BLOCK 5: Integral calculus

Class 1: Introduction to Integral Calculus

Integrals are present in all sciences, since-as surely you know - we need them to calculate, among other things, areas. In economics it is also necessary to calculate areas, but in this case, not of lands or physical issues, but of “concepts” which, in an abstract way, also represent an area. An area will be able to be an accumulated profit, a consumer’s surplus, even a probability. That is why it is important that we handle this instrument well. To do this, you will need:

  • Derive well in a variable
  • Understand the concept of area
  • Have notions of the concept of limit
  • Know how to draw a line, a parable and more general regions in the drawing

The primitive function

Let’s start by defining a key function for integral calculus: the primitive -or antiderivative- of a function \(f(x)\).

A primitive (or antiderivative) of a function \(f(x)\) is another function, for example \(F(x)\) such that if we derive \(F(x)\) we get \(f(x).\) I mean:

\[ F'(x)=f(x). \]

As you can see, we are establishing an inverse operation to the derivative: we look for a function whose derivative is the original function. We also call that it “antiderivative” or “primitive.” For example, if my function is \(f(x)=3\), what will be the primitive or antiderivative? Well, we have than to look for a function \(F(x)\) that, when deriving it, gives us 3. Sure that you have guessed that

\[ F(x)=3x \]

However, there is one detail that you may not have noticed: it is not the the only possible function that, when derived, gives us \(3\). There is more: in fact, there are infinite:

\[ F(x)=3x+C \]

where \(C\) is a constant. As we know, the derivative of a constant is \(0\). In this way,

\[ F'(x)=3 \]

Therefore, if a function has a primitive, it actually has infinite of them!

FIG1: A few primitives

Now, to refer to this operation without the need to have that write the word primitive or antiderivative let’s go to use a symbol:

\[ F(x)=\int f(x) \]

where \(\int\) is the “symbol” of the integral, \(f(x)\) is the integrand (or the function to which we want to calculate the primitive). It is usual (although not always necessary), however, add with respect to what variable do we integrate \(\mathrm{"d}x"\)

\[ F(x)=\int f(x)\mathrm{d}x \]

Therefore, several issues seem interesting, once you have understood the objective we pursue:

  • build a table of antiderivatives (to simplify our work)
  • Study that table
  • practice!

For example, calculate the antiderivative of \(f(x)=3x\)

Now I’m being asked to calculate this:

\[ F(x)=\int3x\mathrm{d}x \]

Enntonces, I have to think of a function that, when derived, gives me \(3x\). Look at this candidate:

\[ F(x)=\frac{3}{2}x^{2}+C \]

Is it well anti-derived? To know, we have to derive it and see that what we are looking for is fulfilled:

\[ F'(x)=\frac{3}{2}{\color{red}2x} \]

simplifying, we get that:

\[ F'(x)=3x=f(x) \]

wow, this method seems unbeatable!

If you notice, we can think of some strategy to get the most famous primitives. These are usually called immediate primitives. To do this, you have to think about how that function is derived and do it. contrary.

IMMEDIATE PRIMITIVES

  • \(\int x^{n}\mathrm{d}x=\frac{x^{n+1}}{n+1}+C\) care, if \(n\neq-1\)
  • \(\int x^{-1}\mathrm{d}x=\int\frac{1}{x}\mathrm{d}x=\ln\left|x\right|+C\) (This is the case above with \(n=-1)\)
  • \(\int e^{x}\mathrm{d}x=e^{x}+C\)
  • \(\int a^{x}\mathrm{d}x=\frac{a^{x}}{\ln a}+C\)
  • \(\int\sin x\mathrm{d}x=-\cos(x)+C\)
  • \(\int\cos x\mathrm{d}x=\sin(x)+C\)$ $

where \(C\in\mathbb{R}\).

In addition, there are two properties of obtaining primitives that are particularly very interesting:

\[ \int kf(x)\mathrm{d}x=k\int f(x)\mathrm{d}x,\:k\in\mathbb{R} \] (i.e., the integral of a constant by a function is equal to the constant multiplied by the function). So, in our example previous, we could have done perfectly \[ F(x)=\int3x\mathrm{d}x=3\int x\mathrm{d}x \] and, using rule 1 in Table 1: \[ F(x)=3\frac{x^{1+1}}{1+1}+C=3\frac{x^{2}}{2}+C \]

The other property, says that

\[ \int\left[f(x)+g(x)\right]\mathrm{d}x=\int f(x)\mathrm{d}x+\int g(x)\mathrm{d}x \]

(That is, the integral of a sum of functions is equal to the sum of the integral of each function).

Example what is the integral of \(\int\left(x^{2}+\frac{1}{x^{2}}+\sqrt{x}\right)\mathrm{d}x\)?

  • First, write it all down more clearly \(\int\left(x^{2}+\frac{1}{x^{2}}+\sqrt{x}\right)\mathrm{d}x=\int\left(x^{2}+x^{-2}+x^{1/2}\right)\mathrm{d}x\)
  • Then, \(\intop x^{2}\mathrm{d}x+\int x^{-2}\mathrm{d}x+\int x^{1/2}\mathrm{d}x=\frac{x^{3}}{3}+\frac{x^{-1}}{-1}+\frac{x^{1/2+1}}{1/2+1}=\frac{x^{3}}{3}-\frac{1}{x}+\frac{2x\sqrt{x}}{3}+C\)

We call these integrals indefinite integrals (because they are made for any value of \(x\) for which these functions make sense) and immediate (because we can use the table of integrals you have done before). With the passing of the classes, we will see that it will not always be possible to have immediate integrals.

As we saw earlier, there is a set of integrals that have easy solution, since knowing how to derive, you immediately get the “reverse” operation. Now, what happens if we modify slightly the table of immediate integrals?

For example, if instead of having \(\int x^{2}dx\), we have a composition of functions? How would we integrate \(\int(2x+3)^{2}dx\)? If you you realize, you can’t follow the above rules to the letter.

If you did this you would NOT get the primitive. Scope:

\[ \int(2x+3)^{2}dx=\frac{(2x+3)^{3}}{3}+C \]

THIS is NOT the primitive. Why? Let’s go to derive it:

\[ \frac{d\frac{(2x+3)^{3}}{3}}{dx}=\frac{3(2x+3)^{2}}{3}2={\color{red}2\left(2x+3\right)^{2}} \]

it is not the original function! If you notice, that “2” is the derivative of \(f(x)=2x+3\), which is the “inner” function. So that if we want to compute the integral, it can be done like this: we should introduce that “2,” following the famous chain rule. As we know that if we multiply and divide by the same number nothing changes, we can do this:

\[ \int\left(2x+3\right)^{2}\mathrm{d}x={\frac{1}{\color{red}2}}\int{\color{red}2}\left(2x+3\right)^{2}\mathrm{d}x=\frac{1}{2}\frac{\left(2x+3\right)^{3}}{3}+C \]

Now, if you derive the primitive, you get the function \(f(x)\). As you can see, we have manipulated on both sides to get it to appear the derivative of the function (we are applying the chain rule). Be careful, these manipulations can only be done with numbers, never multiplying and dividing by the variable \(x\).

Now, let’s reformulate the table of immediate integrals to have composite functionsY

IMMEDIATE PRIMITIVES OF COMPOSITE FUNCTIONS

  • \(\int f'(x)f(x)^{n}\mathrm{d}x=\frac{f(x)^{n+1}}{n+1}+C\) care, if \(n\neq-1\)
  • \(\int f'(x)f(x)^{-1}\mathrm{d}x=\int\frac{f'(x)}{f(x)}\mathrm{d}x=\ln\left|f(x)\right|+C\) (This is the case above with \(n=-1)\)
  • \(\int f'(x)e^{f(x)}\mathrm{d}x=e^{f(x)}+C\)
  • \(\int f'(x)a^{f(x)}\mathrm{d}x=\frac{a^{f(x)}}{\ln a}+C\)
  • \(\int f'(x)\sin f(x)\mathrm{d}x=-\cos(f(x))+C\)
  • \(\int f'(x)\cos f(x)\mathrm{d}x=\sin(f(x))+C\)$ $

where \(C\in\mathbb{R}\).

Important

Many times you will have to decide if an integral is like the first of the list or the second. For example:

  • Case 1

\[ \int\left(\frac{x+2}{\sqrt{x^{2}+4x+10}}\right)\mathrm{d}x \]

  • Case 2

\[ \int\left(\frac{x-8}{x^{2}-16}\right)\mathrm{d}x\label{eq:INT2} \]

In the case (1), we have something similar in the numerator to the derivative of the function that is inside the square root. Then we do not have the derivative of the denominator, but of the inner function in the composition. And, in addition, a square root is a power, so we have, actually:

\[ \int\left(x+2\right)\left(x^{2}+4x+10\right)^{-1/2}\mathrm{d}x \]

so it is of the form (1) of the list. To solve it, you will have to multiply and divide by 2,

\({\frac{1}{\color{red}2}}\int{\color{red}2}\left(x+2\right)\left(x^{2}+4x+10\right)^{-1/2}\mathrm{d}x\).

In the case of the integral given by (2), in the numerator we have “almost” the derivative of the denominator. So it will be of the logarithmic type. Again, you will have to multiply by 2.

\[ {\frac{1}{\color{red}2}}\int{\color{red}2}\left(\frac{x-8}{x^{2}-16}\right)\mathrm{d}x \]

solved exercises

\[ \int\left(e^{x}+1\right)^{2}\mathrm{d}x \]

  • \(\int \left(e^{x}+1\right)^{2}=\int e^{2x}+\int 1+\int2e^{x}\), and therefore,\(\int (e^{2x}+1+2e^{x})\), which is immediate:

\[ \int (e^{2x}+1+2e^{x})={\frac{1}{2}}\int{2}e^{2x}+\int1+2\int e^{x}=\frac{1}{2}e^{2x}+x+2e^{x}+C \]

  • \[ \int\frac{e^{3x-1}}{e^{3x-1}+4}\mathrm{d}x \]

It is of type \(\int\frac{f'(x)}{f(x)}=\ln\left|f(x)\right|+C\), by what \({\frac{1}{3}}\int\frac{{3}e^{3x-1}}{e^{3x-1}+4}\mathrm{d}x={\frac{1}{3}}\ln\left|e^{3x-1}+4\right|+C\)

  • \[ \int\frac{x}{\sqrt{x^{2}+5}}\mathrm{d}x \]

It is of the type \(\int f'(x)f(x)^{n}=\frac{f(x)^{n+1}}{n+1}+C\), so that \(\int x(x^{2}+5)^{-1/2}={\frac{1}{2}}\int{2}x(x^{2}+5)^{-1/2}=\sqrt{x^{2}+5}+C\)

Class 2: The Integration Method “by parts”

Sometimes we have other types of integrals with which it seems that it does not you can do a big deal:

\[ \int xe^{x}\mathbf{d}x \]

are of the type “product of two functions.” These integrals, obviously, they are not- in general-immediate. Let’s see a method that, based on the properties of the derivative of a product, will allow us to integrate some of this style.

Now, to continue working we are going to call \(u\) and \(v\) the image of two functions \(f(x)\) and \(g(x)\)

\[ u=f(x), \] \[ v=g(x) \]

We also know that \(\frac{du}{dx}=f'(x)\Rightarrow du=f'(x)dx\) and that \(\frac{dv}{dx}=g'(x)\Rightarrow dv=g'(x)dx\). In this way (we leave you the proof at the end of this class, in case you are interested), we can say that

\[ \int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)dx \]

which can be rewritten as:

\[ \int u\mathrm{d}v=uv-\int v\mathrm{d}u \]

This also allows us to remember the expression (in spanish words):

\[ \int u\mathrm{d}v={\color{red}u}\mathrm{\mathrm{n\,d\acute{\imath}a\,vi\,una}}\,{\color{red}v}\mathrm{aca}-\int{\color{red}v}\mathrm{estida\:{\color{red}d}e}{\color{red}u}\mathrm{niforme} \]

And now we have to put it into practice. It is advisable to follow the next procedure (which we will apply for the integral \(\int xe^x \mathrm{d}x\))

EXAMPLE OF THE PROCEDURE

  • Let’s choose the function \(u\) and the rest will be \(\mathrm{d}v\). It seems reasonable that we choose to \(u\) a function that is difficult to integrate.

However, we can use as a criterion a mnemonic rule that establishes an order of priorities to choose \(u\): ALPES, by

A: arc functions (arcsine, arctangent, etc….) L: logarithm function P: powers and polynomials E: exponential S: breasts and cosines.

FIG1: One day I saw a cow dressed in uniform in the Alps

In our case, we must choose

\[ u=x, \]

since, in order of priorities “ALPES” is before the “P” of powers and polynomials than the “E” of exponential

  • Once we have \(u\), the rest of the integral we will call \(\mathrm{d}v\).

In our case, \(\mathrm{d}v=e^x\mathrm{d}x\).

  • We get \(\mathrm{d}u=f'(x)\mathrm{d}x\).

In our case, \(\mathrm{d}u=\mathrm{d}x\)

  • We obtain, using that the integral is the inverse of the derivative: \(v=\int \mathrm{d}v\).

In our case \(v=\int e^x\mathrm{d}x\Rightarrow v=e^x\)

  • We apply the final expression, since we have \(u,v,\mathrm{d}u,\mathrm{d}v\)

In our case,

\[ \int xe^x\mathrm{d}x= xe^x-\int e^x\mathrm{d}x=xe^x-e^x+C=e^x(x-1)+C \]

It is important to note that the ALPES rule is a mnemonic rule (no scientific). Therefore, it helps to make decisions. Yes, following the rule, the integral that remains is not immediate, two things can happen:

  • Or that you have chosen the wrong functions and you must undo what you have done
  • That you have to reintegrate in parts (they are usually called integrals by cyclic parts).

These things are learned with practice :)

solved Exercises:

  • \(\int t\sin t\mathrm{d}t\)

\[u=t\Rightarrow\mathrm{d}u=\mathrm{d}t\]

\[\mathrm{d}v=\sin t\mathrm{d}t\Rightarrow v=\int\sin t\mathrm{d}t\Rightarrow v=-\cos t\] \[\int t\sin t\mathrm{d}t=t(-\cos t)-\int-\cos t\mathrm{d}t=-t(\cos t)+\sin t+C\]

  • \(\int\arctan x\mathrm{d}x\)

\[ u=\arctan x\Rightarrow\mathrm{d}u=\frac{1}{1+x^{2}}\mathrm{d}x\]

\[\mathrm{d}v=\mathrm{d}x\Rightarrow v=\int\mathrm{d}x\Rightarrow v=x\] \[ \int\arctan x\mathrm{d}x=x\arctan x-\int\frac{x}{1+x^{2}}\mathrm{d}x\] Notice that \(\int\frac{x}{1+x^{2}}\mathrm{d}x={\frac{1}{2}}\int\frac{2x}{1+x^{2}}\mathrm{d}x=\frac{1}{2}\ln\left|1+x^{2}\right|\) Then \[\int\arctan x\mathrm{d}x=x\arctan x-\frac{1}{2}\ln\left|1+x^{2}\right|+C\]

Where does the expression of the integral by parts come from? Just in case you’re curious

Let’s start from the product of two functions \(f(x)\) and \(g(x)\) and use the rule of the derivative of this

\[ \left(f(x)g(x)\right)'=f'(x)g(x)+f(x)g'(x) \]

As we will want to have a way to integrate these functions, let’s apply integration on both sides:

\[ \int\left(f(x)g(x)\right)'\mathrm{d}x=\int f'(x)g(x)\mathrm{d}x+\int f(x)g'(x)\mathrm{d}x \]

Notice that, taking advantage of the fact that integral and derivative are inverse functions:

\[ f(x)g(x)=\int f'(x)g(x)\mathrm{d}x+\int f(x)g'(x)\mathrm{d}x \]

Now, to continue working we will redefine \[ u=f(x) \] \[ v=g(x) \]

We also know that \(du=f'(x)dx\) and that \(dv=g'(x)dx\). In this way, substituting

\[ uv=\int v\mathrm{d}u+\int u\mathrm{d}v \]

I mean:

\[ uv-\int v\mathrm{d}u=\int u\mathrm{d}v, \] which is the rule of the integral in parts.

Class 3: the rational integral

A type of integrals that can appear, since they arise in many applications, are those that are given as the quotient of two polynomials. For example, we intend to obtain the integral of the following approach \[ \int\frac{P(x)}{Q(x)}\mathrm{d}x \]

where \(P(x)\) and \(Q(x)\) are polynomials with equal or different degree. In fact, we will work differently according to the degree of both:

possible cases

-degree \(P(x)\) \(\geq\) degree \(Q(x)\), then Division of polynomials + Rest

  • grade \(P(x)\) \(<\) degree \(Q(x)\), then factorization of the type \(\frac{A}{x-a}+\frac{B}{x-b}+....\)}

where the roots \(a,b,...\) are real numbers.

We will now focus on the second case. Imagine for example, that you have to integrate this function

\[ \int\frac{x+2}{x^{2}-16}dx \]

where \(P(x)=x+2\) and \(Q(x)=x^{2}-16\). As you can see, the degree of the polynomial of the numerator is lower than that of the denominator. In that case, as you we have said before, you should look for a simpler factorization. Let’s see how:

EXAMPLE OF THE PROCEDURE

  • We set the goal \(\frac{x+2}{x^{2}-16}=\frac{A}{x-a_{1}}+\frac{B}{x-a_{2}}\)
  • We look for the roots of the denominator \(x^{2}-16=0\Rightarrow x=\pm4\)
  • We rewrite \(\frac{x+2}{x^{2}-16}=\frac{A}{x-4}+\frac{B}{x+4}\)
  • With the above expression, we perform the sum:

\[\frac{x+2}{x^{2}-16}=\frac{A(x+4)+B(x-4)}{(x-4)(x+4)}\]

we have, then \(\frac{x+2}{x^{2}-16}=\frac{A(x+4)+B(x-4)}{x^{2}-16}\) we can simplify the denominator

  • \(x+2=A(x+4)+B(x-4)\) how do we continue?

If you notice, this equation must be met for any value of \(x\in\mathbb{R}\). So we could give values to \(x\) to obtain \(A,B\) via a system of equations. But first to put any value, you already know two values that cancel the polynomial \(x=\pm4\). Let’s use them to simplify the process: \[ 4+2=A(8)+B(4-4)\Rightarrow A=\frac{8}{6}=\frac{4}{3} \] \[ -4+2=A(-4+4)+B(-4-4)\Rightarrow-2=-8B\Rightarrow B=\frac{1}{4} \]

  • So, we already have it:

\[ \frac{x+2}{x^{2}-16}=\frac{A}{x-a_{1}}+\frac{B}{x-a_{2}}=\frac{\frac{4}{3}}{x-4}+\frac{\frac{1}{4}}{x+4} \]

  • So the integral will be:

\[ \int\frac{x+2}{x^{2}-16}dx=\int\left[\frac{\frac{4}{3}}{x-4}+\frac{\frac{1}{4}}{x+4}\right]dx \]

I mean

\[ \frac{4}{3}\int\frac{1}{x-4}dx+\frac{1}{4}\int\frac{1}{x+4}dx=\frac{4}{3}\ln\left|x-4\right|+\frac{1}{4}\ln\left|x+4\right|+C \]

where \(C\in\mathbb{R}\).

Solved Exercises

  • \[ \int\frac{10x-2x^{2}}{(x-1)^{2}(x+3)} \]

\[\int\frac{10x-2x^{2}}{(x-1)^{2}(x+3)}\] this has a “trap” since the roots of the denominator polynomial are \(\left\{ 1,1,-3\right\}\), is that is, it has a double root (which is 1). When that happens, we will propose the following schema to solve \[ \frac{10x-2x^{2}}{(x-1)^{2}(x+3)}=\frac{A}{x-1}+{\frac{B}{(x-1)^{2}}}+\frac{C}{x+3} \]

where the denominator associated with the multiple root has to be decomposed in as many sums as the number of times the root is repeated in such a way that the degree of the monomial is increased up to to reach that the degree is equal to the multiplicity of the root. In in this case, the root 1 has multiplicity 2 so we must have a simple fraction divided by \((x-1)\) and the other by \((x-1)^{2}\). Once we have this clear, the rest is standard:

  1. \[ \frac{10x-2x^{2}}{(x-1)^{2}(x+3)}=\frac{A(x-1)(x+3)+B(x+3)+C(x-1)^{2}}{(x-1)^{2}(x+3)} \]

  2. that is, we will have to solve \[ 10x-2x^{2}=A(x-1)(x+3)+B(x+3)+C(x-1)^{2} \]

  3. remember that since we have 3 unknowns, we need three equations and that, in addition, this equation must be verified for all \(x\). Chose such as values of \(x={1,-3}\) and, for example, \(x=0\). Have:

\[ \begin{cases} x=1 & 8=4B\Rightarrow B=2\\ x=-3 & -48=16C\Rightarrow C=-3\\ x=0 & 0=-3A+6-3\Rightarrow A=1 \end{cases} \]

  1. We arrive, then, at the next integral

\[ \int\frac{1}{x-1}+{\frac{2}{(x-1)^{2}}}+\frac{-3}{x+3} \]

that it is immediate and that, surely, you already know how to solve.

  • \[ \int\frac{y+2}{2y^{2}+3y+1} \]
  1. the roots of the polynomial \(Q(x)=2y^{2}+3y+1\) are \(y=\{-1,-1/2\}\)
  2. We rewrite the integral as \(\int\frac{y+2}{2y^{2}+3y+1}\mathrm{d}y=\int\frac{A}{y+1}+\frac{B}{y+\frac{1}{2}}\)
  3. Again, we have to \[ \frac{y+2}{2y^{2}+3y+1}=\frac{A\left(y+\frac{1}{2}\right)+B\left(y+1\right)}{2y^{2}+3y+1} \]
  4. So, we will need two equations to get the two unknowns (\(A,B\)) where we know that the above equation has to be fulfilled for any \(y\).

\[ \begin{cases} y=-1 & 1=\frac{-1}{2}A\Rightarrow A=-2\\ y=-\frac{1}{2} & \frac{3}{2}=-\frac{1}{2}B\Rightarrow B=-3 \end{cases} \] 5) We arrive, then to \(\int\frac{y+2}{2y^{2}+3y+1}\mathrm{d}y=\int\frac{-2}{y+1}+\frac{-3}{y+\frac{1}{2}}=-2\ln\left|y+1\right|-3\ln\left|y+\frac{1}{2}\right|+C.\)

advanced: What if the roots are complex? One possibility: the arctangent

Before we see what happens if the roots are complex, let’s remember -de the tables of derivatives-that

\[ y=\arctan x\Rightarrow y'=\frac{1}{1+x^{2}} \]

and, in the event that the function is compound,

\[ y=\arctan f(x)\Rightarrow y'=\frac{f'(x)}{1+f(x)^{2}} \]

Therefore, it is easy to deduce that, for example, \[ \int\frac{1}{1+x^{2}}\mathrm{d}x=\arctan x+C \] and that, where appropriate, \[ \int\frac{3x^{2}}{1+(x^{3}+1)^{2}}\mathrm{d}x=\arctan\left(x^{3}+1\right)+C \]

Notice that you could not solve them as we have done before because the roots of the polynomial \(Q(x)\) are not real. What is done then, if we find ourselves with such a quotient? For example

\[ \int\frac{1}{x^{2}+6x+10}\mathrm{d}x \]

if you try to get the roots of \(Q(x)=x^{2}+6x+10\) you will see that these they are not real. So, you’ll need to test if you can rewrite this polynomial as a square to see if you can get it to behave like the integral of an arctangent.

PROCEDURE: complete squares

  • We want to have, if possible, \(x^{2}+6x+10=(x+a)^{2}+b\) in this way, will we rewrite the polynomial and stay? something “similar” to what we need for the integral to be of the arctangent type, although then we have to work a little more, of course :)
  1. Matching \(x^{2}+6x+10=x^{2}+a^{2}+2ax+b\)

  2. Now we have to look for the values for \(a.b\). Let’s reorganize

  3. \(x^{2}+6x+10=x^{2}+2ax+a^{2}+b\)

  4. From here, we get that \(2a=6\Rightarrow a=3\)

  5. $a^{2}+b=10b=$1 (since $a=$3)

Then, we come to \(x^{2}+6x+10=(x+3)^{2}+1\)

Given the procedure of completing squares, we solve in a way direct \[ \int\frac{1}{x^{2}+6x+10}\mathrm{d}x=\int\frac{1}{(x+3)^{2}+1}\mathrm{d}x=\arctan(x+3)+C \]

Solved Example

  • \(\int\frac{1}{x^{2}+4}\mathrm{d}x\)

Again, we have complex roots in the denominator (and a 1 in the numerator). Let’s divide everything by 4, so that the denominator looks like \(\int\frac{\frac{1}{4}}{\frac{x^{2}}{4}+1}\mathrm{d}x\). Let’s reorganize \(\int\frac{\frac{1}{4}}{\left(\frac{x}{2}\right)^{2}+1}\mathrm{d}x=\frac{1}{4}\int\frac{1}{\left(\frac{x}{2}\right)^{2}+1}\mathrm{d}x=\frac{{2}}{4}\int\frac{{\frac{1}{2}}}{\left(\frac{x}{2}\right)^{2}+1}\mathrm{d}x=\frac{1}{2}\arctan\left(\frac{x}{2}\right)^{2}+C\)

Class 4: FTC and its applications to economics

In today’s class we introduce a novelty in integral calculus: the definite integral. This is intimately related to the concept of area. It is outside the objectives of the course to give a formal proof of this result, but-as will be seen in the appendix to this class- it can be verified that in a continuous and positive \(f(x)\) function in an interval \(x\in[a,b]\), the area \(A\) between the axis of the \(x\) and the function, and delimited by the points \(a, b\) consists of

\[ A=\int_{a}^{b}f(x)\mathrm{d}x \]

Now, how is this value calculated? We can prove this thanks to the Fundamental Theorem of Calculus: we know that if \(f(x)\) is a continuous function, for any value of \(x\in[a,b]\) is fulfilled. that \(F(x)=\int f(x)\mathrm{d}x\) in such a way that \(F'(x)=f(x).\) say, \(F(x)\) is an antiderivative.

As we already know, a primitive \(F(x)\) satisfies that:

\[ F(x)=\int_{a}^{x}f+C \]

for \(x\in[a,b].\) In fact, notice that \[ F(a)={\color{red}\int_{\color{red}a}^{\color{red}a}}+C\equiv{\color{red}0}+C \]

since at one point we have no area. This tells us that \(F(a)=C\), so

\[ F(x)=\int_{a}^{x}f+C\Rightarrow F(x)=\int_{a}^{x}f+F(a) \]

in such a way that, surprise!

\[ F(x)-F(a)=\int_{a}^{x}f \]

and, particularizing in a specific interval \(x\in[a,b]\)

\[ F(b)-F(a)=\int_{a}^{b}f, \]

This technique is known as “Barrow’s rule” and tells us that solving a definite integral is very easy: only you have to look for the primitive and evaluate it at the extremes of the limits of integration and, finally, subtract them.

For example

\(\int_{1}^{2}x^{2}\mathrm{d}x\)

We get \(F(x)\), as if the integral were \(\int x^{2}\mathrm{d}x\), i.e. \(F(x)=\frac{x^{3}}{3}\)

We evaluated \(F(2)-F(1)=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\).

Generally, everything is done at once, using “brackets” of this way

\[ \int_{1}^{2}x^{2}\mathrm{d}x=\left[\frac{x^{3}}{3}\right]_{1}^{2}=\frac{2^{3}}{3}-\frac{1}{3}=\frac{7}{3} \]

Exercise What is the area of the function \(y=x\) in the range \(x\in[0.1]\)? Try it by the area formula and then by the defined integral. Draw the function first to know which geometric figure it is.

Famous applications of integral calculus

The first thing we are going to stop at is the use of integrals to calculate area of figures. To do this, you have to take into account some instructions that you have already seen, surely, in previous courses.

  • Try to draw the functions involved (it is good to know how they are oriented)
  • Get the cut-off point with the axes that interest you
  • Poses the integral defined in the cut-off points

Let’s look at an example: calculate the area of the region bounded by the functions \(y=x+2\) and \(y=x^{2}\):

FIG1: These are both functions

If you choose (1) as the path:

  1. The cut-off points are found as \((x+1)=x^{2}\). The equation of second degree you have provides as solutions $x=-1,x=$2
  2. The integral that we have to solve will be: \(\int_{-1}^{2}\left[(x+2)-x^{2}\right]\mathrm{d}x\)
  3. The result, with Barrow’s rule:\(\int_{-1}^{2}\left[(x+2)-x^{2}\right]\mathrm{d}x\)=\(\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}=\frac{9}{2}\)

If, instead, you choose (2) as the path, now the functions are \(x=y-2\), \(x=\pm\sqrt{y}\) (think it’s like you’ve rotated now the axes). Therefore

  1. Cut-off points are obtained by matching \(y-2=\pm\sqrt{y}\) implies that \(y=4,y=1.\)
  2. Note that another point must be added $y=$0. That’s why it’s good to draw the functions with precision.
  3. Between the values of \(0<and<1\) the area is given by \(\sqrt{y}\),\(-\sqrt{y}\), i.e. \(\int_{0}^{1}\left[\sqrt{y}-(-\sqrt{y})\right]\mathrm{d}y\)
  4. Between the values \(1<y<4\), the area is given by \(\int_{1}^{4}\left[\sqrt{y}-(y-2)\right]\mathrm{d}y\)

That is:\(\int_{0}^{1}\left[\sqrt{y}-(-\sqrt{y})\right]\mathrm{d}y+\int_{1}^{4}\left[\sqrt{y}-(y-2)\right]\mathrm{d}y\)

Practice (and drawing skill) will allow you to analyze which strategy it’s simpler.

On the other hand, you should take into account what happens if the function changes sign. The integral, as is, could not be interpreted as an area. Look at this example: the integral is 0

FIG2: This is not an area

What you have to do, in that case, is to split the integral and, in the negative part, keep its absolute value.

FIG3: The integral changes sign: this is an area

On the other hand, when you have to calculate the area defined between two functions (we usually say between two curves) what you have to do is draw them to see which one is above. Depending on your orientation, you will be interested, as in the figure, to get \(\int_{a}^{b}\left[f(x)-g(x)\right]\mathrm{d}x\) or, the other way around

FIG4: Area between two curves

Actually, it doesn’t matter where the curves are located on the coordinate axes. You can always calculate the area as a difference from the function that it is above minus the one below.

Applications to finance How is the effect of an interest on an income calculated? If we have an initial amount \(P\), in euros, and we introduce it into the bank and this offers us a fixed interest rate at one year of \(r\), then, after a year, we will have won

\[ B=P(1+r) \]

  • For example, if we have 1000 euros and the interest rate is 10%, then, after a year, we will have $B=1000(1+0.1)=$1100

Of course. Now, what happens if the annual interest rate is 10% but you settle the interest every six months? Here, remunerates the amount paid twice a year \[ B=1000(1+\frac{0.1}{2})(1+\frac{0.1}{2})=1102.5 \] What if interest is settled every three months? \[ B=1000(1+\frac{0.1}{4})(1+\frac{0.1}{4})(1+\frac{0.1}{4})(1+\frac{0.1}{4})=1103.8 \]

As we can see, the formula that we can apply, results in

\[ B=P\left(1+\frac{r}{k}\right)^{k} \]

Let’s do a little bit of math. Weeded in the term \(\frac{r}{k}\) both sides for \(r\), having \(\frac{1}{\frac{k}{r}}\), let’s call \(n=\frac{k}{r}\), so that

\[ B=P\left(1+\frac{1}{n}\right)^{nr} \]

which we can rewrite as

\[ B=P\left[\left(1+\frac{1}{n}\right)^{n}\right]^{r} \]

  • Why do we do this?

Use the calculator to see what you get if you give values to \(n\). See What

\[ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}=2.71...={\color{red}e} \]

In this way, we can rewrite the previous formula in a way dresser:

\[ B=Pe^{r} \]

that will be more accurate, the less it takes the bank to remunerate you for annual interest. For example \[ B=1000e^{0.1}=1105.2 \]

that we can understand it as the maximum that the bank could pay us the interest in a fixed term. Note that if we leave it more than a year, we will have \[ B=Pe^{rt} \]

Well, we are going to need this expression for the question that we we do below.

What happens if, in a period of time \(T\), we periodically deposit money into the account?

Let’s go back to the previous formula. Let’s imagine that we do it in 3 years. In the year \(0\) we enter \(P_{0}\) and that amount is up to the third year \(P_{0}\left(1+\frac{r}{k}\right)^{3k},\) in the second year we enter \(P_{1}\), so we will accumulate \(P_{1}\left(1+\frac{r}{k}\right)^{2k}\) (it will be one year less in the account before taking it out)…. and so, we will have \[ VF=P_{0}\left(1+\frac{r}{k}\right)^{3k}+P_{1}\left(1+\frac{r}{k}\right)^{2k}+P_{2}\left(1+\frac{r}{k}\right)^{k}=\sum_{t=0}^{3}P_{t}\left(1+\frac{r}{k}\right)^{(3-t)k} \]

Now, let us remember that, if we do \(k\) as big as we can, we can use the exponential and the sum, in this case, is transformed in an integral \[ VF=\int_{0}^{T}P(t)e^{(T-t)r}\mathrm{d}t \]

what we will call future value (VF) of an investment flow, and \(T\) the horizon to which we do it. Realize you can write it down how

\[ VF=\int_{0}^{T}P(t)e^{Tr}e^{-rt}\mathrm{d}t=e^{Tr}\int_{0}^{T}P(t)e^{-rt}\mathrm{d}t \]

Now I wonder. How much money will I have to invest in total for have a certain future value? We take it right away from the previous expression, we want to get VF, that is, \[ {\color{green}B}={\color{red}P}{\color{blue}e^{rt}}\:\:\; {\color{green}V\color{green}F}={\color{blue}e^{Tr}}{\color{red}{\int_{0}^{T}P(t)e^{-rt}\mathrm{d}t}} \]

That is, the sum you must invest will be \[ VP=\int_{0}^{T}P(t)e^{-rt}\mathrm{d}t \]

Exercise resolved

The present value of an investment is used to decide TODAY which investment candidates are worth the most and, therefore, which investment I should make. For example

A successful textile entrepreneur is considering two alternative plans to improve your product. Plan A requires immediate disbursement of 350,000 euros, while plan B will need a disbursement immediate of 100,000 euros. It has been estimated that the adoption of the plan A would mean a net stream of income generated at the rate of 730000 Euros. And for plan B it would represent a net stream of income to ratio of 680000 euros over the next three years. If the rate of interest over the next few years was 10% per year Which plan will be the one that is it in the best interest of the entrepreneur?

The data for plan A, are

$P(t)=$730000, $r=$0.1 So the present value will be \[ VP=\int_{0}^{3}730000e^{-0.1t}dt-350000=1542027 \]

The data for plan B, are

$P(t)=$680000, $r=$0.1 So the present value will be \[ VP=\int_{0}^{3}680000e^{-0.1t}dt-100000= \]

The present value of an investment is when we calculate the current value that you will have a certain amount that we will receive or pay in a future, in the agreed period. Future value is the value achieved for a certain capital at the end of the given period.

EXTRA Some solved integrals, to practice

  • \(\int x^{2}\left(3x^{3}+14\right)^{3}\mathrm{d}x\)

\({\frac{1}{9}}\int{9x^2}\left(3x^{3}+14\right)^{3}\mathrm{d}x=\frac{1}{9}\frac{\left(3x^{3}+14\right)^{4}}{4}+C=\frac{\left(3x^{3}+14\right)^{4}}{36}+C\)

  • \(\int\sqrt[5]{5x+6}\mathrm{d}x\)

\({\frac{1}{5}}\int{5}\left(5x+6\right)^{1/5}\mathrm{d}x=\frac{1}{5}\frac{\left(3x^{3}+14\right)^{6/5}}{\frac{6}{5}}+C=\frac{\left(3x^{3}+14\right)^{4}}{6}+C\)

  • \(\int\frac{17x}{\sqrt[3]{6x^{2}+8}}\mathrm{d}x=\)

\(\int17x\left(6x^{2}+8\right)^{-1/3}\mathrm{d}x={\frac{17}{12}}\int{12}x\left(6x^{2}+8\right)^{-1/3}\mathrm{d}x=\frac{17}{12}\frac{\left(6x^{2}+8\right)^{2/3}}{2/3}+C=\frac{51}{24}\left(6x^{2}+8\right)^{2/3}+C\)

  • \(\int\frac{\mathrm{d}x}{\left(3x+1\right)^{4}}\)

\({\frac{1}{3}}\int\frac{\mathrm{{3}d}x}{\left(3x+1\right)^{4}}\mathrm{d}x=\frac{1}{3}\frac{\left(3x+1\right)^{-3}}{-3}+C=\frac{\left(3x+1\right)^{-3}}{-9}+C\)

  • \[ \int x^{2}e^{-\frac{1}{5}x^{3}}\mathrm{d}x \]

This is immediate

\[ \int x^{2}e^{-\frac{1}{5}x^{3}}\mathrm{d}x=-\frac{5}{3}\int-\frac{3}{5}x^{2}e^{-\frac{1}{5}x^{3}}\mathrm{d}x=-\frac{5}{3}e^{-\frac{1}{5}x^{3}}+C \]

-\[ \int\frac{1}{x(x^{2}-1)}\mathrm{d}x \]

The degree of the numerator is less than that of the denominator. The denominator has grade 3, so it has 3 roots \(x(x^{2}-1)=0\) if \(x=0,x=\pm1.\) We factor the denominator and rewrite the integrating as

\[ \frac{1}{x(x^{2}-1)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x+1)} \]

In such a way that:

\[ \frac{1}{x(x^{2}-1)}=\frac{A(x^{2}-1)+Bx(x+1)+Cx(x-1)}{x(x^{2}-1)} \]

So, having the same denominators, we get that:

\[ 1=A(x^{2}-1)+Bx(x+1)+Cx(x-1) \]

We must solve the equation to get \(A,B,C.\) We use the values of the roots as values for \(x\) and we get:

\[ \begin{cases} x=0 & 1=-A\\ x=1 & 1=2B\\ x=-1 & 1=2C \end{cases} \]

so \(A=-1,B=\frac{1}{2},C=\frac{1}{2}.\)

Then \[ \int\frac{-1}{x}dx+\int\frac{1}{2(x-1)}dx+\int\frac{1}{2(x+1)}dx=-ln(x)+\frac{1}{2}ln(x-1)+\frac{1}{2}ln(x+1)+C \]

  • \[ \int\frac{2x^{2}-3x+2}{x(x+5)(2x+1)}dx \]

Since the degree of the numerator is less than that of the denominator, we rewrite:

\[ \frac{2x^{2}-3x+2}{x(x+5)(2x+1)}=\frac{A}{x}+\frac{B}{(x+5)}+\frac{C}{(2x+1)} \]

Where the roots are \(x=0,-5,-\frac{1}{2}\). In this way, operating:

\[ \frac{2x^{2}-3x+2}{x(x+5)(2x+1)}=\frac{A(x+5)(2x+1)+Bx(2x+1)+Cx(x+5)}{x(x+5)(2x+1)} \]

then

\[ 2x^{2}-3x+2=A(x+5)(2x+1)+Bx(2x+1)+Cx(x+5) \]

Where, giving values at $$x, we can get:

\[ \begin{cases} x=0 & \frac{2}{5}=A\\ x=-5 & \frac{67}{45}=B\\ x=-\frac{1}{2} & -\frac{16}{9}=2C \end{cases} \]

So:

\[ \int\frac{2x^{2}-3x+2}{x(x+5)(2x+1)}dx=\frac{2}{5}\int\frac{1}{x}\mathrm{d}x+\frac{67}{45}\int\frac{1}{(x+5)}\mathrm{d}x-\frac{16}{9}\int\frac{1}{(2x+1)}\mathrm{d}x \]

  • \[ \int\frac{10}{(x-1)(x^{2}-9)}dx \]

Since the degree of the numerator is less than that of the denominator, we rewrite:

\[ \frac{10}{(x-1)(x^{2}-9)}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x-3)} \]

Where the ra?ces are \(x=1,-3,+3\). In this way, operating:

\[ \frac{10}{(x-1)(x^{2}-9)}=\frac{A(x+3)(x-3)+B(x-1)(x-3)+C(x-1)(x+3)}{(x-1)(x^{2}-9)} \]

then

\[ 10=A(x+3)(x-3)+B(x-1)(x-3)+C(x-1)(x+3)) \]

Where, giving values at $$x, we can get:

\[ \begin{cases} x=1 & -\frac{5}{4}=A\\ x=-3 & \frac{5}{12}=B\\ x=+3 & \frac{5}{6}=C \end{cases} \]

So:

\[ \int\frac{10}{(x-1)(x^{2}-9)}dx=-\frac{5}{4}\int\frac{1}{(x-1)}\mathrm{d}x+\frac{5}{12}\int\frac{1}{(x+3)}\mathrm{d}x+\frac{5}{6}\int\frac{1}{(x-3)}\mathrm{d}x \]

\[ \int\frac{10}{(x-1)(x^{2}-9)}dx=-\frac{5}{4}\ln(x-1)+\frac{5}{12}\ln(x+3)+\frac{5}{6}\ln(x-3)+C \]

  • \[ \int\frac{x-2}{x^{2}-9x+8}dx \]

Make sure you get:

\[ \int\frac{x-2}{x^{2}-9x+8}dx=\frac{1}{7}\ln(x-1)+\frac{6}{7}\ln(x-8)+C \]

-\[ \int\ln(x)\left(x^{3}+3x^{2}-9\right)\mathrm{d}x \]

Here it is clear, we take \(u=\ln(x)\), and therefore, \(\mathrm{d}v=\left(x^{3}+3x^{2}-9\right)\mathrm{d}x\). Then

\[ \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}\Rightarrow{\mathrm{d}u=\frac{\mathrm{d}x}{x}} \] then

\[ v=\int\left(x^{3}+3x^{2}-9\right)\mathrm{d}x\Rightarrow{\frac{x^{4}}{4}+x^{3}-9x} \]

So, using the expression \(\int u\mathrm{d}v\mathrm{d}x=uv-\int v\mathrm{d}u\)

\[ \int\ln(x)\left(x^{3}+3x^{2}-9\right)\mathrm{d}x=\ln(x)\left(\frac{x^{4}}{4}+x^{3}-9x\right)-{\int\left(\frac{x^{4}}{4}+x^{3}-9x\right)\frac{\mathrm{d}x}{x}} \]

We need to solve \[ \int\left(\frac{x^{4}}{4}+x^{3}-9x\right)\frac{\mathrm{d}x}{x}=\int\left(\frac{x^{3}}{4}+x^{2}-9\right)\mathrm{d}x=\frac{x^{4}}{16}+\frac{x^{3}}{3}-9x \]

So, finally

\[ \int\ln(x)\left(x^{3}+3x^{2}-9\right)\mathrm{d}x={\ln(x)\left(\frac{x^{4}}{4}+x^{3}-9x\right)-\left(\frac{x^{4}}{16}+\frac{x^{3}}{3}-9x\right)+C} \]

-\[ \int x\sqrt{7-x}\mathrm{d}x=\int x\left(7-x\right)^{1/2}\mathrm{d}x \]

It’s not so clear here. However, it seems sensible to take \(u=x\), and therefore, \(\mathrm{d}v=\left(7-x\right)^{1/2}\mathrm{d}x\). Yes You will take the functions upside down, you will be complicated when using the formula. Then

\[ \frac{\mathrm{d}u}{\mathrm{d}x}=1\Rightarrow{\mathrm{d}u=\mathrm{d}x} \]

\[ v=\int\left(7-x\right)^{1/2}\mathrm{d}x=-\int-\left(7-x\right)^{1/2}\mathrm{d}x\Rightarrow{\frac{-2(7-x)^{3/2}}{3}} \]

So, using the expression \(\int u\mathrm{d}v\mathrm{d}x=uv-\int v\mathrm{d}u\):

\[ \int x\left(7-x\right)^{1/2}\mathrm{d}x={ \frac{-2x(7-x)^{3/2}}{3}}-{\frac{2}{3}\int-(7-x)^{3/2}\mathrm{d}x} \]

We need to solve \[ {\frac{2}{3}\int-(7-x)^{3/2}\mathrm{d}x}=\frac{4}{15}\left(7-x\right)^{5/2} \]

So, finally

\[ \int x\left(7-x\right)^{1/2}\mathrm{d}x=\frac{-2x(7-x)^{3/2}}{3}-\frac{4}{15}\left(7-x\right)^{5/2}+C \]

-\[ \int xe^{-\frac{1}{4}x}\mathrm{d}x \]

In this case, again the P (powers, polynomials) prevails over the E (from exponential) so \(u=x\), and therefore, \(\mathrm{d}v=e^{-\frac{1}{4}x}\mathrm{d}x\). Then

\[ \frac{\mathrm{d}u}{\mathrm{d}x}=1\Rightarrow{\mathrm{d}u=\mathrm{d}x}, \]

in this way,

\[ v=\int e^{-\frac{1}{4}x}\mathrm{d}x=-4\int-\frac{1}{4}e^{-\frac{1}{4}x}\mathrm{d}x\Rightarrow-4e^{-\frac{1}{4}x} \]

So, using the expression \(\int u\mathrm{d}v\mathrm{d}x=uv-\int v\mathrm{d}u\):

\[ \int xe^{-\frac{1}{4}x}\mathrm{d}x={ -4xe^{\frac{-1}{4}x}}-{\int-4e^{-\frac{1}{4}x}\mathrm{d}x} \]

We need to solve \[ {\int-4e^{-\frac{1}{4}x}\mathrm{d}x}={4\times4\int-\frac{1}{4}e^{-\frac{1}{4}x}\mathrm{d}x}=16e^{-\frac{1}{4}x} \]

So, finally

\[ \int xe^{-\frac{1}{4}x}\mathrm{d}x={ -4xe^{\frac{-1}{4}x}-16e^{-\frac{1}{4}x}+C}=-4e^{\frac{-1}{4}x}\left(x-16\right)+C \]

EXTRA The Fundamental Theorem of Calculus: A More Detailed Explanation

Below we are going to break down one of the greatest successes of the calculation a couple of centuries ago. The story starts soft but, pay attention, it ends with a fight.

The Greeks

One of the problems that most interested in antiquity was the calculation of areas. (imagine that, in those times, they were very important the land, n o c o m o a h o r a). A way to calculate areas of certain non-obvious figures (such as a rectangle, a square or a triangle) was being exhaustive. That is, filling the figure in question with other figures whose area is known and more easy to get (rectangles, isn’t it?). In fact, the method of the exhaution consists of inscribing and circumscribing rectangulitos in the figure of this way:

FIG1. The idea of the Exhaution method with an area that we know well (in this case it will be \(1/2\))

In such a way that we know that the area we are looking for will be found between the figure on the left (with rectangles inscribed) and that of the right (with circumscribed rectangles). Of course, those registered approximate the area by “default” and the circumscribed, by “excess.” By Example, in this case, let’s put some numbers. Note that this figure, which is a right triangle, we can see it as a function \(f(x)=x,\:x\in[0,1]\), that is, the bisect of the first quadrant bounded on the axis of the \(x\) per the range \([0,1]\). So, as we have inscribed and circumscribed four rectangles, we are presented with the following (FIG 2)

FIG2. Pay attention to numeric values

We will have two “sums” that will lead to the approximation of the area that we are looking for. The lowest sum, \(s\), consisting of $(++)=$0.375, that is, the base of each rectangle is \(\frac{1}{4}\) and the values in parentheses are the different heights. On the other hand, the sum top, denoted with that capitalization, $S=(+++1)=$0.625. We have, therefore, a first approximation to the area of the triangle (which we already know is 0.5). Intuition tells us, then, that it will be reasonable to fill with many more rectangulites: all that Can. Let’s think now in a somewhat more abstract way: Let’s start inscribing and circumscribing \(n\) rectangles. So, the base of each rectangle will be \(\frac{1}{n}\).

FIG3. Partition for “\(n\)” rectangles

We can also deduct the lower and upper sums: if we we realize, the height of the lower sum will reach the height of the rectangle \(n-1\)-th and will be \(\frac{h-1}{n}\). So, the lower sum consist of

\[ s=\frac{\text{1}}{n}\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n-1}{n}\right) \]

while the upper one will reach the rectangle \(n-\)th and, therefore, its sum will be:

\[ S=\frac{1}{n}\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n}{n}\right) \]

Where we have put \(\frac{n}{n}\) to highlight the idea. If you give yourself account, we can rewrite the sums as

\[ \begin{cases} s=\frac{1}{n^{2}}\left(1+2+...+n-1\right)\\ S=\frac{1}{n^{2}}\left(1+2+....+n\right) \end{cases} \]

and there a very famous result that allows us to write the sum of the \(N\) first natural numbers (for example, \(1+2+3+...+N=\frac{N(N+1)}{2}).\) This leads us to:

\[ \begin{cases} s=\frac{1}{n^{2}}\left(\frac{\left[n-1\right]n}{2}\right)=\frac{n-1}{2n}\\ S=\frac{1}{n^{2}}\left(\frac{n\left[n+1\right]}{2}\right)=\frac{n+1}{2n} \end{cases} \]

So, choose how many rectangles you want and you will have an approximation of the area by excess and by default. In the following table, we do it for different partitions (value of \(n\))

\(n\) \(s\) \(S\)
4 \(0.375\) \(0.625\)
10 \(0.45\) \(0.55\)
100 \(0.495\) \(0.505\)
1000 \(0.4995\) \(0.5005\)

FIG4. Partition for “\(n\)” rectangles

as you can see, as it is done \(n\) as big as you want, both sums they look more and more alike. So much so that it is not difficult to calculate:

\[ s_{n\rightarrow\infty}=S_{n\rightarrow\infty}=\frac{1}{2}, \]

which is the area we were looking for. Not bad!

Riemman: a century ago

This German mathematician dedicated himself to shaping the method of exhaution and study its properties. He found that a bounded continuous (or continuous to pieces) function can be calculated using the exhaution method. What he did was to study the properties of the functions to which the area could be calculated with this method:

  • Continuous and upper bounded functions: that is, we must be able to fill with rectangulites and for this, we need a “top”

FIG5. Bounded function >- Continuous functions can also be integrated (in the Riemann sense) in pieces

FIG6. Continuous chunk function Why? As you can see, it can be filled perfectly with the “rectangulito”

FIG7. Continuous chunk function Be careful, because if it is not bounded:

FIG8. Continuous chunk function that is, this function:

\[ \begin{cases} f(x)=\frac{1}{x}\; \; if\;\; 0<x<5\\ f(x)=0 \; \; if \;\; x=0 \end{cases} \]

when we approach zero \((x\rightarrow0)\) we will not go to be able to put the rectangulito, so this function is not integrable according to Riemann.

FIG9. (something like that)

By the way, the function, however, would be integrable in any interval that does not contain zero. For example, \(f(x)=\frac{1}{x}\; if \; x\in(1.5)\).

Riemman, in short, proved that the area covered could be calculated. between the previous continuous (or continuous to pieces) and bounded functions and the axis \(x\) by the sums of inscribed rectangles (or trapezoids) or circumscribed in the figure that forms the function. This was called integral defined Riemann. It basically says that a function \(f(x)\) is Riemann integrable in an interval \((a,b)\) if the lower and upper sums converge in a single number when \(n\rightarrow\infty\) in the expression:

\[ \sum_{n=0}^{\infty}\triangle x_{n}f(x_{n})\;x\in[a,b] \]

and, for short, and not to write so much, it was decided to denote the area in this way. \[ \sum_{n=0}^{\infty}\triangle x_{n}f(x_{n})\;x\in[a,b]=\int_{a}^{b}f(x)\mathrm{d}x \]

Now who dares to calculate areas using Riemann sums?

XVII century: The great fight of calculus and the great theorem

On the other hand, and a century before the work of Riemann, Newton (UK) and Leibniz (Germany) were working on the big ideas of calculus that you have already studied (usually the derivative applied to problems physical). The fact is that, according to the story, both arrived to the same result. They realized that the calculus of the integral as an area it was a trivial task. Let’s see how:

The first thing is to define the “integral function.” If we have a function \(f\:[a,b]\rightarrow\mathbb{R}\), we define the integral function \(A(x)=\int_{a}^{x}f\) as the one that provides you with the area covered by the function and the abscissa axis between point \(a\) and any point \(x\in[a,b].\) That is, something like this, specifically, \(A(x)\):

FIG10. Thus we obtain the area of an integrable function in the Riemann sense \(A(x)=\int_{a}^{x}f\)

Let’s go a little further along the axis of the \(x\) and calculate another area, for example, \(A(x+\Delta x)\).

FIG10. \(A(x+\Delta x)=\int_{a}^{x+\Delta x}f\)

Now, look closely at FIG11, which we are going to analyze the differences between the two

FIG11. \(A(x+\Delta x)=\int_{a}^{x+\Delta x}f\)

The difference \(A(x+\Delta x)-A(x)\) is the yellow piece that, as you can see, we can approximate by a rectangle (as we already know). In fact, we can have one rectangle inscribed and another circumscribed. \[ \Delta xf(x)\leq A(x+\Delta x)-A(x)\leq\Delta xf(x+\Delta x) \]

Now, if we spend dividing \(\Delta x\), we will have to

\[ f(x)\leq\frac{A(x+\Delta x)-A(x)}{\Delta x}\leq f(x+\Delta x) \]

Now, as we know, we must do \(\Delta x\rightarrow0\) what leads us to

\[ f(x)\leq\lim_{\Delta x\rightarrow0}\frac{A(x+\Delta x)-A(x)}{\Delta x}\leq f(x) \]

and, as you know, the limit \(\lim_{\Delta x\rightarrow0}\frac{A(x+\Delta x)-A(x)}{\Delta x}\) defines a derivative : that of the function \(A\) at point \(x\). That is: important result:

\[ A'(x)=f(x). \]

Or, what is the same,

If \(f\) is continuous (or continuous to pieces) in \([a,b]\), then, \[ A(x)=\int_{a}^{x}f \]

is derivable and, in addition, \[ A'(x)=f(x) \]

As you will see, we have omitted for convenience in the integral \(\int_{a}^{x}f\) the term \(f{(x)\mathrm{d}x}\). Actually, it is not common to omit it (unless there is no doubt regarding which variable we are integrating). Of course, we have to draw attention to a matter of notation

EYE!

When in the integral function you have to put the value \(x\) in the upper limit, then, you must change the letter of the function (it is very rare, mathematically, that the same symbol acts as a number and a variable). So, the Fundamental Theorem of Calculus also says:

\[ \frac{\mathrm{d}}{\mathrm{d}x}\int_{a}^{x}f(t)dt=f(x) \]

This result shows, therefore, why we say to calculate a Integral consists of looking for a primitive. Something you’ve done before mechanically. It is the link between the Greeks and mathematics of the nineteenth century. We have one last step left. As we already know, a primitive \(F(x)\) satisfies that:

\[ F(x)=\int_{a}^{x}f(t)\mathrm{d}t+C \]

for \(x\in[a,b].\) In fact, notice that \[ F(a)={\color{red}\int_{\color{red}a}^{\color{red}a}f(t)\mathrm{d}t}+C\equiv{\color{red}0}+C \]

since at one point we have no area. This tells us that \(F(a)=C\), so

\[ F(x)=\int_{a}^{x}f(t)\mathrm{d}t+C\Rightarrow F(x)=\int_{a}^{x}f(t)\mathrm{d}t+F(a) \]

in such a way that, surprise!

\[ F(x)-F(a)=\int_{a}^{x}f(t)\mathrm{d}t \]

and, particularizing in a specific interval \(x\in[a,b]\)

\[ F(b)-F(a)=\int_{a}^{b}f(t)\mathrm{d}t, \]

This technique is known as “Barrow’s rule” and tells us that solving a definite integral is very easy: only you have to look for the primitive and evaluate it at the extremes of the limits of integration and, finally, subtract them.

Reaching this result was not easy and was surrounded by controversy. Both Newton and Leibniz managed to reach similar conclusions. on similar dates. Well, it seems that Newton arrived earlier, but without having published it. Leibniz, however, “discovered” it somewhat. later but published it successfully. They were very quarreling (something which affected, in fact, the relations between the scientists of the Kingdom United and the European continent) and, to this day, there is evidence in favor that both arrived – each on their own – at the same result. Newton he is buried in Westmister Abbey, with full honours. However, Leibniz died abandoned. Although today its notation, since he knew how to give the language of calculation elegance and conciseness. Of course, Newton was granted the appointment of Sir.