Curso de Cálculo II para Ingeniería Comercial - Video 04

Welcome to this Calculus 2 course for the commercial engineering degree. I am the engineer Marcelo Vildoso and this is our fourth video. Next, we will go detailing some methods of integration. For this, it is important that we remember that all the rules that we have been learning for what are the immediate integrals of simple functions are fully applicable.

Let's see the case of the substitution method. But to be able to see this, let's first set an example to see its utility and its application. First,

Let's go with the example number 29. If we have the integral of the function x-2², we, to apply an immediate integral, we would have to have a simple function. And that expression as such is not a simple function, nor the sum of simple functions.

As we can see, there are two expressions that are inside a parenthesis and this one is squared. The alternative that we had initially, is to be able to develop that binomial squared, by the rule that tells us that squared is equal to squared minus 2ab plus b squared. Applying for this particular example, we will say that this is equal to the integral of x squared

minus the double product of x times -2 is -4x plus the square of 2, that is, 2 squared is equal to 4, with which we would have the sum of simple functions, which we could integrate through the rules that we already know. The integral of the sum of functions is the sum of its integrals, it is worth saying that

that each of these elements can be separated into an independent integral and we can solve through the rules already applied previously but what happens as in the example number 30 notice we have the integral of x minus 3 raised to the eighth

to solve that binomial to the eighth would have to do the application of Newton's binomial which for an eighth degree function ends up developing in this way as we can see evidently it is a function too long and complex if we do not know the memory Newton's binomial we could hardly go elaborating each of these terms

are obtained when developing this binomial to the other. Therefore, doing this would mean applying for this particular example as follows: the first term to the eighth minus 8 times the first term to the seventh times the second term plus 28 times the first term, that is, x raised to the sixth

for the second term which is 3 squared, minus 56 for the first term to the fifth, that is, x to the fifth, for the second term to the cube, that is, 3 cubed, plus 70 for the first term to the fourth, for the second term to the fourth, and so on we are developing this expression.

So in these cases in particular, developing could be a very high risk, because having too many arithmetic operations, we could possibly incur error, or perhaps by wrongly substituting some of these values or developing some power, we could find some defective operation.

For this is that the method of substitution has been created. What is the substitution method? Basically, seek to solve an integral that is not direct, by means of a variable change and differential change.

The result of this must be a simpler integral than the one that was raised at the beginning and then it must be integrated by means of direct integrals, it is worth saying that the new expression that has become a simple integral can be solved by means of the rules that we have already known.

Finally, what should be done is restore the initial variable, because it is worth saying, when we make a variable and differential change, we are proposing an exercise that in terms of variables is different. Therefore, it is like solving a question by changing the data of the exercise, therefore it will also change the data of the answer.

result, therefore it is important restore the initial variable to give result to the same. We will exemplify this situation with our example number 30, the example number 30 is raised as we had seen in the previous case is the integral of x minus 3 raised to the eighth and for this case we are not going to apply what is the development through the Newton binomial,

What we are going to do is find a variable change. Why? Because as we can see, this integral described in the terms that is currently, it is not possible to apply any of the known rules. That is, it is not a power only, an exponential function without, it is not an integral of a constant only, or of any trigonometric function that we have seen.

nor the case of 1/x, for example. So, what we are going to do is find an expression to make the variable change. In this case,

Let's look at this particular term, x minus 3. If this expression were simply x, then this integral would become the integral from x to the eighth differential of x, which would be easily solved by applying the integral definition of a power, which was equal to x raised to m plus 1 over m plus 1 more constant.

where m was the exponent of that power. It is worth saying that if we, instead of writing x minus 3, we wrote u, in some way we were changing this expression, making it a simpler expression.

¿Qué nos decía el diferencial? Si recordamos, el diferencial lo que nos decía era cuál es la variable de integración. Si nosotros estuviésemos cambiando en lugar de x menos 3u, estaríamos cambiando la variable. En consecuencia, debe de cambiarse el diferencial.

how do we change the differential? From this equality we can apply the differential rule that tells us that

If I have a function y and it is equal to a function f , the differential of that variable y, will be equal to the derivative of that function x, multiplied by the differential of the derivative variable. What would be the variable that we would derive in this function? Being an f , it will be the variable x, consequently, it must be multiplied by the differential of x. For this particular case then, what will we do?

Derivate x minus 3 and multiply by the differential of the variable. What variable would be being derived in this function? It would be the variable x, consequently I must multiply that result by differential of x. Let's see.

If the derivative of x minus 3 is equal to x derivative is 1, the derivative of minus 3 to be a constant is 0, therefore 1 minus 0 is equal to 1. 1 that would remain here multiplied by the differential of x, I simply have differential of x.

Therefore, I will have to work with the following substitution. Instead of u, I will write x minus 3, and instead of the differential of u, I should write the differential of x. Seen in another way, where I see x minus 3, I will write u, and where I see the differential of x, I will write the differential of u.

Let's see what is going to happen with our exercise when we make the substitution. Substituting in this expression the substitution that are these two elements, it is worth saying u equal to x minus 3 and differential of u equal to differential of x, we will have the following, I will copy the integral, continue doing the integral, but instead of x minus 3 that I have to place, I must place u.

and instead of differential of x I will have to put differential of u. Then the resulting expression will be the integral of u to the eighth differential of u. This expression as it is described, being u the variable and having differential of u, because I can apply directly

the rule of power integration, which tells us that the integral of x raised to m, differential of x is equal to x raised to m plus 1, over m plus 1 plus the constant of integration. It is worth saying that in this case, instead of x we will put u and instead of m we will put 8. So copying this expression, making that analogy between the variables and the number, it will be equal to the following, we say:

above we will have u raised to 8 + 1 and below we will have 8 + 1 also an integral

must come accompanied by an integration constant, but this integration constant we had said that we will reserve it for the result. As there are pending operations, we carry them out, what will this be equal to? This will be equal to "u" raised to 8 plus 1 is equal to 9 divided by 9, but for our case in particular we will say,

1/9 multiplied by u to the ninth, more constant, is totally equivalent to the result of this expression. But notice the following, here we have the variable u and my exercise was asking me to have the differential of x, it is worth saying that here the integration variable is the variable x.

Therefore, my result has to be in terms of the variable x, it cannot be in terms of the variable u. So, what should we do? We had noted it as a step in what is the substitution method, is to restore the variable. Therefore,

Restoring the variable we will have the following: the integral of x minus 3 raised to the eighth differential of x will be equal to, we are copying, only instead of writing u in this expression, we will restore the value of u. What was the value of u? It was equal to x minus 3, it is worth saying that this expression will be equal to 1/9 times u, which is x minus 3 raised to the ninth,

plus the constant of integration. With what we will say that the integral of this expression "x" minus 3 raised to the eighth, had been equal to a ninth of "x" minus 3 to the ninth, plus constant. As we can no longer reduce, we say that this expression is the result of this indefinite integral, by means of the substitution method.

Let's see the following example, example number 31. For example number 31 we have the function to integrate is 5 plus x raised to the cube, that is to say that we have to calculate the indefinite integral of this function. To find it we had seen that we are going to discard developing this as a binomial,

What we want to do is apply the substitution method. Therefore, what do we need to do as a first step? We need to find a variable change. As in the previous exercise, in exercise 30 we had seen that the expression that we, it is worth saying that this made us different to a simple function, well, it was the expression

x minus 3, but in this particular example the expression that we could change would be 5 plus x. What would happen if instead of this I simply had u? What would we have here? The integral of u to the cube, and the integral of u to the cube is easy to solve by means of the integral rule of potential functions, that is,

If "u" is equal to 5 plus "x", how is the differential of "u" calculated? We will remember then that the differential of "u" is calculated by means of the derivative of this expression multiplied by the differential of the variable that has been derived. It is worth saying, if we derive 5, it is equal to 0. The derivative of "x" will be equal to 1. 1 multiplied by the differential of "x" will give us the differential of "x".

with what we already had the two necessary expressions for my substitution. Therefore, by substituting in this example, we will have the following: integral will continue to be integral, only that instead of 5 plus x that we can place, we can place u. Then we will have the integral of u raised to the cube, and instead of the differential of x, which is what we can place, the differential of u.

A substitution is applied appropriately when in the substituted integral I no longer find the original variable. It is worth saying, the original variable in this integral was the variable x, therefore after applying the substitution method, I should not have any x in this expression.

to stay, it would be making an incorrect substitution, we would be substituting only one part. Therefore, we have fulfilled that condition, it is worth saying that in my integral substituted, I have no vestige of the variable x.

Solving this integral is simple because we have that this is the variable u, differential of u, because I can apply the definition of integral of a power, I simply say that u is equal to x and that 3 is equal to m, consequently what will be equal to its integral? It will be equal to u raised to m, which is 3 plus 1, over 3 plus 1, plus the constant that we reserve for the result.

Let's see what happens here, we have "u" raised to 3 plus 1, will be raised to "u" to the fourth over 4, but "u" to the fourth over 4 is the same as writing 1 over 4 times "u" raised to the fourth plus constant. Let's see again what happens, our exercise was proposed in terms of the variable "x",

and my result here is in terms of the variable u therefore does not correspond. It's like if they asked me, "What age are you?" and I would answer 80 kg. It is worth saying that what they have asked me is my age and what they would be waiting for is that I answer in terms of the question. The age is told in years, therefore my answer should be in years.

and not like this, a answer in kilograms, right? So, what should we do? Through the rule, we are going to restore our initial variable. What is the rule that we have applied to substitute this? The one that says that instead of "u", I must write "5 + x",

and instead of differential of "u" I must write differential of "x". So we apply for this exercise and we will say: the integral of 5 + x raised to the cube differential of "x" will be equal to this result, only restoring the variable. It is worth saying, instead of "u" I will place 5 + x.

which will result as 1/4, 1/4, multiplied by 5 plus x, raised to the fourth. Since we are not going to reduce this expression anymore, we simply adhere our constant of integration and we will say that this is my result. Example number 32. For example number 32 we have the following function: the function 2x over x squared minus 1.

When we observe initially this function, we can clearly see that it could be seen as the quotient of both functions. The numerator function would be 2x and the denominator function x squared minus 1.

therefore we had the cosine of two functions. The integral of the cosine of functions, because we have not seen it yet, we do not know if there is a rule, therefore with previous knowledge it is not possible to apply a direct integral for this particular situation. Therefore we have to think about what a variable change is.

What would be the variable that we should change? If we remember, in the previous cases we had seen that the expression that we had here, well we said it was u. As a consequence, what would we have? If u is equal to 2x over x squared minus 1, well here I would have the integral of u,

But just as I would change the variable, I must also change the differential, that is, I must find the differential of u.

But to calculate the differential of "u", I must apply what is the rule of the differential of the function's quotient, which is equal to the denominator by the differential of the numerator, minus the numerator by the denominator differential, and all this divided by the denominator,

If we apply this differential rule for this particular function, we will have the following: 1 would be 2x, b would be equal to x squared minus 1. As a consequence, we will have b, b is x squared minus 1.

by the differential of "u", the differential of "u" is the derivative of 2x by the differential of x, that is, the derivative of 2x is equal to 2, as we have derived the variable x, we multiply by the differential of x.

minus says "u", who would come to be "u"? We had said that it is 2x, we copy 2x and it must be multiplied by the differential of "b", "b" is x squared minus 1, the differential of "b" will be calculated by deriving x squared minus 1, multiplying by the differential of the integrated variable, it is worth saying in this case, by the differential of "x".

with what we will have, derivative of x squared is 2x and derivative of -1 is 0, 2x plus 0, I simply have 2x, differential of x.

all over the denominator squared, the denominator we had said that is x squared minus 1, then we will have x squared minus 1. If we perform the respective multiplications, 2 with all this expression and we take as a common factor 2x, then what we will have will be the following:

2 times x squared, I will have 2x squared, 2 times -1, I will have -2, -2x times 2x will be 4x.

and as both expressions of the denominator have differential of "x", I can extract as common denominator, or as a common factor, the differential of "x", that is to say, that this will be multiplied by the differential of "x". In the denominator we are not going to perform any operation, because basically the denominator is simplified, right?

So we will say the following, this expression can still be operated in the numerator and we say, 2x squared minus 4x squared, I will have minus 2x squared, minus 2, we copy minus 2. So the expression with which I should

and substitute this function, that instead of 2x over x squared minus 1, write u, and instead of the differential of u, I would have to write, minus 2x squared minus 2, over x squared minus 1, squared. Substituting we would have the following then, we had said, that the differential of x, we must replace it by,

This expression we can pass it to divide by the differential of u, we would have as a consequence 1 by differential of u over -2x squared - 2 divided by x squared - 1 squared.

So, what have we done this clearing? We have cleared from this equality of the differential of u equal to this expression, we have cleared the value of the differential of x, sending to divide all this expression to the differential of u, and we have this expression. This expression will be the one that we will replace instead of the differential of x.

Let's see what we have left in the function, in the function we would have, instead of all this we have said that we are going to write "u" and instead of the differential of "x" that we have to write, because all this expression, 1 over -2x squared -2, over x squared -1, squared.

Let's see this substitution. First, this substitution does not comply with what we had mentioned in example 31, that is, that by substituting or changing a variable and a differential within a function, because it should not have the variable x, that is, what I would be doing here is partially substituting, therefore,

this is not useful to be able to reduce or write or be able to solve this function, it is worth saying that what I would be doing is a partial substitution that is not useful to me to apply the method

Therefore, to be able to have "u" instead of "x", what should be done is, from this expression where we have the equality of "u" with respect to this function, clear "x", but that task will be more complicated, right?

The idea of ​​applying the substitution method is that I find simple integrals or integrals that are more reduced than the initial integral. It is worth saying that we would not be fulfilling the purpose of the method by making this change of function, therefore we will discard

the variable change u equal to 2x over x squared minus 1. It is worth saying that this variable change does not help me to solve this exercise. What do we do then? Well, we will have to think about some alternative. What alternatives do we have? This would be the first. As a second option we would have the following, for example, we could choose for the value of u

to the expression 2x, that is, to the numerator of this expression. Therefore, we had instead of u, we are going to write 2x, and instead of the differential of u, 2x. Where do we get this affirmation from? Well, we had simply said that to calculate the differential of a function, what must be done is to derive that function

and multiply it by the differential of the variable that has been derived. It is worth saying, if this function is f , the variable to be derived is x. As a consequence, it must be multiplied by the differential of x. What would happen if it were a function, an f , so to say u? Then, when deriving, it would have to multiply by the differential of u, okay?

then my substitution is given by these two expressions we will try to substitute but to substitute we will remember that for example instead of this expression we will have the integral of

Instead of writing 2x, I can write u. And in the denominator, I have nothing to replace between x²-1. What I would have to do then is find the value that I can substitute instead of x. It is worth saying that from this expression, we can extract the value of x, clearing that x is equal to u medios, or u medios of u, to the middle.

Also, let's see that when we replace, here I have differential of x, I do not have 2 differential of x, consequently, I would also need to have the differential of x clear of this expression, what becomes differential of x would be equal to differential of u over 2, right? So we already have

enough elements to substitute this function. So we will say, the substitution will remain as follows: instead of 2x we had said that we will write u, instead of x I had said that I can write u/2, u/2 squared, -1, it is still -1 times, instead of the differential of x we can write differential of u/2.

This would be my new integral. But I want us to look for a moment in this expression. Is this expression described now in terms of the variable u, has been simplified? That is, it has become simpler than the previous one? Or it seems rather that it has become more complicated, because we can observe that at first I had a differential of x, but now I have a differential of u over 2.

Instead of x here I have u/2 and instead of 2x I have u. It would seem that the expression instead of simplifying, it would have been complicated. Therefore, what can we do? It is again bad choice of the variable to exchange, that is, my change of variable u equal to 2x does not lead me to a solution.

What would be my third alternative? The third alternative we have for this exercise, would be to choose u, because it is the denominator, that is, u is x squared minus 1.

If u would be equal to x squared minus 1, then applying the rule for the calculation of differentials, the differential of u will be equal to the derivative of this multiplied by the differential of x, that is, the derivative of x squared,

Es 2x. Derivada de menos 1 es igual a 0. Y 2x más 0 es igual a 2x. En consecuencia, el diferencial de la variable u será igual a 2x diferencial de x.

x, right? Well, with this expression let's try to change our variables, let's see what is going to happen as we go to replace. First, the integral, because we will simply copy the integral, instead of 2x, well, actually they say 2x differential of x, which would be equal to the differential of u,

Maybe as this expression is described, it is not so easy to identify how we are going to replace, but what happens if we rewrite this integral as this expression? Notice, 2x over x squared minus 1 is equivalent to 1 over x squared minus 1, by 2x differential of x, because what happens if we multiply 1

for 2x I will have 2x and that divided by x squared minus 1 will not have altered the value of this integral. Therefore it is possible to rewrite this integral in terms of this new equivalent integral. If it is equivalent, the result of this will be equal to this one, right? So we say,

Instead of 2x differential of x, what can we write? Differential of 1. Instead of x squared minus 1, what can we write? Well, 1. So, what will we have as an integral replaced? We will have the integral of 1 over, instead of this, 1, and instead of 2x differential of x, differential of 1.

Let's see our substituted integral. The substituted integral, by making the change of variable of u instead of x squared minus 1, we have achieved that my integral raised, first, is described in terms of a simple function. It is worth saying that we are in the case of the integral of 1 over x.

If we remember the integral of 1/x, here it was equal to the natural logarithm of the absolute value of x, plus the constant of integration, it is worth saying that this is already a direct integral, therefore the objective of the substitution method would be being fulfilled. Through substitution we have managed to write a new simpler function, which is already possible to integrate through the rules

So we will say, here it would be equal to the integral of 1 over u, to the natural logarithm of the absolute value of u, plus the constant of integration. Would it be the result? Yes, it is the result of this integral, because in reality the variable of my initial integral, because it is the variable x and not so the variable u. What do we have to do? We had put as a final step,

the restore the variable. So restoring the variable, what do we have? We will have that the integral of 2x over x squared minus 1 differential of x is equal to natural logarithm, only instead of the value of u, that we can place x squared minus 1. That is, we are restoring the value of the variable u. We will then have the natural logarithm of x squared minus 1.

now plus the constant of integration. Can it be reduced? Well, I think it is not worth reducing this expression, if we apply logarithmic properties, for example, to this difference, then we will not be able to apply, to get the expression to become simpler.

We have no way to reduce it, so we say that this expression is the result of this integral. I hope you have been able to understand the application of the substitution method for these specific cases. I am the engineer Marcelo Vildoso Zamorano and we will be seeing each other in future videos. Thank you very much for watching and see you soon.