Computing integrals (1): Anti-derivative

 So far, our discussion on the Riemann integral has been rather abstract. We know its definition and some of its properties. However, we still don't know how to calculate its value for a specific function. There are various techniques to compute integrals. We start from the most basic method based on the notion of anti-derivatives or primitive functions.

Definition (Anti-derivative, primitive function)

For a function $f(x)$ on an open interval $I$, a differentiable function $F(x)$ on $I$ is said to be an anti-derivative or primitive function of $f(x)$ (on $I$) if $F^{\prime }(x)=f(x)$ holds.

Remark.   If $F(x)$ is an anti-derivative of $f(x)$, then for any constant $C$, $F(x)+C$ is also an anti-derivative of $f(x)$. The following lemma shows these are the only anti-derivatives. □

Lemma

Let $f(x)$ be a function on an open interval $I$ with its anti-derivative $F(x)$. Then any anti-derivative of $f(x)$ is given as $F(x)+C$ where $C$ is a constant.

Proof. Suppose $G(x)$ is also an anti-derivative of $f(x)$. Then

$$(G(x)-F(x){)}^{\prime }=f(x)-f(x)=0$$

for all $x\in I$. Therefore $G(x)-F(x)=C$ (constant). ■

Remark. The collection of all anti-derivatives of $f(x)$ is called the indefinite integral of $f(x)$, denoted $\int f(x)dx$. The constant term in an anti-derivative is called the constant of integration. We most often omit the constant of integration. □

Theorem

Let $f(x)$ be a continuous function on an open interval $I$. Then $f(x)$ has an anti-derivative on $I$. More precisely, with an arbitrary constant $a\in I$, $F(x)={\int }_a^{x}f(t)dt$ is an anti-derivative of $f(x)$.

Proof. Trivial from the theorem that continuous functions are integrable and the fundamental theorem of calculus. ■

Remark. In this theorem, we could choose another constant $b\in I$ to define another anti-derivative $G(x)={\int }_b^{x}f(t)dt$. In fact,

$$G(x)={\int }_b^{x}f(t)dt={\int }_b^{a}f(t)dt+{\int }_a^{x}f(t)dt=F(x)+C$$

where $C={\int }_b^{a}f(t)dt$ is a constant. □

The following corollary shows that we can compute the integral of a function from its anti-derivative.

Corollary

Let $f(x)$ be a continuous function on an open interval $I$ and $F(x)$ be an anti-derivative of $f(x)$. For any $a,b\in I$, we have

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

Proof. Since we can write $F(x)={\int }_a^{x}f(t)dt+C$ where $C$ is a constant,

$$F(b)-F(a)=({\int }_a^{b}f(t)dt+C)-({\int }_a^{a}f(t)dt+C)={\int }_a^{b}f(t)dt.$$ ■

Example. Let $F(x)=x^2$ and $f(x)=2x$ Then, $F(x)$ is an anti-derivative of $f(x)$ because $F^{\prime }(x)=(x^2{)}^{\prime }=2x=f(x)$. Thus, for example,

$${\int }_1^{2}2xdx=2^2-1^2=3.$$ □

Example. Let $f(x)=\cos x$. Then, its anti-derivative is $F(x)=\sin x$. Thus, for example,

$${\int }_0^{\frac{\pi }{2}}\cos xdx=\sin \frac{\pi }{2}-\sin 0=1.$$

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