Transcription
Hello and welcome to another lecture of this course, called Mathematics for Economics—Part I. And we are discussing the topic called Integration presently.
We started with a lecture in the previous class, and there we introduced the idea that integration can be thought of as the reverse of differentiation. Let us see how far we can go from there. We also talked about the fact that integration can be thought of as an area under a curve. And we introduced the idea of the indefinite integral; so this is the idea. The anti-derivative of the function is small fx, as defined as capital Fx such that capital F dashed x is equal to small fx. So, this capital Fx is the indefinite integral, but we also have to notice that there is this constant term, capital C, which has to be added because if you take the derivative of capital C, it will become 0. So, this is what we write: Integration of small fx dx is equal to capital Fx plus C, where C is the constant of integration. So, in this case, capital Fx is the anti-derivative, and it is capital Fx plus C is the indefinite integral of small fx.
And then we took one example where, let us suppose x cube, we have to find the integral of that, and we found that to be one-fourth of x to the power of 4 plus C. And here is the conventional way we write it: First comes the integral sign, then the function that we are trying to integrate dx, dx because x is the variable with respect to which the integration has been done. Here fx is called the integrand, and on the right-hand side, we have capital Fx plus C. C is the constant of integration.
Now we are talking about certain rules of integration. We said that if you integrate x to the power n with respect to the dx, then what we get is x to the power n plus 1 divided by n plus 1 plus C. C is the constant of integration. However, this rule is valid if n is not equal to minus 1. The reason being if n is equal to minus 1, then the denominator becomes 0, and so this quantity on the right-hand side becomes undefined. So, the question that naturally arises is: If you have to integrate x to the power n and n is equal to minus 1, then how do you do it? This above rule cannot be used, but we have another formula, which is x to the power minus 1 as we know is 1 divided by x, and this can be written as log of modulus of x plus C. Remember why we are writing it because we know if I take the derivative of log of x, then it becomes 1 divided by x. So, if I integrate 1 divided by x, I should get back log of x, and there is this plus C, which is the constant of integration.
Now how is the modulus sign making an appearance here? The modulus sign is appearing because log natural of a negative number is undefined. So, that is why you have to take the positive; if x happens to be negative, you cannot just say log of x because x is negative. So, you take the modulus of that number, and then you have log of modulus of x plus C.
Similarly, it can be verified that integration of e to the power ax, dx is equal to e to the power ax divided by a plus c, provided, of course, small a is not equal to 0 because small a, if it is equal to 0, then again the right-hand side becomes undefined. So, we have to make sure that small a is not equal to 0. And this relationship can be verified; again, you take the derivative of this e to the power ax divided by a plus c, what do you get? d dx of e to the power ax is a multiplied by e to the power ax plus 0, and this is simply equal to e to the power ax, which is on the left-hand side as the integrand. So, this rule is all right.
So, instead of the exponential function like this, if I take a to the power x—A is a constant—then what happens? Then it should be equal to e to the power ax divided by log of a plus c, and of course, I have to make sure that a is not equal to 1. If a is equal to 1, then log of 1, as we know, is 0. Again, there is a problem on the right-hand side, so a cannot be equal to 1 and a cannot be equal to 0 or negative either because in either case, you have either minus infinity or it is undefined. So, these numbers—these conditions—have to be satisfied. And I can again check whether this formula is all right or not. What I do is I take the derivative of the right-hand side, which is a to the power x; this and this we know is—this is a constant term; it will come out. d dx of ax; what is that? It is log of a, a to the power x and plus 0. And so log a, log a will get canceled, so you are going to get simply a to the power x, and which is an integrand on the left-hand side. So, this formula is in order.
Some other properties; what is known as the constant multiple property. So, here what is there in the integrand? You have a constant, which is small a, multiplied by a function of x, which is fx. So, if that is the integrand, then I can write this simply as a multiplied by integration of fxdx, where a is a real constant. So, essentially what is happening: If in the integrand there is a constant term, real constant term, then I can take the real constant term out, and whatever is left of the integrand remains inside. Now this makes the estimation of the integral simple because now you do not have to deal with the constant.
And what happens if you have two functions? So, here on the left-hand side, that is exactly what you have; the integrand is the summation of 2 separate functions, fx and gx. Now this can be written as the summation of 2 integrals. So, integration of fx dx plus integration of gx dx. So, essentially what is being said is that the integral of a sum is the sum of the integrals.
And if we combine this property with this property, we get a third property, which is integration of a1, f1 x plus a2, f2 x plus, and so on. Let us say the last term is an fn x and the whole thing dx. This becomes equal to a1 integration of f1 x dx plus a2 integration of f2x dx plus so many terms and the last term is an integration of fn x dx. So, essentially, I am first breaking this term down into n different parts and then taking the constant terms outside the integral in each of these n terms.
Here is an application of what we have just seen. Suppose you have an expression like this: a polynomial 3x square minus 6x plus 1 is what we have to integrate, and what we do is that we apply the formula before where I write this as separate integrals. So, 3 multiplied by x square dx because 3 will come out minus 6 multiplied by integration of x dx because minus 6 is coming out plus just integration of dx. And from here I get this expression 3 multiplied by x cube divided by 3 minus 6, x square divided by 2 plus x plus the constant term C.
Now at this stage, you might be wondering that since you have three integrals, so there will be 3 constants. From each of these integrals, you get one constant. There are three operations of integration, so 3 constants, so how come there is just a single constant? What we have done is that I have assumed that the single constant a is the summation of all these 3 constants. Since these are arbitrary constants, so I can do that because c is undefined. It is just a constant. All right, in the next step, I am simplifying this; 3 and 3 will get canceled, so only x cube minus 3 x square plus x plus the constant term of integration. So, the constant term integration C cannot be deciphered unless some additional information is given. This is termed as the initial value problem. Right now, C is just an arbitrary constant. We cannot say anything about the value of C, so to understand what will be the value of C, we need certain conditions, some additional information, and this is termed as the initial value problem.
Here is an application of that. Find all the functions fx—capital Fx—such that f prime of x is equal to minus of x minus 1 whole square. Find the particular function, which passes through this point 1, 1. So, there are two parts to this question. We have to find fx such that f prime of x is a particular function of x. And secondly, there will be many such functions which will satisfy this condition. We have to find that particular function which passes through 1 and 1. So, first we have to basically integrate this minus of x minus 1 whole square because this is the anti-derivative. Now so here we are trying to find the indefinite integral of minus x minus 1 whole square dx because x is the variable of integration. One can take the minus 1 outside so it becomes a minus of integration of x minus 1 whole square dx. Now this is a minus b whole square formula. So, this will be x square minus 2x plus 1, and then I can break it up; it will become minus integral x square dx plus 2 integral x dx minus just integral of dx. And if you simplify this, so this term if you integrate this, it will become x square because x square divided by 2, 2 and 2 will get canceled. So, I am writing this as the first term, and then I am writing the integration of this as the second term. This will be x cube divided by 3, and integration of 1 is just x plus the constant of integration. So, this is the indefinite integral that we have found. So, y is equal to x square minus 1 divided by 3 x cube minus x plus c represents the required family of functions with C taking all possible values. So, instead of y, I could have written it as fx as well because that is the form that is written in the question.
Now we have to find that particular function, which passes through the point 1, 1. It is written there, find the particular function which passes through 1, 1. So, fx is fine, fx is this, but here capital C is undefined. C is just a constant. We do not know the value of C. So, I have to give a particular value to C only then the function becomes a definite function. Otherwise, the function might have different positions. So, in terms of a diagram, you can imagine this: so suppose this is the x and this is fx that is y, and suppose the function I do not know looks like this, it is just an arbitrary figure. Now since c is undefined, this function could as well have this position depending on what the value of C is because as you can see if I put x is equal to 0 here then the value of the function becomes C. So, this, this is C. But C is unknown so depending on the value of c you have different positions of the function. But our problem is that we want to find that particular function, that specific function which passes through a given point which is 1, 1. So, suppose 1 is here and 1 is here. So, this is the 1, 1 point. So, we want to find this function which is represented by this continuous line, not the dotted lines. So, what will be the equation of this particular line? So, how do I do that? This particular function that we are after passes through this point 1, 1. So, therefore, on the left-hand side, you are going to get 1, if you put x is equal to 1, so this is the equation you will get, and therefore I get c is equal to 4 divided by 3 after simplifying, and I put this 4 divided by 3 here and I get this particular function. Y is equal to x square minus 1 divided by 3 x cube minus x plus 4 divided by 3. So, this is the required form.
Here is an example from economics. The marginal cost of producing x units of a commodity is 3 plus x plus 2x square. The fixed cost is 100. What is the total cost function? So, this is what we want to find, the total cost function. Now let us assume that C, which is a function of x, is the total cost function. What we know is that if I take the derivative of Cx, then I get C prime x. This C prime x is equal to 3 plus x plus 2x square. This is the first information we know. And why am I writing this? Because marginal cost. If you have noticed my previous lectures when we introduced the idea of derivatives, then we met this point that the marginal cost is nothing but the derivative of the total cost function, and marginal cost is given here in the problem, so C prime x is equal to 3 plus x plus 2.x square. Remember the marginal cost is a function of x. And further, we also know that fixed cost is equal to 100. What does it mean? That if the producer does not produce anything, then also he has to bear certain costs; the value of that cost is 100. In terms of our mathematical language, the cost function at 0, that is x is equal to 0, that means C is equal to C of 0 is equal to 100. So, these two information we know, and from these two information, I have to find Cx. So, that is the problem. Now C prime x is given. Let us try to find the indefinite integral of this, and that will give me the total cost function, but there will be a constant term, and then we shall deal with the constant term. So, the total cost will be equal to integration of 3 plus x plus 2x square, so this becomes integration of 3dx plus integration of x dx plus integration of 2x square dx. And this turns out to be 3x plus x square divided by 2 plus 2x cube divided by 3 plus C, the constant of integration. So, this is the cost function. But here obviously there is this plus C terms, and I have to get rid of that, and here I am using this information that is C of 0 is known to be 100. So, if I put—there are 2 equal signs here; one is redundant—so if I put x is equal to 0, then the right-hand side becomes C, and the left-hand side is known to be 100. So, therefore, C is equal to 100. And this is substituted back in this function, so the total cost function is found to be 3x plus x square divided by 2 plus 2x cube divided by 3 plus 100. Done.
Now we come to the idea of definite integrals, and let us see what the definition is saying. Let small f be a continuous function defined over the closed interval a to b. Let capital F be a function also continuous over the closed interval a to b such that capital F prime x is equal to small fx for all x belonging to the open interval a to b. Then the difference capital Fb minus capital Fa is the definite integral of small f over the closed interval a to b. This is the definition of a definite integral. Importantly, this difference does not depend on which of the infinitely many indefinite integrals of small f we choose as capital F. The point that is being made is this: since we know capital F prime x is equal to small fx, one can take the indefinite integral of small fx, and that will give me capital fx plus capital C, where C is the constant of integration. So, depending on the value of c, you are going to get a host of functions which will satisfy this condition, but the interesting point of definite integrals is that it is immaterial which particular function that satisfies this you use; the value of the definite integral will remain the same; this value remains the same. The definite integral of small f is a number. The definite integral is a number, which only depends on the function f and the interval a to b. And it is denoted by this notation: so first you have the usual integral sign, and you note that here I have written a and b. On top, I have written b, and on bottom, I have written a; then comes the integrand, which is small fx, and then comes dx. Small fx is the integrand; we have noted this; a to b, the closed interval is the interval of integration. So, this interval of integration is a new concept. It was not there when we talked about indefinite integrals. The numbers a and b are called the lower and the upper limit of integration. So, from where to where the integration is being done. The definite integral is defined. Those 2 things a and b are called the limits. One is called the lower limit, the lower value, and the other higher value is called the upper limit. Small x is a dummy variable here. Since its label does not affect the definite integral’s value, so it is just a placeholder. It doesn't really matter what variable you use; the value of the definite integral remains the same. So, to make it clear, I have written this: integration of a to b fxdx is the same as integration of a to b fy dy. So, as you can see on the left-hand side, I have used that variable x. On the right-hand side, everything remains the same; only the variable has been changed from x to y, and the claim is that the left-hand side is equal to the right-hand side. The reason being x is a dummy variable. It doesn't really matter what variable I take; the value of the definite integral remains the same. What is the value of the definite integral? It is capital Fb minus Fa. It is also denoted by, in short, fx a to b, so here b and a they denote the lower limit and upper limit. So, in other words, I can write fx; this is equal to—and this is what we are going to see later on repeatedly. So, this is what is being written here: integration of a to b fxdx is equal to capital Fb minus capital Fa is equal to capital fx a to b where capital f dashed of x is equal to small fx for all x belonging to the open interval a to b. Thus, the notations of definite and indefinite integrals are the same. The notations are the same; the integral sign is there; dx is there. There is a small difference where in the definite integral you have the lower and the upper limits; in the case of indefinite integrals, the limits are not there; they are the same, but they are different concepts. Conceptually, these two things are very much different. The definite integral is a single number, so integration of a to b fxdx denotes a single number. On the other hand, if I talk about the indefinite integral, integration of fxdx, it represents one of the infinitely many functions, all of which have fx as their derivative. We have talked about this before, so that is how the indefinite integral is defined. It is not a single number; it is one of the many functions, and all these functions satisfy a common property. The relationship between definite and indefinite integrals can be spelled out as follows. So, I take the indefinite integral of fx dx. A is equal to capital Fx plus C over an interval I if and only if integration of a to b fx dx, a is equal to capital Fb minus capital Fa where a and b belong to this interval I. The thing to note here—whatever the statement that is written here—is that here capital fx, this function is used, and that same function is making an appearance here also. And when I am talking about the definite integral, then this constant term is not there.
We now talk about properties of definite integrals. If small f is a function, continuous in the interval I which contains 3 points, let us say small a, small b, small c, then integration of a to b fx dx. A is equal to the minus of integration of b to a fx dx. This is not very difficult to see because what we are going to get from here is f of b minus f of a, assuming that f prime of x is equal to small fx. And you can write this as Fa minus Fb, and what is this? This is b to a fx dx. So, that is why you are going to get this particular property. Secondly, integration of fx dx defined over the interval a to a, a is equal to 0. And this is easy to see if you imagine integration as an area under a curve, so suppose you have x here on this axis and you have fx on the y-axis and the curve looks like this. You have a here and let us say b here. Now this entire area is f of b minus f of a. This entire shaded area and that is the same as saying this is equal to a to b fx dx. So, these three things are equivalent. This shaded area is capital Fb minus capital Fa, and thirdly, integration of small fx dx a to b. These three things are the same. Now if I take the limits to be a and a, so the same point, so there is no area that we are enclosing; therefore, this left-hand side is equal to 0. Thirdly, integration of a to b alpha f x dx is equal to alpha multiplied by integration of a to b fx dx where alpha is an arbitrary number. Now this is equivalent to what we have seen in the case of indefinite integrals. There also, if you have a constant term in the integrand, then it can be brought out. Here the same thing is occurring; the difference is that we are talking about definite integrals here. The fourth property is integration of a to b fx dx is equal to integration of fx from a to c plus integration of fx from c to b. The last property mentioned above can be easily seen when the definite integral is interpreted as the area under a curve. This can be seen from this diagram itself: so a and b are 2 numbers; a is less than b, and I take, suppose c somewhere in between; here is c. Now I know the left-hand side, that is a to b fx dx, is the entire shaded region. Now this entire shaded region can be subdivided into two separate regions. One is from a to c, and the other is from c to b. So, this a to b can be seen as the area between a and c, and that will be denoted by this plus the area between c to b, which is denoted by this. So, essentially, I am talking about this particular area on the left plus this particular area on the right. So, their summation is the total area. If f and g are continuous functions in the interval, closed interval a to b, and alpha and beta are arbitrary real numbers, then we have this property that integration of alpha fx plus beta gx dx from a to b, a is equal to alpha of integration of fx from a to b plus beta integration of gx from a to b. Again, this is similar to the property of indefinite integrals that we have seen before. The difference being here there are limits to the integral because we are talking about definite integrals here.
Now it is clear that if I take the derivative of the integral—basically these 2 things cancel each other because integration is anti-derivative. So, if you take the derivative of the anti-derivative, what you get is there—these 2 things are nullifying each other—and you are going to get the integrand itself. So, that is what is being written here. Also, if I take the limit from a to t, t is let us suppose a particular variable, then we get f of t minus f of a, a is a constant. Now I take this relationship and I differentiate this with respect to t because, as I told you, t is a variable whereas a is a constant, then what do we get? If I differentiate this, then the left-hand side will be this, d dt of the integration of a to t fx dx. And on the right-hand side, this term will drop out because this is constant; a differentiation of capital Ft is what we know; it is equal to small t—small of ft. So, what does this relationship mean? The derivative of the definite integral with respect to the upper limit of integration is equal to the integrand as a function evaluated at the upper limit. So, this is the integrand right, a small.
fx that is the integrand. We are evaluating this at the upper limit; the upper limit is small, and that is what we have got on the right-hand side. So, that is the same thing; that is written in words here.
By a similar logic, if I just reverse it, if I take the upper limit to be a and the lower limit to be t, that is what we have done here. Then the derivative of this with respect to t will come out to be minus of small ft. And again, what does it mean? The derivative of the definite integral with respect to the lower limit of integration is equal to minus the integrand as a function evaluated at the lower limit. These two properties might be used in some exercises.
Continuous functions are integrable. So, if you have a continuous function, if small f is a continuous function in the closed interval a to b, then there exists a continuous function capital Fx in the same interval a to b such that capital F prime x is equal to small fx for all x in the open interval a to b, which basically means that small fx is integrable; it is possible to integrate small fx.
Now here is a clarification, or sort of disclaimer: it must be clear that the definite integral has a general definition. Its geometric interpretation is one of the many interpretations. By geometric interpretation, what we mean is the area under a curve. So, that is the interpretation we are often invoking, but the definition, the general definition of definite integral as such, is more than just this interpretation, the geometric interpretation. So, here are some other sorts of interpretations that one can adduce to.
If small fx is a probability distribution—this is coming from the subject of statistics—so suppose small fx is a probability distribution, then integration of fx from a to b can be interpreted as the probability of the variable taking a value between small a and small b. So, here one is interpreting the definite integral as the probability of a particular event. And then there is this other sort of interpretation from the subject of economics. If fx is an income—this should be an—is an income density function, then integration of a to b fxdx can be interpreted as the proportion of the people having income between a to b. So, this interpretation is somewhat similar to the statistical interpretation, but you can see what is going on: fx is the income density function, and we are going to talk more about that later on when we deal with income distribution and the application of definite integrals there. Now, fx is the income density function; then if I take the definite integral of fx within a particular range, a particular limit a to b, then how do we interpret that number, that definite integral number? It can be interpreted as the proportion of people having income between these two values, that is, a and b.
So, here are some examples; enough theory. We have talked about many theories so far. Now let us take some examples to fix our ideas. The profit of a firm as a function of its output is given by this fx, and what is fx? It is this: 4000 minus x minus 3 million divided by x. So, this function fx is the profit, and what is x? x is the output. So, as you can see, the function is not a kind of a very monotonic function; as x rises, it is not sure how fx will behave; it might go up, it might go down also.
What is the output level where profit is maximized? That is the first part: at what output level is profit maximized? Second: the actual output varies between 1000 and 3000 units; find the average profit, which is given by this form. So, there are two parts to this question. First is: at what output level is the profit maximized? Now, this first part is simply a problem of optimization with a single variable. The second part is the part which is relevant to integration. So, what is being said in the second part? The second part is saying that the actual output level, which might not be maximizing the profit, mind you, the actual output varies between 1000 and 3000; find the average profit, and helpfully they have given us the form of the average profit. How does it define integration of 1000 to 3000 fx dx? So, this is what? This is the total profit, because fx is the profit, but profit can vary depending on the output level. So, if I integrate toward this range 1000 to 3000, I get the total profit, and what is the average profit? It is the total profit divided by the midpoint between 1000 and 3000. So, this is the midpoint, 2000, so that is how this is the average profit.
Now, first thing first, we take this function and try to see where it is getting maximized. So, I use the first-order necessary condition. So, I take the derivative and set that equal to 0. From this, I get minus 1 plus 3 million divided by x square, because there is a—what is the power of x? It is minus 1. So, if I take the derivative of that, minus 1 will come first, and then minus, minus become plus, and x to the power minus 1, minus 1, that is minus 2, and therefore I have divided by x square. And then this gives me this solution: x is equal to the root over 3 million. I take the positive root because output cannot be negative, and the root over of 3 million is 1000 multiplied by the root over 3. Now this was coming from the first-order condition, so I have to see whether the second-order condition is satisfied or not. For that, I take the derivative of this, that is the second derivative of the profit function, and this comes out to be clearly negative because here the power of x is minus 2, so minus 2 will come first. And so the second-order condition is satisfied because the second derivative is negative. So, this indeed is the profit-maximizing output level. So, the first part is solved.
Now, as we know, the actual output does not remain fixed at 1000 multiplied by the root over 3. The actual output varies between 1000 and 3000, so the average profit is this much. So, we have to find the integration of this. I write the profit function here, fx, which is this: 4000 minus x minus 3 million divided by x. And I am dividing this whole thing by 2000, so 2000 is appearing here, and integration of this is 4000x minus x square divided by 2 minus 3 million; integration of 1 divided by x, what is that? It is log of x. Now I have to take the value of this at 3000, and from that I have to subtract the value of this at x is equal to 1000. So, that is what is written here, and if I do that, actually I am going to get this figure 352.08. So, I have not shown this in the slide itself; it is a tedious mathematical procedure, so at the end of the day I am going to get 352.08. So, this is the average profit of the firm in reality. So, this may not be equal to the maximized profit, mind you.
I think I will stop here and take up the next topics in the next lecture. So, thank you for joining us, and I will see you in the next lecture. Thank you. Have a good day.