{"id":4486,"date":"2026-09-01T12:04:03","date_gmt":"2026-09-01T12:04:03","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4486"},"modified":"2026-09-01T14:06:26","modified_gmt":"2026-09-01T14:06:26","slug":"definite-integral","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/definite-integral.html","title":{"rendered":"Definite Integral: Geometric Meaning and Properties"},"content":{"rendered":"

The definite integral is one of the fundamental concepts of calculus. It assigns a specific numerical value to a function over a given interval. This value can be obtained as the limit of approximations constructed using a large number of narrow rectangles. But what does this number mean geometrically, and which properties help us work with it?<\/p>\n

To answer these questions, we will first look at how an integral sum is formed and how the definite integral arises from it. Then we will move on to its geometric meaning, main properties, methods of estimation, and the special features of integration over symmetric intervals.<\/p>\n

Definite Integral: From an Integral Sum to a Number<\/h2>\n

Suppose a function \\( f(x) \\) is defined on the interval \\( [a,b] \\). Divide this interval into smaller parts using the points<\/p>\n

\\[
\na=x_0<x_1<x_2<\\dots<x_n=b.
\n\\]<\/p>\n

As a result, we obtain the subintervals<\/p>\n

\\[
\n[x_0,x_1],\\ [x_1,x_2],\\ \\dots,\\ [x_{n-1},x_n].
\n\\]<\/p>\n

The length of each subinterval is<\/p>\n

\\[
\n\\Delta x_i=x_i-x_{i-1}.
\n\\]<\/p>\n

In each subinterval \\( [x_{i-1},x_i] \\), choose a point \\( \\xi_i \\). The value of the function at this point is \\( f(\\xi_i) \\). If<\/p>\n

\\[
\nf(x)\\geq0,
\n\\]<\/p>\n

then \\( f(\\xi_i) \\) can be taken as the height of a small rectangle, while \\( \\Delta x_i \\) is its width. The area of this rectangle is therefore<\/p>\n

\\[
\nf(\\xi_i)\\cdot\\Delta x_i.
\n\\]<\/p>\n

We do this for all parts of the interval and add the resulting areas:<\/p>\n

\\[
\n\\sum_{i=1}^{n}f(\\xi_i)\\cdot\\Delta x_i.
\n\\]<\/p>\n

This sum is called an integral sum<\/a>.<\/p>\n

For a nonnegative function, the integral sum can be interpreted geometrically as the total area of narrow rectangles used to approximate the curved region under the graph of the function. As long as the rectangles have a finite width, the resulting value remains an approximation.<\/p>\n

\"An<\/p>\n

How can we make this approximation more accurate? To do this, the interval \\( [a,b] \\) is divided into smaller and smaller parts. Accordingly, the rectangles become narrower and narrower.<\/p>\n

Let the maximum length of a subinterval be denoted by<\/p>\n

\\[
\n\\lambda=\\max\\Delta x_i.
\n\\]<\/p>\n

Now consider the case when<\/p>\n

\\[
\n\\lambda\\to0.
\n\\]<\/p>\n

This means that the greatest width among all parts of the partition approaches zero.<\/p>\n

If, at the same time, the integral sum approaches the same finite number regardless of how the interval is divided and how the points \\( \\xi_i \\) are chosen, then this number is called the definite integral of the function \\( f(x) \\) over the interval \\( [a,b] \\):<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx=
\n\\lim_{\\lambda\\to0}
\n\\sum_{i=1}^{n}f(\\xi_i)\\cdot\\Delta x_i.
\n\\]<\/p>\n

In other words, the integral sum gives an approximate value, while the definite integral is the number that these sums approach as the widths of the subintervals approach zero.<\/p>\n

In the notation<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx,
\n\\]<\/p>\n

\\( a \\) is called the lower limit of integration, and \\( b \\) is called the upper limit of integration. The function \\( f(x) \\) is the integrand, while \\( dx \\) indicates the variable of integration.<\/p>\n

If the function \\( f(x) \\) is continuous on the interval \\( [a,b] \\), then its definite integral over this interval exists.<\/p>\n

Thus, the definite integral is obtained as the limit of integral sums. Now we can determine what exactly this number means geometrically.<\/p>\n

Geometric Meaning of the Definite Integral: Area and Sign<\/h2>\n

Suppose that throughout the entire interval \\( [a,b] \\),<\/p>\n

\\[
\nf(x)\\geq0.
\n\\]<\/p>\n

Then the graph of the function lies on or above the \\( x \\)-axis.<\/p>\n

The region bounded by the graph<\/p>\n

\\[
\ny=f(x),
\n\\]<\/p>\n

the \\( x \\)-axis, and the vertical lines<\/p>\n

\\[
\nx=a,\\qquad x=b,
\n\\]<\/p>\n

is called a curvilinear trapezoid.<\/p>\n

If the function is nonnegative over the entire interval, then the definite integral is equal to the area of this region:<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx=A.
\n\\]<\/p>\n

So, in this case, the geometric meaning of the definite integral is simple: its numerical value is equal to the area of the region under the graph of the function.<\/p>\n

However, the graph does not always lie above the \\( x \\)-axis. If throughout the entire interval<\/p>\n

\\[
\nf(x)\\leq0,
\n\\]<\/p>\n

then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx\\leq0.
\n\\]<\/p>\n

An ordinary geometric area cannot be negative. Therefore, in the general case, the definite integral should not simply be identified with area.<\/p>\n

To interpret it geometrically, we use the concept of signed, or algebraic, area. The part of the region above the \\( x \\)-axis is counted with a positive sign, while the part below it is counted with a negative sign.<\/p>\n

\"Geometric<\/p>\n

Suppose the graph lies above the \\( x \\)-axis on one part of the interval and encloses an area \\( A_1 \\), while on another part it lies below the axis and encloses an area \\( A_2 \\). Then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx=A_1-A_2.
\n\\]<\/p>\n

If we need to find the ordinary geometric area of the entire region between the graph and the \\( x \\)-axis, then<\/p>\n

\\[
\nA=A_1+A_2.
\n\\]
\nThis is why the definite integral and the geometric area can have different values.<\/p>\n

For example, if<\/p>\n

\\[
\nA_1=A_2,
\n\\]<\/p>\n

then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx=0,
\n\\]<\/p>\n

even though the geometric area itself is not zero.<\/p>\n

Thus, the value of the definite integral depends not only on the size of the regions between the graph and the \\( x \\)-axis, but also on which side of the axis they lie.<\/p>\n

Now that the geometric meaning of the definite integral is clear, we can move on to the rules that help us work with it.<\/p>\n

Main Properties of the Definite Integral: Limits, Sums, and Constants<\/h2>\n

Let us begin with the limits of integration. If the lower and upper limits are the same, then<\/p>\n

\\[
\n\\int_a^a f(x)\\,dx=0.
\n\\]<\/p>\n

The interval \\( [a,a] \\) has zero length, so the value of the integral over it is also zero.<\/p>\n

If we reverse the limits, we obtain<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n=
\n-\\int_b^a f(x)\\,dx.
\n\\]<\/p>\n

So, when the limits of integration are reversed, the sign of the integral changes.<\/p>\n

The next property concerns splitting the interval. Suppose that the point \\( c \\) belongs to the interval \\( [a,b] \\). Then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n=
\n\\int_a^c f(x)\\,dx+
\n\\int_c^b f(x)\\,dx.
\n\\]<\/p>\n

This property is called the additivity of the definite integral over the interval of integration.<\/p>\n

Its meaning is simple. If the interval \\( [a,b] \\) is divided into two parts at the point \\( c \\), then the integral over the entire interval is equal to the sum of the integrals over the two resulting parts.<\/p>\n

Now let us consider properties related to the functions being integrated.<\/p>\n

For the sum of two integrable functions,<\/p>\n

\\[
\n\\int_a^b\\left(f(x)+g(x)\\right)\\,dx
\n=
\n\\int_a^b f(x)\\,dx+
\n\\int_a^b g(x)\\,dx.
\n\\]<\/p>\n

For the difference of two functions, we similarly have<\/p>\n

\\[
\n\\int_a^b\\left(f(x)-g(x)\\right)\\,dx
\n=
\n\\int_a^b f(x)\\,dx-
\n\\int_a^b g(x)\\,dx.
\n\\]<\/p>\n

If a function is multiplied by a constant \\( k \\), this constant can be taken outside the integral sign:<\/p>\n

\\[
\n\\int_a^b k\\cdot f(x)\\,dx
\n=
\nk\\cdot\\int_a^b f(x)\\,dx.
\n\\]<\/p>\n

In general, these properties can be combined into one formula:<\/p>\n

\\[
\n\\int_a^b\\left(\\alpha\\cdot f(x)+\\beta\\cdot g(x)\\right)\\,dx
\n=
\n\\alpha\\cdot\\int_a^b f(x)\\,dx+
\n\\beta\\cdot\\int_a^b g(x)\\,dx,
\n\\]<\/p>\n

where \\( \\alpha \\) and \\( \\beta \\) are constants.<\/p>\n

This property is called the linearity of the definite integral.<\/p>\n

Thus, a sum or difference of functions can be integrated separately, while constant factors can be taken outside the integral sign. In addition, the interval of integration can be divided into parts whenever necessary.<\/p>\n

These properties show how to work with definite integrals and simplify their notation. They also make it possible to compare integral values and establish certain bounds.<\/p>\n

Definite Integral: Comparison and Estimation<\/h2>\n

Suppose that throughout the interval \\( [a,b] \\), where \\( a<b \\),<\/p>\n

\\[
\nf(x)\\leq g(x).
\n\\]<\/p>\n

Then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n\\leq
\n\\int_a^b g(x)\\,dx.
\n\\]<\/p>\n

So, if one function does not exceed another anywhere on the interval, then its definite integral over the same interval will also not be greater.<\/p>\n

An important estimate follows directly from this property.<\/p>\n

Suppose that for every \\( x\\in[a,b] \\),<\/p>\n

\\[
\nm\\leq f(x)\\leq M,
\n\\]<\/p>\n

where \\( m \\) and \\( M \\) are constants.<\/p>\n

This means that all values of the function \\( f(x) \\) on the interval lie between \\( m \\) and \\( M \\). Therefore, its integral can be compared with the integrals of constant functions:<\/p>\n

\\[
\n\\int_a^b m\\,dx
\n\\leq
\n\\int_a^b f(x)\\,dx
\n\\leq
\n\\int_a^b M\\,dx.
\n\\]<\/p>\n

For constant functions, we have<\/p>\n

\\[
\n\\int_a^b m\\,dx=m\\cdot(b-a)
\n\\]<\/p>\n

and<\/p>\n

\\[
\n\\int_a^b M\\,dx=M\\cdot(b-a).
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nm\\cdot(b-a)
\n\\leq
\n\\int_a^b f(x)\\,dx
\n\\leq
\nM\\cdot(b-a).
\n\\]<\/p>\n

This inequality allows us to determine the range in which the value of the integral lies.<\/p>\n

For example, suppose that<\/p>\n

\\[
\n2\\leq f(x)\\leq5,\\qquad x\\in[1,4].
\n\\]<\/p>\n

Then<\/p>\n

\\[
\n2\\cdot(4-1)
\n\\leq
\n\\int_1^4 f(x)\\,dx
\n\\leq
\n5\\cdot(4-1),
\n\\]<\/p>\n

which gives<\/p>\n

\\[
\n6\\leq
\n\\int_1^4 f(x)\\,dx
\n\\leq15.
\n\\]<\/p>\n

So, even without calculating the integral exactly, we can determine that its value lies between \\( 6 \\) and \\( 15 \\).<\/p>\n

Another important estimate involves the absolute value of the function:<\/p>\n

\\[
\n\\left|\\int_a^b f(x)\\,dx\\right|
\n\\leq
\n\\int_a^b |f(x)|\\,dx.
\n\\]<\/p>\n

Why does this inequality hold? In a definite integral, positive and negative parts can partially cancel each other out. For the function \\( |f(x)| \\), all values are nonnegative, so this cancellation no longer occurs.<\/p>\n

Thus, comparison properties allow us to estimate the possible value of a definite integral even before finding it exactly.<\/p>\n

In some cases, we can obtain an even simpler result. To do this, it is enough to consider the symmetry of both the interval and the function.<\/p>\n

Symmetry of the Definite Integral: Even and Odd Functions<\/h2>\n

Consider the symmetric interval \\( [-a,a] \\). On such an interval, the result depends on whether the function is even or odd.<\/p>\n

\"Symmetry<\/p>\n

If the function \\( f(x) \\) is even, then<\/p>\n

\\[
\nf(-x)=f(x).
\n\\]<\/p>\n

The graph of an even function is symmetric about the \\( y \\)-axis. Therefore, the parts of the graph on the intervals \\( [-a,0] \\) and \\( [0,a] \\) have the same shape.<\/p>\n

It follows that<\/p>\n

\\[
\n\\int_{-a}^{a}f(x)\\,dx
\n=
\n2\\cdot\\int_0^a f(x)\\,dx.
\n\\]<\/p>\n

So, for an even function, it is enough to find the integral over half of the symmetric interval and then double the result.<\/p>\n

\"Symmetry<\/p>\n

If the function \\( f(x) \\) is odd, then<\/p>\n

\\[
\nf(-x)=-f(x).
\n\\]<\/p>\n

Its graph is symmetric about the origin. The function values at the points \\( x \\) and \\( -x \\) have the same absolute value but opposite signs.<\/p>\n

Therefore, the corresponding parts of the integral over a symmetric interval cancel each other out:<\/p>\n

\\[
\n\\int_{-a}^{a}f(x)\\,dx=0.
\n\\]<\/p>\n

So, before evaluating a definite integral over the interval \\( [-a,a] \\), it is useful to check whether the integrand is even or odd. For an even function, the integration can be reduced to half of the interval, while for an odd function, the value of the integral over the entire symmetric interval is immediately equal to zero.<\/p>\n

Definite Integral in Practice: Definition and Main Properties<\/h2>\n

The definition of the definite integral and its properties are easier to understand when they are applied to specific problems. Let us consider examples in which we will use an integral sum, function symmetry, linearity, additivity, reversal of the limits of integration, and estimation of the integral value.<\/p>\n

Example 1. Evaluate using the definition
\n\\[
\n\\int_1^4 3\\,dx.
\n\\]<\/h3>\n

Divide the interval \\( [1,4] \\) into \\( n \\) equal parts.<\/p>\n

The length of the entire interval is<\/p>\n

\\[
\n4-1=3.
\n\\]<\/p>\n

Therefore, the length of each subinterval is<\/p>\n

\\[
\n\\Delta x=\\frac{3}{n}.
\n\\]<\/p>\n

The integrand is a constant function:<\/p>\n

\\[
\nf(x)=3.
\n\\]<\/p>\n

So, regardless of which point \\( \\xi_k \\) is chosen in each part of the interval,<\/p>\n

\\[
\nf(\\xi_k)=3.
\n\\]<\/p>\n

Now form the integral sum:<\/p>\n

\\[
\nS_n=
\n\\sum_{k=1}^{n}f(\\xi_k)\\cdot\\Delta x.
\n\\]<\/p>\n

Substitute the known values:<\/p>\n

\\[
\nS_n=
\n\\sum_{k=1}^{n}
\n3\\cdot\\frac{3}{n}.
\n\\]<\/p>\n

The sum contains \\( n \\) identical terms, so<\/p>\n

\\[
\nS_n=
\nn\\cdot3\\cdot\\frac{3}{n}.
\n\\]<\/p>\n

Simplifying gives<\/p>\n

\\[
\nS_n=9.
\n\\]<\/p>\n

In this case, the integral sum does not depend on the number of parts in the partition. Therefore, as \\( n\\to\\infty \\), its value remains equal to \\( 9 \\).<\/p>\n

By the definition of the definite integral,<\/p>\n

\\[
\n\\int_1^4 3\\,dx
\n=
\n\\lim_{n\\to\\infty}S_n.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_1^4 3\\,dx=9.
\n\\]<\/p>\n

This result can also be checked geometrically. The graph of the function \\( y=3 \\), the \\( x \\)-axis, and the lines \\( x=1 \\) and \\( x=4 \\) form a rectangle. Its width is<\/p>\n

\\[
\n4-1=3,
\n\\]<\/p>\n

and its height is \\( 3 \\). Therefore, its area is<\/p>\n

\\[
\n3\\cdot3=9.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int_1^4 3\\,dx=9.
\n\\]<\/p>\n

Example 2. The function \\( f(x) \\) is odd. Evaluate
\n\\[
\n\\int_{-4}^{4}f(x)\\,dx.
\n\\]<\/h3>\n

By assumption, the function \\( f(x) \\) is odd. Therefore,<\/p>\n

\\[
\nf(-x)=-f(x).
\n\\]<\/p>\n

The limits of integration are \\( -4 \\) and \\( 4 \\), so the interval \\( [-4,4] \\) is symmetric about zero.<\/p>\n

For an odd function over a symmetric interval, the following property holds:<\/p>\n

\\[
\n\\int_{-a}^{a}f(x)\\,dx=0.
\n\\]<\/p>\n

In our case, \\( a=4 \\). Therefore,<\/p>\n

\\[
\n\\int_{-4}^{4}f(x)\\,dx=0.
\n\\]<\/p>\n

On the intervals \\( [-4,0] \\) and \\( [0,4] \\), the values of an odd function at symmetric points have the same absolute values but opposite signs. Therefore, the corresponding parts of the definite integral cancel each other out.<\/p>\n

Thus,<\/p>\n

\\[
\n\\int_{-4}^{4}f(x)\\,dx=0.
\n\\]<\/p>\n

Example 3. It is known that
\n\\[
\n\\int_0^2 f(x)\\,dx=5,
\n\\qquad
\n\\int_0^2 g(x)\\,dx=3.
\n\\]
\nEvaluate
\n\\[
\n\\int_0^2\\left(2\\cdot f(x)-3\\cdot g(x)\\right)\\,dx.
\n\\]<\/h3>\n

Under the integral sign, we have the difference of two functions multiplied by constants. Therefore, we use the linearity property of the definite integral:<\/p>\n

\\[
\n\\int_a^b
\n\\left(
\n\\alpha\\cdot f(x)+\\beta\\cdot g(x)
\n\\right)\\,dx
\n=
\n\\alpha\\cdot\\int_a^b f(x)\\,dx
\n+
\n\\beta\\cdot\\int_a^b g(x)\\,dx.
\n\\]<\/p>\n

In our case, the coefficient of \\( f(x) \\) is \\( 2 \\), while the coefficient of \\( g(x) \\) is \\( -3 \\). Therefore,<\/p>\n

\\[
\n\\int_0^2
\n\\left(
\n2\\cdot f(x)-3\\cdot g(x)
\n\\right)\\,dx
\n=
\n2\\cdot\\int_0^2 f(x)\\,dx
\n–
\n3\\cdot\\int_0^2 g(x)\\,dx.
\n\\]<\/p>\n

Substitute the known values:<\/p>\n

\\[
\n\\int_0^2
\n\\left(
\n2\\cdot f(x)-3\\cdot g(x)
\n\\right)\\,dx
\n=
\n2\\cdot5-3\\cdot3.
\n\\]<\/p>\n

Now calculate:<\/p>\n

\\[
\n2\\cdot5-3\\cdot3=10-9=1.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_0^2
\n\\left(
\n2\\cdot f(x)-3\\cdot g(x)
\n\\right)\\,dx=1.
\n\\]<\/p>\n

Example 4. It is known that
\n\\[
\n\\int_1^3 f(x)\\,dx=4,
\n\\qquad
\n\\int_3^6 f(x)\\,dx=7.
\n\\]
\nEvaluate
\n\\[
\n\\int_6^1 f(x)\\,dx.
\n\\]<\/h3>\n

First, find the value of the integral over the interval from \\( 1 \\) to \\( 6 \\).<\/p>\n

The point \\( 3 \\) divides this interval into two parts, \\( [1,3] \\) and \\( [3,6] \\). By the additivity property of the definite integral,<\/p>\n

\\[
\n\\int_1^6 f(x)\\,dx
\n=
\n\\int_1^3 f(x)\\,dx+
\n\\int_3^6 f(x)\\,dx.
\n\\]<\/p>\n

Substitute the known values:<\/p>\n

\\[
\n\\int_1^6 f(x)\\,dx=4+7.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_1^6 f(x)\\,dx=11.
\n\\]<\/p>\n

However, the problem asks us to evaluate the integral with the limits written in reverse order:<\/p>\n

\\[
\n\\int_6^1 f(x)\\,dx.
\n\\]<\/p>\n

When the limits of integration are reversed, the sign of the integral changes:<\/p>\n

\\[
\n\\int_6^1 f(x)\\,dx
\n=
\n-\\int_1^6 f(x)\\,dx.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_6^1 f(x)\\,dx=-11.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int_6^1 f(x)\\,dx=-11.
\n\\]<\/p>\n

Example 5. For every \\( x\\in[-2,2] \\), we have \\( 1\\leq f(x)\\leq3 \\). Determine the bounds for
\n\\[
\n\\int_{-2}^{2}f(x)\\,dx.
\n\\]<\/h3>\n

By assumption, throughout the entire interval \\( [-2,2] \\),<\/p>\n

\\[
\n1\\leq f(x)\\leq3.
\n\\]<\/p>\n

If one function does not exceed another throughout the entire interval, the same order is preserved for their definite integrals. Therefore,<\/p>\n

\\[
\n\\int_{-2}^{2}1\\,dx
\n\\leq
\n\\int_{-2}^{2}f(x)\\,dx
\n\\leq
\n\\int_{-2}^{2}3\\,dx.
\n\\]<\/p>\n

The length of the interval \\( [-2,2] \\) is<\/p>\n

\\[
\n2-(-2)=4.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_{-2}^{2}1\\,dx
\n=
\n1\\cdot4
\n=
\n4.
\n\\]<\/p>\n

Similarly,<\/p>\n

\\[
\n\\int_{-2}^{2}3\\,dx
\n=
\n3\\cdot4
\n=
\n12.
\n\\]<\/p>\n

Substitute these values into the inequality:<\/p>\n

\\[
\n4
\n\\leq
\n\\int_{-2}^{2}f(x)\\,dx
\n\\leq
\n12.
\n\\]<\/p>\n

Therefore, the value of the definite integral lies within the bounds<\/p>\n

\\[
\n4
\n\\leq
\n\\int_{-2}^{2}f(x)\\,dx
\n\\leq
\n12.
\n\\]<\/p>\n

Next Step in Integral Calculus: Topics to Explore Further<\/h2>\n

The definite integral opens the way to many important problems in calculus. The next step is to learn how to evaluate it and how to apply it to practical problems in geometry.<\/p>\n

    \n
  1. Newton\u2013Leibniz Formula: How to Evaluate a Definite Integral<\/a> \u2014 This article will explain how to use an antiderivative to find the value of a definite integral and how to work correctly with the limits of integration.<\/li>\n
  2. Finding Area Using Integrals: Formulas and Examples<\/a> \u2014 This material will explain how the definite integral is used to find the areas of regions bounded by graphs of functions and coordinate axes.<\/li>\n
  3. Volume of a Solid of Revolution: Applying the Definite Integral<\/a> \u2014 This article will focus on finding the volumes of solids formed by rotating plane figures around the coordinate axes.<\/li>\n<\/ol>\n

    Definite Integral in Your Own Program: Practice for Programmers<\/h2>\n

    If you enjoy programming, try turning the definition of the definite integral into a working algorithm. The flowchart below shows how to use an integral sum to approximate the definite integral of the function \\( f(x)=x^2 \\) over the interval \\( [0,2] \\):<\/p>\n

    \\[
    \n\\int_0^2 x^2\\,dx.
    \n\\]<\/p>\n

    Follow the sequence of steps in the flowchart, and then implement the algorithm in your favorite programming language. Try running the program with different numbers of subintervals and compare how the resulting approximate value of the integral changes.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

    The definite integral is one of the fundamental concepts of calculus. It assigns a specific numerical value to a function<\/p>\n","protected":false},"author":1,"featured_media":4514,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[562],"tags":[539,197,589,587,588],"class_list":["post-4486","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-integral-calculus","tag-calculus","tag-definite-integral","tag-geometric-meaning","tag-integral-sum","tag-properties-of-integrals"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4486","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=4486"}],"version-history":[{"count":21,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4486\/revisions"}],"predecessor-version":[{"id":4513,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4486\/revisions\/4513"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/4514"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=4486"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=4486"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=4486"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}