Antiderivative
Definition antiderivative function
- Function y=F(x) is called the antiderivative of the function y=f(x) at a given interval X, if for everyone X ∈X equality holds: F′(x) = f(x)
Can be read in two ways:
- f derivative of a function F
- F antiderivative of a function f
Property of antiderivatives
- If F(x)- antiderivative of a function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written in the form F(x) + C, where C is an arbitrary constant.
Geometric interpretation
- Graphs of all antiderivatives of a given function f(x) are obtained from the graph of any one antiderivative parallel transfers along the O axis at.
Rules for calculating antiderivatives
- The antiderivative of the sum is equal to the sum of the antiderivatives. If F(x)- antiderivative for f(x), and G(x) is an antiderivative for g(x), That F(x) + G(x)- antiderivative for f(x) + g(x).
- Constant multiplier can be taken out of the sign of the derivative. If F(x)- antiderivative for f(x), And k- constant, then k·F(x)- antiderivative for k f(x).
- If F(x)- antiderivative for f(x), And k, b- constant, and k ≠ 0, That 1/k F(kx + b)- antiderivative for f(kx + b).
Remember!
Any function F(x) = x 2 + C , where C is an arbitrary constant, and only such a function is an antiderivative for the function f(x) = 2x.
- For example:
F"(x) = (x 2 + 1)" = 2x = f(x);
f(x) = 2x, because F"(x) = (x 2 – 1)" = 2x = f(x);
f(x) = 2x, because F"(x) = (x 2 –3)" = 2x = f(x);
Relationship between the graphs of a function and its antiderivative:
- If the graph of a function f(x)>0 F(x) increases over this interval.
- If the graph of a function f(x)<0 on the interval, then the graph of its antiderivative F(x) decreases over this interval.
- If f(x)=0, then the graph of its antiderivative F(x) at this point changes from increasing to decreasing (or vice versa).
To denote an antiderivative, the sign of an indefinite integral is used, that is, an integral without indicating the limits of integration.
Indefinite integral
Definition:
- The indefinite integral of the function f(x) is the expression F(x) + C, that is, the set of all antiderivatives of a given function f(x). The indefinite integral is denoted as follows: \int f(x) dx = F(x) + C
- f(x)- called the integrand function;
- f(x)dx- called the integrand;
- x- called the variable of integration;
- F(x)- one of the antiderivatives of the function f(x);
- WITH- arbitrary constant.
Properties of the indefinite integral
- The derivative of the indefinite integral is equal to the integrand: (\int f(x) dx)\prime= f(x) .
- The constant factor of the integrand can be taken out of the integral sign: \int k \cdot f(x) dx = k \cdot \int f(x) dx.
- The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions: \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx.
- If k, b are constants, and k ≠ 0, then \int f(kx + b) dx = \frac(1)(k) \cdot F(kx + b) + C.
Table of antiderivatives and indefinite integrals
| Function f(x) | Antiderivative F(x) + C | Indefinite integrals \int f(x) dx = F(x) + C |
| 0 | C | \int 0 dx = C |
| f(x) = k | F(x) = kx + C | \int kdx = kx + C |
| f(x) = x^m, m\not =-1 | F(x) = \frac(x^(m+1))(m+1) + C | \int x(^m)dx = \frac(x^(m+1))(m+1) + C |
| f(x) = \frac(1)(x) | F(x) = l n \lvert x \rvert + C | \int \frac(dx)(x) = l n \lvert x \rvert + C |
| f(x) = e^x | F(x) = e^x + C | \int e(^x )dx = e^x + C |
| f(x) = a^x | F(x) = \frac(a^x)(l na) + C | \int a(^x )dx = \frac(a^x)(l na) + C |
| f(x) = \sin x | F(x) = -\cos x + C | \int \sin x dx = -\cos x + C |
| f(x) = \cos x | F(x) =\sin x + C | \int \cos x dx = \sin x + C |
| f(x) = \frac(1)(\sin (^2) x) | F(x) = -\ctg x + C | \int \frac (dx)(\sin (^2) x) = -\ctg x + C |
| f(x) = \frac(1)(\cos (^2) x) | F(x) = \tg x + C | \int \frac(dx)(\sin (^2) x) = \tg x + C |
| f(x) = \sqrt(x) | F(x) =\frac(2x \sqrt(x))(3) + C | |
| f(x) =\frac(1)( \sqrt(x)) | F(x) =2\sqrt(x) + C | |
| f(x) =\frac(1)( \sqrt(1-x^2)) | F(x)=\arcsin x + C | \int \frac(dx)( \sqrt(1-x^2))=\arcsin x + C |
| f(x) =\frac(1)( \sqrt(1+x^2)) | F(x)=\arctg x + C | \int \frac(dx)( \sqrt(1+x^2))=\arctg x + C |
| f(x)=\frac(1)( \sqrt(a^2-x^2)) | F(x)=\arcsin \frac (x)(a)+ C | \int \frac(dx)( \sqrt(a^2-x^2)) =\arcsin \frac (x)(a)+ C |
| f(x)=\frac(1)( \sqrt(a^2+x^2)) | F(x)=\arctg \frac (x)(a)+ C | \int \frac(dx)( \sqrt(a^2+x^2)) = \frac (1)(a) \arctg \frac (x)(a)+ C |
| f(x) =\frac(1)( 1+x^2) | F(x)=\arctg + C | \int \frac(dx)( 1+x^2)=\arctg + C |
| f(x)=\frac(1)( \sqrt(x^2-a^2)) (a \not= 0) | F(x)=\frac(1)(2a)l n \lvert \frac (x-a)(x+a) \rvert + C | \int \frac(dx)( \sqrt(x^2-a^2))=\frac(1)(2a)l n \lvert \frac (x-a)(x+a) \rvert + C |
| f(x)=\tg x | F(x)= - l n \lvert \cos x \rvert + C | \int \tg x dx =- l n \lvert \cos x \rvert + C |
| f(x)=\ctg x | F(x)= l n \lvert \sin x \rvert + C | \int \ctg x dx = l n \lvert \sin x \rvert + C |
| f(x)=\frac(1)(\sin x) | F(x)= l n \lvert \tg \frac(x)(2) \rvert + C | \int \frac (dx)(\sin x) = l n \lvert \tg \frac(x)(2) \rvert + C |
| f(x)=\frac(1)(\cos x) | F(x)= l n \lvert \tg (\frac(x)(2) +\frac(\pi)(4)) \rvert + C | \int \frac (dx)(\cos x) = l n \lvert \tg (\frac(x)(2) +\frac(\pi)(4)) \rvert + C |
Newton–Leibniz formula
Let f(x) this function F its arbitrary antiderivative.
\int_(a)^(b) f(x) dx =F(x)|_(a)^(b)= F(b) - F(a)
Where F(x)- antiderivative for f(x)
That is, the integral of the function f(x) on an interval is equal to the difference of antiderivatives at points b And a.
Area of a curved trapezoid
Curvilinear trapezoid is a figure bounded by the graph of a function that is non-negative and continuous on an interval f, Ox axis and straight lines x = a And x = b.
The area of a curved trapezoid is found using the Newton-Leibniz formula:
S= \int_(a)^(b) f(x) dx
Previously, given a given function, guided by various formulas and rules, we found its derivative. The derivative has numerous uses: it is the speed of movement (or, more generally, the speed of any process); the angular coefficient of the tangent to the graph of the function; using the derivative, you can examine a function for monotonicity and extrema; it helps solve optimization problems.
But along with the problem of finding the speed according to a known law of motion, there is also an inverse problem - the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.
Example 1. A material point moves in a straight line, its speed at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). This means that to solve the problem you need to select a function s = s(t), the derivative of which is equal to gt. It is not difficult to guess that \(s(t) = \frac(gt^ 2)(2) \).In fact
\(s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t = gt\)
Answer: \(s(t) = \frac(gt^2)(2) \)
Let us immediately note that the example is solved correctly, but incompletely. We got \(s(t) = \frac(gt^2)(2) \). In fact, the problem has infinitely many solutions: any function of the form \(s(t) = \frac(gt^2)(2) + C\), where C is an arbitrary constant, can serve as a law of motion, since \(\left (\frac(gt^2)(2) +C \right)" = gt \)
To make the problem more specific, we had to fix the initial situation: indicate the coordinate of a moving point at some point in time, for example at t = 0. If, say, s(0) = s 0, then from the equality s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0. Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0.
In mathematics, mutually inverse operations are given different names, special notations are invented, for example: squaring (x 2) and square root (\(\sqrt(x) \)), sine (sin x) and arcsine (arcsin x) and etc. The process of finding the derivative of a given function is called differentiation, and the inverse operation, i.e. the process of finding a function from a given derivative, is integration.
The term “derivative” itself can be justified “in everyday terms”: the function y = f(x) “gives birth” to a new function y" = f"(x). The function y = f(x) acts as a “parent”, but mathematicians, naturally, do not call it a “parent” or “producer”; they say that it is, in relation to the function y" = f"(x) , primary image, or primitive.
Definition. The function y = F(x) is called antiderivative for the function y = f(x) on the interval X if the equality F"(x) = f(x) holds for \(x \in X\)
In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).
Let's give examples.
1) The function y = x 2 is antiderivative for the function y = 2x, since for any x the equality (x 2)" = 2x is true
2) The function y = x 3 is antiderivative for the function y = 3x 2, since for any x the equality (x 3)" = 3x 2 is true
3) The function y = sin(x) is antiderivative for the function y = cos(x), since for any x the equality (sin(x))" = cos(x) is true
When finding antiderivatives, as well as derivatives, not only formulas are used, but also some rules. They are directly related to the corresponding rules for calculating derivatives.
We know that the derivative of a sum is equal to the sum of its derivatives. This rule generates the corresponding rule for finding antiderivatives.
Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives.
We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.
Rule 2. If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).
Theorem 1. If y = F(x) is an antiderivative for the function y = f(x), then the antiderivative for the function y = f(kx + m) is the function \(y=\frac(1)(k)F(kx+m) \)
Theorem 2. If y = F(x) is an antiderivative for the function y = f(x) on the interval X, then the function y = f(x) has infinitely many antiderivatives, and they all have the form y = F(x) + C.
Integration methods
Variable replacement method (substitution method)
The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. There are no general methods for selecting substitutions. The ability to correctly determine substitution is acquired through practice.
Let it be necessary to calculate the integral \(\textstyle \int F(x)dx \). Let's make the substitution \(x= \varphi(t) \) where \(\varphi(t) \) is a function that has a continuous derivative.
Then \(dx = \varphi " (t) \cdot dt \) and based on the invariance property of the integration formula for the indefinite integral, we obtain the integration formula by substitution:
\(\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt \)
Integration of expressions of the form \(\textstyle \int \sin^n x \cos^m x dx \)
If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.
Integration by parts
Integration by parts - applying the following formula for integration:
\(\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du \)
or:
\(\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx \)
Table of indefinite integrals (antiderivatives) of some functions
$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$This lesson is the first in a series of videos on integration. In it we will analyze what an antiderivative of a function is, and also study the elementary methods of calculating these very antiderivatives.
In fact, there is nothing complicated here: essentially it all comes down to the concept of derivative, which you should already be familiar with. :)
I will immediately note that since this is the very first lesson in our new topic, today there will be no complex calculations and formulas, but what we will learn today will form the basis for much more complex calculations and constructions when calculating complex integrals and areas.
In addition, when starting to study integration and integrals in particular, we implicitly assume that the student is already at least familiar with the concepts of derivatives and has at least basic skills in calculating them. Without a clear understanding of this, there is absolutely nothing to do in integration.
However, here lies one of the most common and insidious problems. The fact is that, when starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, stupid and offensive mistakes are made during exams and independent work.
Therefore, now I will not give a clear definition of an antiderivative. In return, I suggest you see how it is calculated using a simple specific example.
What is an antiderivative and how is it calculated?
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is calculated simply:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look carefully at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can write it this way, according to the definition of a derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of an antiderivative. But to write it correctly, you need to write the following:
Let us write the following expression in the same way:
If we generalize this rule, we can derive the following formula:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now we can formulate a clear definition.
An antiderivative of a function is a function whose derivative is equal to the original function.
Questions about the antiderivative function
It would seem a fairly simple and understandable definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, okay, this formula is correct. However, in this case, with $n=1$, we have problems: “zero” appears in the denominator, and we cannot divide by “zero”.
- The formula is limited to degrees only. How to calculate the antiderivative, for example, of sine, cosine and any other trigonometry, as well as constants.
- Existential question: is it always possible to find an antiderivative? If yes, then what about the antiderivative of the sum, difference, product, etc.?
I will answer the last question right away. Unfortunately, the antiderivative, unlike the derivative, is not always considered. There is no universal formula by which from any initial construction we will obtain a function that will be equal to this similar construction. As for powers and constants, we’ll talk about that now.
Solving problems with power functions
\[((x)^(-1))\to \frac(((x)^(-1+1)))(-1+1)=\frac(1)(0)\]
As you can see, this formula for $((x)^(-1))$ does not work. The question arises: what works then? Can't we count $((x)^(-1))$? Of course we can. Let's just remember this first:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has studied this topic at least a little will remember that this expression is equal to the derivative of the natural logarithm:
\[((\left(\ln x \right))^(\prime ))=\frac(1)(x)\]
Therefore, we can confidently write the following:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
You need to know this formula, just like the derivative of a power function.
So what we know so far:
- For a power function - $((x)^(n))\to \frac(((x)^(n+1)))(n+1)$
- For a constant - $=const\to \cdot x$
- A special case of a power function is $\frac(1)(x)\to \ln x$
And if we start multiplying and dividing the simplest functions, how then can we calculate the antiderivative of a product or quotient. Unfortunately, analogies with the derivative of a product or quotient do not work here. There is no standard formula. For some cases, there are tricky special formulas - we will get acquainted with them in future video lessons.
However, remember: there is no general formula similar to the formula for calculating the derivative of a quotient and a product.
Solving real problems
Task No. 1
Let's calculate each of the power functions separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Problem No. 2
As I already said, prototypes of works and particulars “to the point” are not considered. However, here you can do in the following way:
We broke down the fraction into the sum of two fractions.
Let's do the math:
The good news is that knowing the formulas for calculating antiderivatives, you can already calculate more complex structures. However, let's go further and expand our knowledge a little more. The fact is that many constructions and expressions, which, at first glance, have nothing to do with $((x)^(n))$, can be represented as a power with rational indicator, namely:
\[\sqrt(x)=((x)^(\frac(1)(2)))\]
\[\sqrt[n](x)=((x)^(\frac(1)(n)))\]
\[\frac(1)(((x)^(n)))=((x)^(-n))\]
All these techniques can and should be combined. Power expressions can be
- multiply (degrees add);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving power expressions with rational exponent
Example #1
Let's calculate each root separately:
\[\sqrt(x)=((x)^(\frac(1)(2)))\to \frac(((x)^(\frac(1)(2)+1)))(\ frac(1)(2)+1)=\frac(((x)^(\frac(3)(2))))(\frac(3)(2))=\frac(2\cdot (( x)^(\frac(3)(2))))(3)\]
\[\sqrt(x)=((x)^(\frac(1)(4)))\to \frac(((x)^(\frac(1)(4))))(\frac( 1)(4)+1)=\frac(((x)^(\frac(5)(4))))(\frac(5)(4))=\frac(4\cdot ((x) ^(\frac(5)(4))))(5)\]
In total, our entire construction can be written as follows:
Example No. 2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore we get:
\[\frac(1)(((x)^(3)))=((x)^(-3))\to \frac(((x)^(-3+1)))(-3 +1)=\frac(((x)^(-2)))(-2)=-\frac(1)(2((x)^(2)))\]
In total, collecting everything into one expression, we can write:
Example No. 3
To begin with, we note that we have already calculated $\sqrt(x)$:
\[\sqrt(x)\to \frac(4((x)^(\frac(5)(4))))(5)\]
\[((x)^(\frac(3)(2)))\to \frac(((x)^(\frac(3)(2)+1)))(\frac(3)(2 )+1)=\frac(2\cdot ((x)^(\frac(5)(2))))(5)\]
Let's rewrite:
I hope I will not surprise anyone if I say that what we have just studied is only the simplest calculations of antiderivatives, the most elementary constructions. Let's now look at slightly more complex examples, in which, in addition to the tabular antiderivatives, you will also need to remember the school curriculum, namely, abbreviated multiplication formulas.
Solving more complex examples
Task No. 1
Let us recall the formula for the squared difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the prototype of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
Let's put everything together into a common structure:
Problem No. 2
In this case, we need to expand the difference cube. Let's remember:
\[((\left(a-b \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Taking this fact into account, we can write it like this:
Let's transform our function a little:
We count as always - for each term separately:
\[((x)^(-3))\to \frac(((x)^(-2)))(-2)\]
\[((x)^(-2))\to \frac(((x)^(-1)))(-1)\]
\[((x)^(-1))\to \ln x\]
Let us write the resulting construction:
Problem No. 3
At the top we have the square of the sum, let's expand it:
\[\frac(((\left(x+\sqrt(x) \right))^(2)))(x)=\frac(((x)^(2))+2x\cdot \sqrt(x )+((\left(\sqrt(x) \right))^(2)))(x)=\]
\[=\frac(((x)^(2)))(x)+\frac(2x\sqrt(x))(x)+\frac(x)(x)=x+2((x) ^(\frac(1)(2)))+1\]
\[((x)^(\frac(1)(2)))\to \frac(2\cdot ((x)^(\frac(3)(2))))(3)\]
Let's write the final solution:
Now attention! A very important thing, which is associated with the lion's share of errors and misunderstandings. The fact is that until now, counting antiderivatives using derivatives and bringing transformations, we did not think about what the derivative of a constant is equal to. But the derivative of a constant is equal to “zero”. This means that you can write the following options:
- $((x)^(2))\to \frac(((x)^(3)))(3)$
- $((x)^(2))\to \frac(((x)^(3)))(3)+1$
- $((x)^(2))\to \frac(((x)^(3)))(3)+C$
This is very important to understand: if the derivative of a function is always the same, then the same function has an infinite number of antiderivatives. We can simply add any constant numbers to our antiderivatives and get new ones.
It is no coincidence that in the explanation of the problems that we just solved, it was written “Write down the general form of antiderivatives.” Those. It is already assumed in advance that there is not one of them, but a whole multitude. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks we will correct what we did not complete.
Once again we rewrite our constructions:
In such cases, you should add that $C$ is a constant - $C=const$.
In our second function we get the following construction:
And the last one:
And now we really got what was required of us in the original condition of the problem.
Solving problems of finding antiderivatives with a given point
Now that we know about constants and the peculiarities of writing antiderivatives, it is quite logical that the next type of problem arises when, from the set of all antiderivatives, it is required to find the one and only one that would pass through a given point. What is this task?
The fact is that all antiderivatives of a given function differ only in that they are shifted vertically by a certain number. And this means that no matter what point on the coordinate plane we take, one antiderivative will definitely pass, and, moreover, only one.
So, the problems that we will now solve are formulated as follows: not just find the antiderivative, knowing the formula of the original function, but choose exactly the one that passes through the given point, the coordinates of which will be given in the problem statement.
Example #1
First, let’s simply count each term:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((x)^(3))\to \frac(((x)^(4)))(4)\]
Now we substitute these expressions into our construction:
This function must pass through the point $M\left(-1;4 \right)$. What does it mean that it passes through a point? This means that if instead of $x$ we put $-1$ everywhere, and instead of $F\left(x \right)$ - $-4$, then we should get the correct numerical equality. Let's do this:
We see that we have an equation for $C$, so let's try to solve it:
Let's write down the very solution we were looking for:
Example No. 2
First of all, it is necessary to reveal the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original construction will be written as follows:
Now let's find $C$: substitute the coordinates of point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As a final touch to what we have just discussed, I propose to consider two more complex problems that involve trigonometry. In them, in the same way, you will need to find antiderivatives for all functions, then select from this set the only one that passes through the point $M$ on the coordinate plane.
Looking ahead, I would like to note that the technique that we will now use to find antiderivatives of trigonometric functions is, in fact, a universal technique for self-test.
Task No. 1
Let's remember the following formula:
\[((\left(\text(tg)x \right))^(\prime ))=\frac(1)(((\cos )^(2))x)\]
Based on this, we can write:
Let's substitute the coordinates of point $M$ into our expression:
\[-1=\text(tg)\frac(\text( )\!\!\pi\!\!\text( ))(\text(4))+C\]
Let's rewrite the expression taking this fact into account:
Problem No. 2
This will be a little more difficult. Now you'll see why.
Let's remember this formula:
\[((\left(\text(ctg)x \right))^(\prime ))=-\frac(1)(((\sin )^(2))x)\]
To get rid of the “minus”, you need to do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Let's substitute the coordinates of point $M$:
In total, we write down the final construction:
That's all I wanted to tell you about today. We studied the very term antiderivatives, how to calculate them from elementary functions, and also how to find an antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you understand this complex topic at least a little. In any case, it is on antiderivatives that indefinite and indefinite integrals are constructed, so it is absolutely necessary to calculate them. That's all for me. See you again!
Lesson and presentation on the topic: "An antiderivative function. Graph of a function"
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Antiderivative function. Introduction
Guys, you know how to find derivatives of functions using various formulas and rules. Today we will study the inverse operation of calculating the derivative. The concept of derivative is often used in real life. Let me remind you: the derivative is the rate of change of a function at a specific point. Processes involving motion and speed are well described in these terms.Let's look at this problem: “The speed of an object moving in a straight line is described by the formula $V=gt$. It is required to restore the law of motion.
Solution.
We know the formula well: $S"=v(t)$, where S is the law of motion.
Our task comes down to finding a function $S=S(t)$ whose derivative is equal to $gt$. Looking carefully, you can guess that $S(t)=\frac(g*t^2)(2)$.
Let's check the correctness of the solution to this problem: $S"(t)=(\frac(g*t^2)(2))"=\frac(g)(2)*2t=g*t$.
Knowing the derivative of the function, we found the function itself, that is, we performed the inverse operation.
But it’s worth paying attention to this point. The solution to our problem requires clarification; if we add any number (constant) to the found function, then the value of the derivative will not change: $S(t)=\frac(g*t^2)(2)+c,c=const$.
$S"(t)=(\frac(g*t^2)(2))"+c"=g*t+0=g*t$.
Guys, pay attention: our problem has an infinite number of solutions!
If the problem does not specify an initial or some other condition, do not forget to add a constant to the solution. For example, our task may specify the position of our body at the very beginning of the movement. Then it is not difficult to calculate the constant; by substituting zero into the resulting equation, we obtain the value of the constant.
What is this operation called?
The inverse operation of differentiation is called integration.
Finding a function from a given derivative – integration.
The function itself will be called an antiderivative, that is, the image from which the derivative of the function was obtained.
It is customary to write the antiderivative with a capital letter $y=F"(x)=f(x)$.
Definition. The function $y=F(x)$ is called the antiderivative of the function $у=f(x)$ on the interval X if for any $хϵХ$ the equality $F’(x)=f(x)$ holds.
Let's make a table of antiderivatives for various functions. It should be printed out as a reminder and memorized.
In our table, no initial conditions were specified. This means that a constant should be added to each expression on the right side of the table. We will clarify this rule later.
Rules for finding antiderivatives
Let's write down a few rules that will help us in finding antiderivatives. They are all similar to the rules of differentiation.Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives. $F(x+y)=F(x)+F(y)$.
Example.
Find the antiderivative for the function $y=4x^3+cos(x)$.
Solution.
The antiderivative of the sum is equal to the sum of the antiderivatives, then we need to find the antiderivative for each of the presented functions.
$f(x)=4x^3$ => $F(x)=x^4$.
$f(x)=cos(x)$ => $F(x)=sin(x)$.
Then the antiderivative of the original function will be: $y=x^4+sin(x)$ or any function of the form $y=x^4+sin(x)+C$.
Rule 2. If $F(x)$ is an antiderivative for $f(x)$, then $k*F(x)$ is an antiderivative for the function $k*f(x)$.(We can easily take the coefficient as a function).
Example.
Find antiderivatives of functions:
a) $y=8sin(x)$.
b) $y=-\frac(2)(3)cos(x)$.
c) $y=(3x)^2+4x+5$.
Solution.
a) The antiderivative of $sin(x)$ is minus $cos(x)$. Then the antiderivative of the original function will take the form: $y=-8cos(x)$.
B) The antiderivative of $cos(x)$ is $sin(x)$. Then the antiderivative of the original function will take the form: $y=-\frac(2)(3)sin(x)$.
C) The antiderivative for $x^2$ is $\frac(x^3)(3)$. The antiderivative of x is $\frac(x^2)(2)$. The antiderivative of 1 is x. Then the antiderivative of the original function will take the form: $y=3*\frac(x^3)(3)+4*\frac(x^2)(2)+5*x=x^3+2x^2+5x$ .
Rule 3. If $у=F(x)$ is an antiderivative for the function $y=f(x)$, then the antiderivative for the function $y=f(kx+m)$ is the function $y=\frac(1)(k)* F(kx+m)$.
Example.
Find antiderivatives of the following functions:
a) $y=cos(7x)$.
b) $y=sin(\frac(x)(2))$.
c) $y=(-2x+3)^3$.
d) $y=e^(\frac(2x+1)(5))$.
Solution.
a) The antiderivative of $cos(x)$ is $sin(x)$. Then the antiderivative for the function $y=cos(7x)$ will be the function $y=\frac(1)(7)*sin(7x)=\frac(sin(7x))(7)$.
B) The antiderivative of $sin(x)$ is minus $cos(x)$. Then the antiderivative for the function $y=sin(\frac(x)(2))$ will be the function $y=-\frac(1)(\frac(1)(2))cos(\frac(x)(2) )=-2cos(\frac(x)(2))$.
C) The antiderivative for $x^3$ is $\frac(x^4)(4)$, then the antiderivative of the original function $y=-\frac(1)(2)*\frac(((-2x+3) )^4)(4)=-\frac(((-2x+3))^4)(8)$.
D) Slightly simplify the expression to the power $\frac(2x+1)(5)=\frac(2)(5)x+\frac(1)(5)$.
The antiderivative of the exponential function is itself exponential function. The antiderivative of the original function will be $y=\frac(1)(\frac(2)(5))e^(\frac(2)(5)x+\frac(1)(5))=\frac(5)( 2)*e^(\frac(2x+1)(5))$.
Theorem. If $y=F(x)$ is an antiderivative for the function $y=f(x)$ on the interval X, then the function $y=f(x)$ has infinitely many antiderivatives, and all of them have the form $y=F( x)+С$.
If in all the examples considered above it was necessary to find the set of all antiderivatives, then the constant C should be added everywhere.
For the function $y=cos(7x)$ all antiderivatives have the form: $y=\frac(sin(7x))(7)+C$.
For the function $y=(-2x+3)^3$ all antiderivatives have the form: $y=-\frac(((-2x+3))^4)(8)+C$.
Example.
By given law changes in the speed of a body over time $v=-3sin(4t)$ find the law of motion $S=S(t)$, if in starting moment time the body had a coordinate equal to 1.75.
Solution.
Since $v=S’(t)$, we need to find the antiderivative for a given speed.
$S=-3*\frac(1)(4)(-cos(4t))+C=\frac(3)(4)cos(4t)+C$.
In this problem it is given additional condition- initial moment of time. This means that $t=0$.
$S(0)=\frac(3)(4)cos(4*0)+C=\frac(7)(4)$.
$\frac(3)(4)cos(0)+C=\frac(7)(4)$.
$\frac(3)(4)*1+C=\frac(7)(4)$.
$C=1$.
Then the law of motion is described by the formula: $S=\frac(3)(4)cos(4t)+1$.
Problems to solve independently
1. Find antiderivatives of functions:a) $y=-10sin(x)$.
b) $y=\frac(5)(6)cos(x)$.
c) $y=(4x)^5+(3x)^2+5x$.
2. Find antiderivatives of the following functions:
a) $y=cos(\frac(3)(4)x)$.
b) $y=sin(8x)$.
c) $y=((7x+4))^4$.
d) $y=e^(\frac(3x+1)(6))$.
3. According to the given law of change in the speed of a body over time $v=4cos(6t)$, find the law of motion $S=S(t)$ if at the initial moment of time the body had a coordinate equal to 2.