13. Indefinite integrals - Mathematics for Business

13.1Introduction¶

The central problem of this part of the course is the following: how to calculate the area under the graph of a function?

In order to give a full answer to this question, we need to understand the following two questions:

why would we care about the area under a graph of a function?
how to calculate the area under the graph of a function?

The following chart illustrates the way in which we will answer these questions.

This entire chapter is devoted to the second question - what new tools do we need in order to be able to find the area under the graph of a function? The answer is given in the chart above - we need to know how to compute indefinite integrals.

So, in some sense integration is the opposite operation of differentiation - given some function $f$ , we want to find a function $F$ whose derivative is exactly the function $f$ .

Remark

Suppose that $F$ is an integral of the function $f$ . By definition, that means that

F'(x) = f(x).

But also the following identities hold:

\begin{split} (F(x) + 2)' &= F'(x) + 2' = f(x) + 0 = f(x) \\ (F(x) + 17)' &= F'(x) + 17' = f(x) + 0 = f(x) \\ (F(x) + 1013)' &= F'(x) + 1013' = f(x) + 0 = f(x) \end{split}

What we can see from this is that adding a constant to the integral of the function $f$ does not change the result, since the derivative of a constant is equal to zero. This is indicated by “ $+C$ ” that you will see in the formulas in the following section, where that “ $+C$ ” simply means that we can add any constant to an integral without affecting the end result.

13.2Direct integration¶

Basic rules for integration

\begin{split} \int f + g \, dx &= \int f \, dx + \int g \, dx \\ \int c \cdot f \, dx &= c \cdot \int f \, dx \end{split}

Table of integrals

\begin{split} \int c \, dx &= cx + C, \quad \quad \int x^n \, dx = \frac{1}{n+1}\cdot x^{n+1} + C \\ \int a^x \, dx &= \frac{1}{\ln(a)}\cdot a^x + C, \quad \quad \int e^x \, dx = e^x + C, \quad \quad \int \frac{1}{x} \, dx = \ln(x) + C \end{split}

13.3Substitution method¶

As it was with the derivative, not all integrals we would like to know how to compute can be calculated using just the most basic formulas for integration, so we need a new method that will allow us to find integrals of more complicated functions. One such method is called substitution method.

Solution to Problem 13.5

We use the substitution methond in order to compute this integral:

\begin{split} \int \frac{2x-2}{x^2-2x+9} \, dx &= \left\{ \begin{align*} t &= x^2 -2x + 9 \\ dt &= (2x-2)dx \implies dx = \frac{dt}{2x-2} \end{align*} \right\} \\ &= \int \frac{2x-2}{t} \cdot \frac{dt}{2x-2} \\ &= \int \frac{1}{t} \, dt \\ &= \ln(t) \\ &= \ln(x^2-2x+9) + C. \end{split}

Solution to Problem 13.8

We use the substitution method in order to compute this integral:

\begin{split} \int \frac{1}{x \cdot \ln(x)} \, dx &= \left\{ \begin{align*} t &= \ln(x) \\ dt &= \frac{1}{x} \, dx \implies dx = x \cdot dt \end{align*} \right\} \\ &= \int \frac{1}{x \cdot t} \cdot x \, dt \\ &= \int \frac{1}{t} \, dt \\ &= \ln(t) \\ &= \ln(\ln(x)) + C. \end{split}

13.4Integration by parts¶

As we will see in this section, not even relying solely on the substituion method will be enough to compute all the integrals we will need. Therefore, we are introducing yet another method of integration that we will use in order to find integrals - integration by parts.

Solution to Problem 13.9

We use the formula for the integration by parts:

\begin{split} \int (x+5) \cdot e^{x} \, dx &= \left\{ \begin{align*} u &= x+5 &\implies du &= dx \\ dv &= e^x \, dx &\implies v&= \int dv = \int e^x \, dx = e^x \end{align*} \right\} \\ &= (x+5) \cdot e^x - \int e^{x} \, dx \\ &= (x+5) \cdot e^x - e^x + C. \end{split}

Lastly, what would’ve happend if we reversed the roles of $u$ and $dv$ when applying the formula for integration by parts? Well, we would’ve got something more complicated than we started with - namely, the following:

\begin{split} \int (x+5) \cdot e^x \, dx &= \left\{ \begin{align*} u &= e^x &\implies du &= e^x \, dx\\ dv &= (x+5) \, dx &\implies v &= \int \, dv = \int (x+5) \, dx = \frac{1}{2} \cdot (x+5)^2 \end{align*} \right\} \\ &= \frac{1}{2}(x+5)^2 \cdot e^x - \int \frac{1}{2}(x+5)^2 \cdot e^x \, dx. \end{split}

Notice that with this choice we end up with having to calculate something more complicated than we originally had, so that’s how you know whether or not you have made the correct choice when deciding what will be $u$ and what will be $dv$ . Remember: the main point is get something simpler!

Solution to Problem 13.11

First of all, because the ingral of a sum is the sum of the integrals, we have

\int \frac{x}{e^x} -3x^2 \, dx = \int \frac{x}{x^x} \, dx - \int 3x^2 \, dx.

So, in order to compute the integral in this question we simply need to compute the two integrals above. The second integral is easy, since we can use the formula given in the table of integrals:

\int 3x^2 \, dx = x^3.

As for the first one, we use the integration by parts:

\begin{split} \int \frac{x}{e^x} \, dx &= \int x \cdot e^{-x} \, dx \\ &= \left\{ \begin{align*} u &= x &\implies du &= dx\\ dv &= e^{-x} \, dx &\implies v &= \int e^{-x} \, dx \, = -e^{-x} \end{align*} \right\} \\ &= -x \cdot e^{-x} - \int -e^{-x} \, dx \\ &= -x \cdot e^{-x} + \int e^{-x} \, dx \\ &= -x \cdot e^{-x} -e^{-x} + C \end{split}

Part 3. Functions of several variables

12. Optimization of functions of two variables

Part 4. Integrals

14. Definite integrals and area