Integrals or Antiderivatives

We have learned how to find the derivative $$F'(x)$$ of a given function $$F(x)$$ in a variety of cases, including all those of the commonest occurrence. The operation is indicated by $\frac{d}{dx}F(x)=F'(x),$ or, if we are using differentials, by $dF(x)=F'(x)dx.\tag{a}$

The next step is to consider the inverse operation, namely: to find a function $$F(x)$$ whose derivative $F'(x)=f(x)$ is given. The function $$F(x)$$ is called an integral, or an integral function, or an antiderivative of $$f(x)$$, the process of finding it is called integration, and the operation is indicated by writing the integral sign $$\int$$ in front of the given differential expression: $\int f(x)dx=F(x),\tag{b}$ read an integral of $$f(x)dx$$ equals $$F(x)$$. The differential $$dx$$ indicates that $$x$$ is the variable of integration.

Now some examples
1. If $$F(x)=x^{3}$$, then $F'(x)dx=\left(\frac{dx^{3}}{dx}\right)dx=3x^{2}dx,$ thus $\int3x^{2}dx=x^{3}.$
2. If $$F(x)=\sin x$$, then $F'(x)dx=\cos x\ dx,$ thus $\int\cos xdx=\sin x.$
3. If $$F(x)=\cos x,$$ then $F'(x)dx=-\sin xdx$ thus $\int-\sin xdx=\cos x$
4. If $$F(x)=\arctan x$$ [Recall that $$\arctan x$$ is sometimes denoted by $$\tan^{-1}x$$], then $F'(x)dx=\frac{1}{1+x^{2}}dx,$ thus $\int\frac{1}{1+x^{2}}dx=\arctan x.$

Definition 1. If $$f(x)$$ is the derivative of $$F(x)$$, $$F'(x)=f(x)$$, the function $$F(x)$$ is called an integral of $$f(x)$$, and the operation of forming $$F(x)$$ from $$f(x)$$ is called integration. We shall use the notation $F(x)=\int f(x)dx.\tag{c}$ $$f(x)$$ is called the integrand or subject of integration.

• Differentiation and integration are inverse operations: $d\int f(x)dx=dF(x)=F'(x)dx=f(x)dx.\tag{d}$ Conversely, substituting the value of $$dF(x)=F'(x)dx$$ from (a) into (b), we get $\int dF(x)=F(x).\tag{e}$ Therefore, considered as symbols of operation, $$\frac{d}{dx}$$ and $$\int\cdots dx$$ are inverse to each other, or if we are using differentials $$d$$ and $$\int$$ are inverse to each other. When $$d$$ is followed by $$\int$$, they annul each other, as in (d), but when $$\int$$ is followed by $$d$$, as in (e), that will not, in general, be the case unless we ignore the constant of integration. We will see the reason for this in the following.

Constant of Integration

Let’s go back to the examples of the proceeding section.

Because $$d(x^{3})=3x^{2}dx$$, we have $$\int3x^{2}dx=x^{3}$$.

Because $$d(x^{3}+1)=3x^{2}dx,$$ we have $$\int3x^{2}dx=x^{3}+1$$.

Because $$d(x^{3}-5)=3x^{2}dx,$$ we have $$\int3x^{2}dx=x^{3}-5$$.

In general, because $$d(x^{3}+C)=3x^{2}dx,$$ where $$C$$ is an arbitrary number, we have $\int3x^{2}dx=x^{3}+C.$ Here $$C$$ is independent of $$x$$ and is called the constant of integration. Because we can give $$C$$ as many values as we please, it follows that if a function has one integral function (antiderivative), it has infinitely many integral functions (antiderivatives) differing only by constants. Hence $\int F'(x)dx=F(x)+C.$ Because $$C$$ is unknown and indefinite, the expression $$F(x)+C$$ is called an indefinite integral of $$F'(x)$$. From this discussion, we conclude that if $$F(x)$$ is a function whose derivative is $$f(x)$$, then $$F(x)+C$$ where $$C$$ is a constant is likewise a function whose derivative is $$f(x)$$. Hence

Theorem 1. If two functions differ by a constant, they have the same derivative.

It is, however, not obvious that if $$F(x)$$ is a function whose derivative is $$f(x)$$, then all functions having the same derivative $$f(x)$$ are of the form $$F(x)+C$$, where $$C$$ is a constant. In other words, there remains to be proved the converse theorem.

Theorem 2. Let $$F(x)$$ and $$G(x)$$ be two functions such that $F'(x)=G'(x),$ on a certain interval, then there exist a constant $$C$$ such that $F(x)=G(x)+C,$ for all $$x$$ in the interval.

Show the proof …

Let $$\phi(x)=F(x)-G(x)$$. Then \begin{aligned} \phi'(x) & =\frac{d}{dx}[F(x)-G(x)]\\ & =F'(x)-G'(x)=0.\end{aligned} It follows that $$\phi(x)$$ must have a constant value. The reasoning is: If the rate of a function is always zero, it does not change at all, so it must be constant. Equivalently, if every tangent line to the graph of a function is horizontal, then the graph of the function cannot go up or down and therefore must be a horizontal line.

To prove this theorem rigorously, consider two arbitrary numbers $$x_{1}$$ and $$x_{2}$$ in the given interval. From the Mean Value Theorem, we have $\phi(x_{2})-\phi(x_{1})=\phi'(c)(x_{2}-x_{1})$ for some $$c$$ between $$x_{1}$$ and $$x_{2}$$. Because $$\phi'(x)=0$$ for all values of $$x$$, $\phi'(c)=0\Rightarrow\phi(x_{2})=\phi(x_{1}).$ Because $$x_{1}$$ and $$x_{2}$$ are arbitrary, the above calculations shows that $$\phi(x)=F(x)-G(x)$$ is a constant function $$\phi(x)=C$$. That is, $$F(x)$$ and $$G(x)$$ differ only by a constant: $F(x)=G(x)+C.$

• Although the formulas $\int3x^{2}dx=x^{3}\quad\text{and}\quad\int3x^{2}dx=x^{3}+C,$ are both correct, the first one gives one integral of $$x^{2}$$, the second one gives all possible integrals of $$3x^{2}$$.
• The above theorem depends on the domain of definition being an interval. For example, let $F(x)=\begin{cases} x^{2} & \text{if }-1<x<1\\ x^{2}-1 & \text{if }2<x<5 \end{cases}$ and $G(x)=\begin{cases} x^{2}-3 & \text{if }-1<x<1\\ x^{2}+12 & \text{if }2<x<5 \end{cases}.$ Then $$F'(x)=G'(x)=2x$$ for all $$x\in(-1,1)\cup(2,5)$$, but $$F(x)$$ is not of the form $$G(x)+C$$.

In any given case, the value of $$C$$ can be found when we know the value of the integral for some value of the variable, and this will be discussed later. For the present, we content ourselves with first learning how to find the indefinite integrals of given functions.

• In what follows, we assume that every continuous function has an indefinite integral, a statement the rigorous proof of which is beyond the scope of this book.