Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

Objectives

When you have completed this unit, you should be able to

1. find antiderivative functions.
2. evaluate indefinite integrals.
3. evaluate definite integrals.
4. state the Mean Value Theorem and apply it to prove the Fundamental Theorem of Calculus.
5. explain and use the inverse relationship between differentiation and integration when applying the Fundamental Theorem of Calculus.

Integration is the inverse of the operation of differentiation, and as does the derivative, the integral has two different interpretations. In this unit, we consider the process of integration, and in Unit 7, we study the applications of integration that arise from these two different interpretations.

Antiderivatives

In this section, we do the opposite of differentiation; that is, given a function $f(x)$, we want to find a function $F(x)$ such that ${\displaystyle F^{\prime }\left(x\right)=f\left(x\right)\text{.}}$

Definition 6.1. A function $F(x)$ is the antiderivative of a function $f(x)$ defined on an interval $J$ if ${\displaystyle F^{\prime }\left(x\right)=f\left(x\right)}$ for all $x$ in the interval ${\displaystyle J\text{.}}$

Note: Observe that the antiderivative of the function $\color{#384877}{\! f'(x)}$ is the function $\color{#384877}{\! f(x)}$, and the antiderivative of $\color{#384877}{\! f''(x)}$ is the function $\color{#384877}{\! f'(x)}$.

Since we know how to differentiate, we can use trial and error to find antiderivatives of several functions.

Example 6.1.

1. The antiderivative of a zero constant function ${\displaystyle f\left(x\right)=0}$ is any constant function ${\displaystyle F\left(x\right)=C}$, because ${\displaystyle F^{\prime }\left(x\right)=0=f\left(x\right)\text{.}}$ Later, we prove that $F(x)$ is the only antiderivative of the zero constant function. That is, if the derivative of a function $F$ is zero, then ${\displaystyle F\left(x\right)=C}$ for some constant ${\displaystyle C\text{.}}$

   Since $C$ is any constant, this example shows that the antiderivative is not unique, there are many antiderivatives for a given function.

2. The antiderivative of any constant function ${\displaystyle f\left(x\right)=m}$ is the function ${\displaystyle F\left(x\right)=mx}$, because ${\displaystyle F^{\prime }\left(x\right)=m=f\left(x\right)}$, but it also the function ${\displaystyle F_1\left(x\right)=mx+6}$, because its derivative is equal to $m$, as is the derivative ${\displaystyle F_2\left(x\right)=mx-123\text{.}}$ As before, we have many antiderivatives, but the only difference among them is a constant. [We prove this statement later.] Hence, the most general antiderivative of ${\displaystyle f\left(x\right)=m}$ is the function ${\displaystyle F\left(x\right)=mx+C}$, for any constant $C$.

We use rules of differentiation to find derivative functions, but there are no rules for finding antiderivatives. Instead, we must rely on what we know about differentiation, paying attention to particular cases, to generalize whenever possible.

Example 6.2.

1. The antiderivative of the function ${\displaystyle f\left(x\right)=2x}$ is a function $F(x)$ such that ${\displaystyle F^{\prime }\left(x\right)=2x\text{.}}$

   If we make a search among all the functions whose derivatives we know, we realize that the derivative of ${\displaystyle F\left(x\right)=x^2}$ is ${\displaystyle f\left(x\right)=2x\text{.}}$

   Hence, the most general antiderivative of ${\displaystyle f\left(x\right)=2x}$ is ${\displaystyle F\left(x\right)=x^2+C\text{.}}$

2. The antiderivative of ${\displaystyle f\left(x\right)=x}$ is a function $F(x)$ such that ${\displaystyle F^{\prime }\left(x\right)=x\text{.}}$

   The derivative of ${\displaystyle F\left(x\right)=x^2}$ is ${\displaystyle 2x}$, which is not $x$, but is close.

   To obtain $x$, we can consider the function $F_1(x)={\displaystyle \frac{1}{2}}{x}^2$; its derivative is ${\displaystyle f\left(x\right)=x\text{.}}$ Hence, the most general antiderivative is ${\displaystyle F_1\left(x\right)={\displaystyle \frac{1}{2}}{x}^2+C\text{.}}$

3. What would be the antiderivative of ${\displaystyle f\left(x\right)=x^2}$? We know that the derivative of ${\displaystyle F\left(x\right)=x^3}$ is ${\displaystyle 3x^2}$, again close, and we must consider the function $F_2(x)={\displaystyle {\displaystyle \frac{1}{3}}}{x}^3+C$ as the most general antiderivative.

We can make a general statement based on these examples. We see that the (general) antiderivative of the function ${\displaystyle f\left(x\right)=x^r}$ is the function

${\displaystyle F\left(x\right)={\displaystyle \frac{x^{r+1}}{r+1}}+C\text{.}}$

Let us check this conclusion with differentiation:

${\displaystyle \frac{d}{dx}}{x}^{r+1}=\left(r+1\right)x^r$   so   ${\displaystyle \frac{d}{dx}}{\displaystyle \frac{x^{r+1}}{r+1}}=x^r\text{.}$

However, this statement is not true if ${\displaystyle r=-1\text{.}}$ [Why?]

So our final conclusion is that the (general) antiderivative of ${\displaystyle f\left(x\right)=x^r}$ for any ${\displaystyle r\ne -1}$ is the function

${\displaystyle F\left(x\right)={\displaystyle \frac{x^{r+1}}{r+1}}+C\text{.}}$

Note: The antiderivative of the function $\color{#384877}{\! f(x) = \dfrac{1}{x}}$ is not covered in this course.

Example 6.3. We know that the derivative of the sine function is the cosine function; hence, the antiderivative of the cosine function is the sine function. Similarly the derivative of the cosine function is the negative sine function; hence, the antiderivative of the sine function is the negative cosine function. Moreover, we know that

${\displaystyle \frac{d}{dx}}\sin \left(ax\right)=a\cos \left(ax\right)\text{.}$

Hence,

${\displaystyle \frac{d}{dx}}{\displaystyle \frac{\sin \left(ax\right)}{a}}=\cos \left(ax\right)$.

We conclude that the antiderivative of the function $f\left(x\right)=\cos \left(ax\right)$ is the function

$F\left(x\right)={\displaystyle \frac{\sin \left(ax\right)}{a}}+C$.

You can use differentiation to see that the antiderivative of the function $g(x)=\sin (ax)$ is the function

$G\left(x\right)=-{\displaystyle \frac{\cos \left(ax\right)}{a}}+C$.

Note: In what follows, by “antiderivative,” we mean the most general antiderivative. The letter $\color{#384877}{C}$ will be any constant.

We know that

${\displaystyle \frac{d}{dx}}kf\left(x\right)=k{\displaystyle \frac{d}{dx}}f\left(x\right)$;

therefore, the antiderivative of the function ${\displaystyle g\left(x\right)=kf\left(x\right)}$ is ${\displaystyle G\left(x\right)=kF\left(x\right)}$, where $F(x)$ is the antiderivative of the function ${\displaystyle f\left(x\right)\text{.}}$

We also know that

${\displaystyle {\displaystyle \frac{d}{dx}}f\left(x\right)+g\left(x\right)=f^{\prime }\left(x\right)+g^{\prime }\left(x\right);}$

therefore, the antiderivative of the function ${\displaystyle h\left(x\right)=f\left(x\right)+g\left(x\right)}$ is the function ${\displaystyle H\left(x\right)=F\left(x\right)+G\left(x\right)}$, where ${\displaystyle F^{\prime }\left(x\right)=f\left(x\right)}$ and ${\displaystyle G^{\prime }\left(x\right)=g\left(x\right)\text{.}}$

Example 6.4. To find the antiderivative of the function

${\displaystyle f\left(x\right)={\displaystyle \frac{6x^3-2x+6}{\sqrt{x}}},}$

we simplify first:

${\displaystyle f\left(x\right)=6x^{5∕2}-2x^{1∕2}+6x^{-1∕2}\text{.}}$

We then apply our general conclusion from Example 6.2, and we find that the antiderivative is

$F\left(x\right)=6\left({\displaystyle \frac{x^{7/2}}{7/2}}\right)-2\left({\displaystyle \frac{x^{3/2}}{3/2}}\right)+6\left({\displaystyle \frac{x^{1/2}}{1/2}}\right)$.

So,

${\displaystyle F\left(x\right)={\displaystyle \frac{12x^{7∕2}}{7}}-{\displaystyle \frac{4x^{3∕2}}{3}}+12\sqrt{x}+C\text{.}}$

To make a unique determination of the constant in a general antiderivative, we need to know at least one exact value of the antiderivative function. These values are referred to as “extra conditions.”

Example 6.5. If ${\displaystyle f^{\prime }\left(x\right)=\sin \left(3x\right)+x^4}$ and we know that ${\displaystyle f\left(0\right)=-1}$, then we can find that the antiderivative is

${\displaystyle F\left(x\right)=-{\displaystyle \frac{\cos \left(3x\right)}{3}}+{\displaystyle \frac{x^5}{5}}+C\text{.}}$

Since ${\displaystyle -1=f\left(0\right)=-{\displaystyle \frac{1}{3}}+C,}$ we conclude that ${\displaystyle C=-{\displaystyle \frac{2}{3}}\text{.}}$

So, the constant is determined, and the antiderivative is

${\displaystyle f\left(x\right)=F\left(x\right)=-{\displaystyle \frac{\cos \left(3x\right)}{3}}+{\displaystyle \frac{x^5}{5}}-{\displaystyle \frac{2}{3}}\text{.}}$

Example 6.6. If ${\displaystyle f^{″}\left(x\right)=\cos \left(4x\right)-6\sqrt{x}-1}$, ${\displaystyle f^{\prime }\left(0\right)=1}$ and ${\displaystyle f\left(0\right)=-1}$, what is ${\displaystyle f\left(x\right)\text{?}}$

The antiderivative of the function ${\displaystyle f^{″}}$ is ${\displaystyle f^{\prime }\text{.}}$ So,

${\displaystyle f^{\prime }\left(x\right)={\displaystyle \frac{\sin \left(4x\right)}{4}}-{\displaystyle \frac{12x^{3∕2}}{3}}-x+C\text{.}}$

Since ${\displaystyle 1=f^{\prime }\left(0\right)=C}$,

${\displaystyle f^{\prime }\left(x\right)={\displaystyle \frac{\sin \left(4x\right)}{4}}-4x^{3∕2}-x+1\text{.}}$

The antiderivative of ${\displaystyle f^{\prime }}$ is $f$. So,

${\displaystyle f\left(x\right)=-{\displaystyle \frac{\cos \left(4x\right)}{16}}-{\displaystyle \frac{8x^{5∕2}}{5}}-{\displaystyle \frac{x^2}{2}}+x+C\text{.}}$

Since ${\displaystyle -1=f\left(0\right)=-{\displaystyle \frac{1}{16}}+C\text{,}}$ we conclude that ${\displaystyle C=-{\displaystyle \frac{15}{16}}\text{,}}$ and

${\displaystyle f\left(x\right)=-{\displaystyle \frac{\cos \left(4x\right)}{16}}-{\displaystyle \frac{8x^{5∕2}}{5}}-{\displaystyle \frac{x^2}{2}}+x-{\displaystyle \frac{15}{16}}\text{.}}$

If $F(x)$ is the antiderivative of ${\displaystyle f\left(x\right)\text{,}}$ then we write

${\displaystyle \int f\left(x\right)dx=F\left(x\right)+C}$

and we say that the indefinite integral of the integrand function $f(x)$ is the function ${\displaystyle F\left(x\right)\text{.}}$ We justify this notation later. The constant $C$ is called the constant of integration. See Figure 6.1, below.

Figure 6.1: The antiderivative function

With this notation, we can state that

(6.1) ${\displaystyle \int x^{r}dx={\displaystyle \frac{x^{r+1}}{r+1}}+C\text{ for }r\ne -1\text{.}}$

(6.2) ${\displaystyle \int \sin \left(ax\right)dx=-{\displaystyle \frac{\cos \left(ax\right)}{a}}+C\text{ for }a\ne 0\text{.}}$

(6.3) ${\displaystyle \int \cos \left(ax\right)dx={\displaystyle \frac{\sin \left(ax\right)}{a}}+C\text{ for }a\ne 0\text{.}}$

(6.4) ${\displaystyle \int kf\left(x\right)dx=k\int f\left(x\right)dx\text{.}}$

(6.5) ${\displaystyle \int f\left(x\right)+g\left(x\right)=\int f\left(x\right)dx+\int g\left(x\right)dx\text{.}}$

A table of basic indefinite integrals is given on page 273 of the textbook. You must know all of them. Use what you know about differentiation to help you remember them.

Example 6.7. From Example 6.4 we know that

${\displaystyle \int {\displaystyle \frac{6x^3-2x+6}{\sqrt{x}}}dx={\displaystyle \frac{12x^{7∕2}}{7}}-{\displaystyle \frac{4x^{3∕2}}{3}}+12\sqrt{x}+C\text{.}}$

The process of finding the antiderivative of a function is called integration. Hence, when you are asked to integrate, what you need to do is to find the antiderivative of the integrand function.

Example 6.8. To see if the indefinite integral

${\displaystyle \int {\displaystyle \frac{1}{{\left(x^2+4\right)}^{3∕2}}}dx={\displaystyle \frac{x}{\sqrt{x^2+4}}}+C}$

is correct, we differentiate, and we see that

${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{x}{\sqrt{x^2+4}}}={\displaystyle \frac{4}{{\left(x^2+4\right)}^{3∕2}}}\text{.}}$

So, the antiderivative is not correct; instead, it should be

$F\left(x\right)={\displaystyle \frac{x}{4\sqrt{x^2+4}}}$.

To use the basic indefinite integrals, we must identify the given integrand function with the integrand function of one of the basic integrals. Algebraic operations and trigonometric identities are used to achieve this identification.

Example 6.9. As it stands, the integrand function in the indefinite integral

${\displaystyle \int {\left(x+{\displaystyle \frac{4}{x}}\right)}^{2}dx}$

does not fit any of the types of the integrand functions of the basic integrals. But, if we multiply out, we get

${\displaystyle \int x^2+8+16x^{-2}dx\text{.}}$

This integrand function fits the basic integral (see Equation 6.1), and we have

${\displaystyle \int x^2+8+16x^{-2}dx={\displaystyle \frac{x^3}{3}}+8x-{\displaystyle \frac{16}{x}}+C\text{.}}$

Example 6.10. The integrand function in the indefinite integral

${\displaystyle \int {\displaystyle \frac{\sin \left(2x\right)}{\cos x}}dx}$

does not fit any of the integrand functions of the basic integrals. But, if we use a trigonometric identity, we find that

${\displaystyle \frac{\sin \left(2x\right)}{\cos x}}={\displaystyle \frac{2\sin x\cos x}{\cos x}}=2\sin x$.

Hence, we can apply Equation 6.2 with ${\displaystyle a=1\text{:}}$

${\displaystyle \int {\displaystyle \frac{\sin \left(2x\right)}{\cos x}}dx=2\int \sin xdx=-2\cos x+C\text{.}}$

Example 6.11. To find the antiderivative of $\sin^2 x$ we use the double angle formula and we apply Equation 6.3, with $a = 2$:

\begin{align*} \int \sin^2 x \, dx & = \int \dfrac{1 - \cos(2x)}{2}\,dx \\ & = \dfrac{1}{2} \int 1 - \cos(2x) \,dx \\ & = \dfrac{1}{2}\left( x - \dfrac{\sin(2x)}{2}\right) + C \\ & = \dfrac{2x - \sin(2x)}{4} + C \end{align*}