x 2 +302x–1.875 is one of the antiderivatives of the function y= f (x). Find the area of the shaded figure.
As has already been said about the geometric meaning of the integral, this is the area of the figure limited by the graph of the function f (x), the straight lines x = a and x = b and the ox axis.
Theorem (Newton–Leibniz):
Thus, the task comes down to calculating the definite integral of a given function on the interval from –11 to –9, or in other words, we need to find the difference in the values of the antiderivatives calculated at the indicated points:
Answer: 6
323080. The figure shows a graph of some function y = f (x).
Function F (x) = –x 3 –27x 2 –240x– 8 is one of the antiderivatives of the function f (x). Find the area of the shaded figure.
Theorem (Newton–Leibniz):
The problem comes down to calculating the definite integral of a given function over the interval from –10 to –8:
Answer: 4
Another solution to this problem, from the site.
Derivatives and differentiation rules are also in . It is necessary to know them, not only to solve such tasks.
You can also look at the help information on the website and.
Watch a short video, this is an excerpt from the film “The Blind Side”. We can say that this is a film about education, about mercy, about the importance of supposedly “random” meetings in our lives... But these words will not be enough, I recommend watching the film itself, I highly recommend it.
I wish you success!
Best regards, Alexander Krutitskikh
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