Lecture. Definite integral: Riemann integral.
We conclude the section "Integral calculus" with the topic "Definite integral", which is also called Riemann integral. The problem that leads to the concept of a definite integral is the problem of the area of a curvilinear trapezoid.
Let's see what this concept is. So, we are given a function y=f(x) defined on the interval [a, b], it is continuous, non-negative. This means that the graph is located above the x-axis. The resulting planar figure (its periphery is made by the graph of f function, the x-axis, and right lines parallel to the y-axis x=a, x=b) is a figure and is called a curvilinear trapezoid.
How to find its area and what to start from? We will assume that we know the area of the rectangle. How, knowing of the rectangle area, to calculate the area of this figure, the upper periphery of which, as you can see, is not at all a horizontal line.
Divide interval [a, b] on the x axis with points x1, x2, and so on. We will mark points a and b: a is x0, b is xn. We got n partial intervals on the Ox axis. Draw vertical lines through these points and build rectangles. What do we take as the height of these rectangles? We randomly select points ξi on each partial interval. Note that points xi are also taken randomly.
This is not obligatory. The lengths of the partial intervals are equal, as in this picture. So, for each partial interval selected as the rectangle base, the value of the rectangle height is equal to the function value in point ξi. We got a stepped figure made up of rectangles.
The area of such a figure is equal to the sum of the areas of n rectangles, and in general, it is clear that the area of this figure is approximately close to the area of a curvilinear trapezoid. If we mark the length of this interval as Δxi, then the area of the step figure is approximately equal to the sum of the areas of n rectangles.
After the sum sign, there is the product f(ξi)Δxi which is the area of this rectangle. Let's denote λ the largest of the lengths of partial intervals. This value is called the subinterval.
The smaller λ, the smaller the partitioning of interval [a, b] becomes, and the more accurate the area value of the step figure is to the area value of the curvilinear trapezoid. The limit value is the area of the curvilinear trapezoid. So, S is the limit of the compiled sum with subinterval tending to zero.
Well, now let's move on to the definition of a definite integral. Function f is defined on interval [a, b]. We partition the interval just as when solving problem A for the area of a curvilinear trapezoid. We will reason - we will make a sum, in this case it is no longer the area of a curvilinear trapezoid, but the term "integral sum of function f".
It is composed for the given partition and for the given selection of points ξi. And for a different selection of points, the value of this integral sum will also be different.
Λ is a again subinterval. Let's aim λ to zero. If the limit of integral sums is finite and does not depend on the choice of points ξi and the choice of points xi, on the partitioning of interval [a, b], then the value of this limit is called the definite integral of function f on interval [a, b] and is denoted by the symbol.
We see that this symbol is very similar to the symbol of an indefinite integral with the only difference – there appear values a and b, which are called integration limits. a is the lower limit of integration, and b is the upper limit of integration.
What should be noted? The definite integral is a number, so the variable name doesn't matter the horizontal axis can be named differently, so it can be letter t, it can be letter z, and so on. If the limits of integration are equal, then the definite integral of the function on this interval (numerical) is zero.
Important theorems that relate the concept of integrability of a function to other properties of a function. A necessary condition for integrability of a function on an interval is boundedness, if the function is integrable, then it is bounded on interval [a, b]. The converse theorem is not true. There are functions that are bounded, but not integrable.
Another important theorem that will immediately allow us to distinguish a huge class of functions for which there is a definite integral is continuous functions. Sufficient condition for integrability. If a function is continuous on an interval, then it is integrable on that interval. An important class of integrable functions is continuous functions.
Let's move on to the properties of a definite integral. We will use these properties later to calculate definite integrals. The first property states that if you interchange the position of integration limits, the integral changes its sign to the opposite. Linearity property follow the linearity property written for an indefinite integral, we only do not forget to write down integration limits.
The third property is often used to calculate the areas of planar figures. If a function is integrable on an interval, then it is integrable on any other interval contained in this one, and the integral is divided into two summands.
If the function is non-negative, this property is easily illustrated by the area of the curvilinear trapezoid constructed for function f on interval [a, b], which is equal to the sum of the areas of the curvilinear trapezoids of the other two S1 and S2. But it should also be remembered that this highlighted formula is valid not only in the case when the point c lies between intervals, between points a and b, but also for any other mutual arrangement of points a, b and c.
Properties of the integral for even and uneven functions. For a symmetric interval [-a, a], the integral can be narrowed to be considered on the interval from [0, a]. For an even function, we get a doubled integral, and for an uneven function, the integral is zero.
Well, the illustrations show that for an uneven function, the area is divided into two symmetrical parts and the same is done for an even function. For an uneven function, the summands differ by a sign. For an even function, the sum of the summands is twice as large as each of them.
Properties associated with inequality. If the function is non-negative, then the integral of this function also takes a non-negative value. If it is less than zero, then the integral will also be less than 0 or equal to 0. You can easily get from this property, if function f1 and f2 are connected by inequality (more than or equal to), then the integrals of these functions on interval [a, b] are also connected by the same inequality.
The sixth property. The integral module does not exceed the integral of the function module.
The seventh property. The average value of an integrable function. This property states that if a function is integrable on interval [a, b], then there is such a value c belonging to the interval [a, b] that the integral is equal to the product of the interval length by the function value at this point. The value, as we see in the picture, μ, equal to f (c), is called the average function value on interval [a, b].
In order to proceed to the calculation of definite integrals, we need to consider the concept of an integral with a variable upper limit. So, let function f be continuous on interval [a, b]. By the properties of a definite integral, it is integrable on any interval contained in this interval. We will consider interval from [a, x], where x is a variable that takes the value of [a, b]. In this case, each value of x is assigned its own value of integral Ф(x). Ф(x) becomes a function of variable x running through values [a, b]. This function is called an integral with a variable upper limit. If function f is non-negative, then f(x) is the area of the curvilinear trapezoid for interval [a, x]. If you change x values f(x) changes.
A theorem about this integral. If the function is continuous on interval [a, b], then function f(x) of the integral with a variable upper limit has the following property. The integral with a variable upper limit is a continuous function on interval [a, b], a differentiable function on interval (a, b), and the derivative of this function is equal to integrand function f(x) at any point x of interval (a, b).
We get very important conclusions from this theorem. Since functions Ф and f are connected by differentiation, function f(x) is primitive of the function f (x). And since it is a primitive one, it follows from the definition of the indefinite integral that the indefinite integral of function f(x) is Ф (x) +C.
One last thing. The formula written here shows the relation between indefinite and definite integrals. It is this relation that we will use later to derive Newton-Leibniz formula.
Let F(x) (F large of x) be some primitive function f(x), then function f(x) of an integral with a variable upper limit, being also the primitive of functions f(x) differs from F(x) by a constant. This means we can write the formula. We will substitute x equal to a in this formula. We get the integral with upper and lower integration limits equal to a, and it equals to zero.
This gives us value c equal to minus f large from a (F(a)). We place the selected formula instead of the upper limit x x=b and get a definite integral [a, b] equal to F(b)+c. But we found value of c earlier and get the formula.
So, F(b)-F(a), taking into account the remarks that there may be any variable in a definite integral, we derive Newton-Leibniz formula as a result. So, the definite integral of function f on interval [a, b] is equal to the difference between the values of primitive F at points b and a, at the interval ends.
To solve problems related to calculating a definite integral, the difference F(b)-F(a) is often written as F(x), then the vertical line and integration limits of a and b are written after the line. In this case, the Newton-Leibniz formula is written as follows.
This calculation has some properties that are useful further in solving practical problems. This linearity property - calculation from the sum of functions develops into the sum of calculations. And a constant multiplier can also be taken off the sign of this calculation.
A simple example. We don't do integral calculations yet, but see how you can use these properties. You can, of course, calculate the value of the function at point 2, and then calculate at point -1, and write the difference between these two numbers.
But we use the property that we formulated above. The linearity property is used as follows. Then we multiply 1/3 by the difference of the values of x in the cube at points 2 and -1, and so on. And the calculation comes to this conclusion. Thus is the answer.
Let's move on to geometric applications of a definite integral. The first application is the area of a curvilinear trapezoid, as we have already discussed. If the planar figure is limited to graphs of two functions f and g, and the graph of f is above the graph of function g, regardless whether axis Oh is above or below this figure, boundaries are also made by lines x=a and x=b - vertical lines, the area of this figure is equal to the integral of the difference between functions f and g (subtract the lower function from the top one) at the interval [a, b].
Volume of the solid of revolution. What is a solid of revolution? Let the curvilinear trapezoid rotate about ox axis. The solid obtained during rotation, we call the solid of revolution. x index indicates that the trapezoid rotated about ox axis. You can see the formula.
Also, a definite integral allows you to calculate the surface area for this solid of revolution. Here is the formula.
If a curvilinear trapezoid rotates about Oу axis, we also get a solid of revolution. To calculate the volume, you can also use the previous formula, only re-assigning variables x and y, or you can use the formula that you see on the frame.
The next application is calculating the length of the curve. If the curve is a graph of function f at interval [a, b], then the length is expressed by the following formula. If the curve is set parametrically (x is a function of variable t and y is a function of te variable t) at a definite interval [a, b], then the formula for the length of the curve has the following form.
Let's move on to calculating a definite integral. To calculate the definite integral, the same methods are used which we considered for the indefinite integral. To illustrate some of the features, we will move on to calculating, to solving practical problems.
Let's illustrate the method of replacing a variable. Let's introduce a new variable, denoting the root expression t^2. Find dx - this is differential t^2 - derivative 2t and do not forget to multiply by the differential variable - dt. The peculiarity of calculating a definite integral is that we must also change integration limits.
In the original integral, we are given the bounds of changes in variable x, the new integral will contain variable t. Find out how variable t changes, if x were equal to zero, then look at the formula - x is t^2, we are to find out what t is equal to. Well, let's see how t is expressed. x takes a positive value, and t is root of x.
If x were zero, the root of x takes value 0. If x (upper value, upper limit value) were 4, then t is two. We get a new integral. We write 2tdt in the numerator instead of dx, in the denominator - the root of t^2 (generally speaking, this is module of t, but t takes a positive value so it is t), plus 1.
The limits of integration vary from 0 to 2. We got a fractionally rational function. Let's start by taking 2 off the integral sign. We will write dt multiplier here, in the denominator - t+1, in the numerator - t. A fractionally rational function is under the sign of an integral.
We begin to analyze the situation with the fact whether this fraction is proper or not. This fraction is improper, so we must select the quotient (divide with excess). It is convenient to perform identical transformations here
Add a unit and subtract a unit and divide with the excess. So under the integral sign we get this (so do not forget to write integration limits - this is a definite integral, t varies from 0 to 2), so we see that when dividing the numerator by the denominator we get, the quotient as a unit and the next summand is a fraction.
We use the linearity property and write two integrals dt from 0 to 2 minus, do not forget to multiply the second integral by 2, dt divided by (t+1), t changes from 0 to 2. The first integral is immediately tabular. In the second, under the sign of the differential, we add (t+1) and then apply the table of integrals. The first integral is 2t, the second integral is minus two natural logarithms of module (t+1).
Now we either put a vertical line for the entire expression, but using the calculation properties, we can write the integration limits after the vertical line for the first summand from 0 to 2 and for the second summand also from 0 to 2. Perform the calculation. Let's write in detail - the upper value of t is 2, lower values of t – zero) by (2-0), multiply the natural logarithm of the upper value by minus 2 (so I will write - 2 multiplied by the upper value (substitute 2 instead of t) is 3) minus substituted the lower value of the logarithmic unit. The logarithm of unit is zero, so the next record will be the final answer. We got two multiplied by two is four, minus two natural logarithms of three (4-2ln3).