5.5: Antiderivatives (Primitives, Integrals)

Given $f : E^{1} \rightarrow E,$ we often have to find a function $F$ such that $F^{\prime}=f$ on $I$, or at least on $I-Q.$ We also require $F$ to be relatively continuous and finite on $I.$ This process is called antidifferentiation or integration.

Definition 1

We call $F : E^{1} \rightarrow E$ a primitive, or antiderivative, or an indefinite integral, of $f$ on $I$ iff

(i) $F$ is relatively continuous and finite on $I,$ and

(ii) $F$ is differentiable, with $F^{\prime}=f,$ on $I-Q$ at least.

We then write

$$F=\int f, \text { or } F(x)=\int f(x) dx, \text { on } I.$$

(The latter is classical notation.)

If such an $F$ exists (which is not always the case), we shall say that $\int f$ exists on $I,$ or that $f$ has a primitive (or antiderivative) on $I,$ or that $f$ is primitively integrable (briefly integrable) on $I$.

If $F^{\prime}=f$ on a set $B \subseteq I,$ we say that $\int f$ is exact on $B$ and call $F$ an exact primitive on $B.$ Thus if $Q=\emptyset, \int f$ is exact on all of $I.$

Note 1. Clearly, if $F^{\prime}=f,$ then also $(F+c)^{\prime}=f$ for a finite constant c. Thus the notation $F=\int f$ is rather incomplete; it means that $F$ is one of many primitives. We now show that all of them have the form $F+c$ (or
$\int f+c ).$

Theorem $\PageIndex{1}$

If $F$ and G are primitive to $f$ on $I$, then $G-F$ is constant on $I$.

Proof

By assumption, $F$ and $G$ are relatively continuous and finite on $I$; hence so is $G-F.$ Also, $F^{\prime}=f$ on $I-Q$ and $G^{\prime}=f$ on $I-P. (Q$ and $P$ are countable, but possibly $Q \neq P. )$

Hence both $F^{\prime}$ and $G^{\prime}$ equal $f$ on $I-S,$ where $S=P \cup Q,$ and $S$ is countable itself by Theorem 2 of Chapter 1, §9.

Thus by Corollary 3 in §4, $F^{\prime}=G^{\prime}$ on $I-S$ implies $G-F=c$ (constant) on each $[x, y] \subseteq I;$ hence $G-F=c$ (or $G=F+c)$ on $I. \quad \square$

Definition 2

If $F=\int f$ on $I$ and if $a, b \in I$ (where $a \leq b$ or $b \leq a),$ we define

$$\int_{a}^{b} f=\int_{a}^{b} f(x) d x=F(b)-F(a), \text { also written } F\left.(x)\right|_{a} ^{b}.$$

This expression is called the definite integral of $f$ from $a$ to $b.$

The definite integral of $f$ from $a$ to $b$ is independent of the particular choice of the primitive $F$ for $f,$ and thus unambiguous, for if $G$ is another primitive, Theorem 1 yields $G=F+c,$ so

$$G(b)-G(a)=F(b)+c-[F(a)+c]=F(b)-F(a),$$

and it does not matter whether we take $F$ or $G.$

Note that $\int_{a}^{b} f(x) d x,$ or $\int_{a}^{b} f,$ is a constant in the range space $E$ (a vector if $f$ is vector valued). The "$x$" in $\int_{a}^{b} f(x) dx$ is a "dummy variable" only, and it may be replaced by any other letter. Thus

$$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(y) d y=F(b)-F(a).$$

On the other hand, the indefinite integral is a function: $F : E^{1} \rightarrow E$.

Note 2. We may, however, vary $a$ or $b$ (or both) in (1). Thus, keeping $a$ fixed and varying $b,$ we can define a function

$$G(t)=\int_{a}^{t} f=F(t)-F(a), \quad t \in I.$$

Then $G^{\prime}=F^{\prime}=f$ on $I,$ and $G(a)=F(a)-F(a)=0.$ Thus if $\int f$ exists on $I, f$ has $a$ (unique) primitive $G$ on $I$ such that $G(a)=0.$ (It is unique by Theorem 1. Why?)

Examples

(a) Let

$$f(x)=\frac{1}{x} \text { and } F(x)=\ln |x|, \text { with } F(0)=f(0)=0.$$

Then $F^{\prime}=f$ and $F=\int f$ on $(-\infty, 0)$ and on $(0,+\infty)$ but not on $E^{1},$ since $F$ is discontinuous at $0,$ contrary to Definition 1. We compute

$$\int_{1}^{2} f=\ln 2-\ln 1=\ln 2.$$

(b) On $E^{1},$ let

$$f(x)=\frac{|x|}{x} \text { and } F(x)=|x|, \text { with } f(0)=1.$$

Here $F$ is continuous and $F^{\prime}=f$ on $E^{1}-\{0\}.$ Thus $F=\int f$ on $E^{1}$, exact on $E^{1}-\{0\}.$ Here $I=E^{1}, Q=\{0\}$.

We compute

$$\int_{-2}^{2} f=F(2)-F(-2)=2-2=0$$

(even though $f$ never vanishes on $E^{1})$.

Basic properties of integrals follow from those of derivatives. Thus we have the following.

Corollary $\PageIndex{1}$ (linearity)

If $\int f$ and $\int g$ exist on $I,$ so does $\int(p f+q g)$ for any scalars $p, q$ (in the scalar field of $E).$ Moreover, for any $a, b \in I,$ we obtain

(i) $\int_{a}^{b}(p f+q g)=p \int_{a}^{b} f+q \int_{a}^{b} g$;

(ii) $\int_{a}^{b}(f \pm g)=\int_{a}^{b} f \pm \int_{a}^{b} g;$ and

(iii) $\int_{a}^{b} p f=p \int_{a}^{b} f$.

Proof

By assumption, there are $F$ and $G$ such that

$$F^{\prime}=f \text { on } I-Q \text { and } G^{\prime}=g \text { on } I-P.$$

Thus, setting $S=P \cup Q$ and $H=p F+q G,$ we have

$$H^{\prime}=p F^{\prime}+q G^{\prime}=p f+q g \text { on } I-S,$$

with $P, Q,$ and $S$ countable. Also, $H=p F+q G$ is relatively continuous and finite on $I,$ as are $F$ and $G.$

Thus by definition, $H=\int(p f+q g)$ exists on $I,$ and by (1),

$$\int_{a}^{b}(p f+q g)=H(b)-H(a)=p F(b)+q G(b)-p F(a)-q G(a)=p \int_{a}^{b} f+q \int_{a}^{b} g,$$

proving (i*).

With $p=1$ and $q=\pm 1,$ we obtain (ii*).

Taking $q=0,$ we get (iii*). $\quad \square$

Corollary $\PageIndex{2}$

If both $\int f$ and $\int|f|$ exist on $I=[a, b],$ then

$$\left|\int_{a}^{b} f\right| \leq \int_{a}^{b}|f|.$$

Proof

As before, let

$$F^{\prime}=f \text { and } G^{\prime}=|f| \text { on } I-S(S=Q \cup P, \text { all countable),}$$

where $F$ and $G$ are relatively continuous and finite on $I$ and $G=\int|f|$ is real. Also, $\left|F^{\prime}\right|=|f|=G^{\prime}$ on $I-S.$ Thus by Theorem 1 of §4,

$$|F(b)-F(a)| \leq G(b)-G(a)=\int_{a}^{b}|f|. \quad \square$$

Corollary $\PageIndex{3}$

If $\int f$ exists on $I=[a, b],$ exact on $I-Q,$ then

$$\left|\int_{a}^{b} f\right| \leq M(b-a)$$

for some real

$$M \leq \sup _{t \in I-Q}|f(t)|.$$

This is simply Corollary 1 of §4, when applied to a primitive, $F=\int f$

Corollary $\PageIndex{4}$

If $F=\int f$ on I and $f=g$ on $I-Q,$ then $F$ is also a primitive of $g,$ and

$$\int_{a}^{b} f=\int_{a}^{b} g \quad \text { for } a, b \in I.$$

(Thus we may arbitrarily redefine $f$ on a countable $Q.)$

Proof

Let $F^{\prime}=f$ on $I-P.$ Then $F^{\prime}=g$ on $I-(P \cup Q).$ The rest is clear. $\quad \square$

Corollary $\PageIndex{5}$ (integration by parts)

Let $f$ and $g$ be real or complex (or let $f$ be scalar valued and $g$ vector valued), both relatively continuous on I and differentiable on $I-Q.$ Then if $\int f^{\prime} g$ exists on $I,$ so does $\int f g^{\prime},$ and we have

$$\int_{a}^{b} f g^{\prime}=f(b) g(b)-f(a) g(a)-\int_{a}^{b} f^{\prime} g \quad \text { for any } a, b \in I.$$

Proof

By assumption, $f g$ is relatively continuous and finite on $I,$ and

$$(f g)^{\prime}=f g^{\prime}+f^{\prime} g \text { on } I-Q.$$

Thus, setting $H=f g,$ we have $H=\int\left(f g^{\prime}+f^{\prime} g\right)$ on $I.$ Hence by Corollary 1 if $\int f^{\prime} g$ exists on $I,$ so does $\int\left(\left(f g^{\prime}+f^{\prime} g\right)-f^{\prime} g\right)=\int f g^{\prime},$ and

$$\int_{a}^{b} f g^{\prime}+\int_{a}^{b} f^{\prime} g=\int_{a}^{b}\left(f g^{\prime}+f^{\prime} g\right)=H(b)-H(a)=f(b) g(b)-f(a) g(a).$$

Thus (2) follows. $\quad \square$

The proof of the next three corollaries is left to the reader.

Corollary $\PageIndex{6}$ (additivity of the integral)

If $\int f$ exists on $I$ then, for $a, b, c \in I$, we have

(i) $\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f$;

(ii) $\int_{a}^{a} f=0;$ and

(iii) $\int_{b}^{a} f=-\int_{a}^{b} f$.

Corollary $\PageIndex{7}$ (componentwise integration)

A function $f : E^{1} \rightarrow E^{n}\left(^{*} C^{n}\right)$ is integrable on $I$ iff all its components $\left(f_{1}, f_{2}, \ldots, f_{n}\right)$ are, and then by Theorem 5 in §1)

$$\int_{a}^{b} f=\left(\int_{a}^{b} f_{1}, \ldots, \int_{a}^{b} f_{n}\right)=\sum_{k=1}^{n} \vec{e}_{k} \int_{a}^{b} f_{k} \text { for any } a, b \in I.$$

Hence if $f$ is complex,

$$\int_{a}^{b} f=\int_{a}^{b} f_{\mathrm{re}}+i \cdot \int_{a}^{b} f_{\mathrm{im}}$$

(see Chapter 4, §3, Note 5).

Examples (continued)

(c) Define $f : E^{1} \rightarrow E^{3}$ by

$$f(x)=(a \cdot \cos x, a \cdot \sin x, 2 c x), \quad a, c \in E^{1}.$$

Verify that

$$\int_{0}^{\pi} f(x) d x=\left.\left(a \cdot \sin x,-a \cdot \cos x, c x^{2}\right)\right|_{0} ^{\pi}=\left(0,2 a, c \pi^{2}\right)=2 a \vec{j}+c \pi^{2} \vec{k}.$$

(d) $\int_{0}^{\pi} e^{i x} d x=\int_{0}^{\pi}(\cos x+i \cdot \sin x) d x=\left.(\sin x-i \cdot \cos x)\right|_{0} ^{\pi}=2i.$

Corollary $\PageIndex{8}$

If $f=0$ on $I-Q,$ then $\int f$ exists on $I,$ and

$$\left|\int_{a}^{b} f\right|=\int_{a}^{b}|f|=0 \quad \text { for } a, b \in I.$$

Theorem $\PageIndex{2}$ (change of variables)

Suppose $g : E^{1} \rightarrow E^{1}$ (real) is differentiable on $I,$ while $f : E^{1} \rightarrow E$ has a primitive on $g[I],$ exact on $g[I-Q]$.

Then

$$\int f(g(x)) g^{\prime}(x) d x \quad\left(i . e ., \int(f \circ g) g^{\prime}\right)$$

exists on $I,$ and for any $a, b \in I,$ we have

$$\int_{a}^{b} f(g(x)) g^{\prime}(x) d x=\int_{p}^{q} f(y) d y, \text { where } p=g(a) \text { and } q=g(b).$$

Thus, using classical notation, we may substitute $y=g(x),$ provided that we
also substitute $dy=g^{\prime}(x) dx$ and change the bounds of integrals (3). Here we treat the expressions $dy$ and $g^{\prime}(x) dx$ purely formally, without assigning them any separate meaning outside the context of the integrals.

Proof

Let $F=\int f$ on $g[I],$ and $F^{\prime}=f$ on $g[I-Q].$ Then the composite function $H=F \circ g$ is relatively continuous and finite on $I.$ (Why?) By Theorem 3 of §1,

$$H^{\prime}(x)=F^{\prime}(g(x)) g^{\prime}(x) \text { for } x \in I-Q;$$

i.e.,

$$H^{\prime}=\left(F^{\prime} \circ g\right) g^{\prime} \text { on } I-Q.$$

Thus $H=\int(f \circ g) g^{\prime}$ exists on $I,$ and

$$\int_{a}^{b}(f \circ g) g^{\prime}=H(b)-H(a)=F(g(b))-F(g(a))=F(q)-F(p)=\int_{p}^{q} f. \quad \square$$

Note 3. The theorem does not require that $g$ be one to one on $I,$ but if it is, then one can drop the assumption that $\int f$ is exact on $g[I-Q].$ (See Problem 4.)

Examples (continued)

(e) Find $\int_{0}^{\pi / 2} \sin ^{2} x \cdot \cos x dx$.

Here $f(y)=y^{2}, y=g(x)=\sin x, d y=\cos x d x, F(y)=y^{3} / 3, a=0,$ $b=\pi / 2, p=\sin 0=0,$ and $q=\sin (\pi / 2)=1,$ so (3) yields

$$\int_{0}^{\pi / 2} \sin ^{2} x \cdot \cos x d x=\int_{0}^{1} y^{2} d y=\left.\frac{y^{3}}{3}\right|_{0} ^{1}=\frac{1}{3}-0=\frac{1}{3}.$$

For real functions, we obtain some inferences dealing with inequalities.

Theorem $\PageIndex{3}$

If $f, g : E^{1} \rightarrow E^{1}$ are integrable on $I=[a, b],$ then we have the following:

(i) $f \geq 0$ on $I-Q$ implies $\int_{a}^{b} f \geq 0$.

(i') $f \leq 0$ on $I-Q$ implies $\int_{a}^{b} f \leq 0$.

(ii) $f \geq g$ on $I-Q$ implies

$$\int_{a}^{b} f \geq \int_{a}^{b} g \text { (dominance law).}$$

(iii) If $f \geq 0$ on $I-Q$ and $a \leq c \leq d \leq b,$ then

$$\int_{a}^{b} f \geq \int_{c}^{d} f \text { (monotonicity law).}$$

(iv) If $\int_{a}^{b} f=0,$ and $f \geq 0$ on $I-Q,$ then $f=0$ on some $I-P, P$ countable.

Proof

By Corollary 4, we may redefine $f$ on $Q$ so that our assumptions in (i)-(iv) hold on all of $I$. Thus we write "$I$" for "$I-Q.$"

By assumption, $F=\int f$ and $G=\int g$ exist on $I.$ Here $F$ and $G$ are relatively continuous and finite on $I=[a, b],$ with $F^{\prime}=f$ and $I-P,$ for another countable set $P$ (this $P$ cannot be omitted). Now consider the cases (i)-(iv). ($P$ is fixed henceforth.)

(i) Let $f \geq 0$ on $I;$ i.e., $F^{\prime}=f \geq 0$ on $I-P.$ Then by Theorem 2 in §4, $F \uparrow$ on $I=[a, b].$ Hence $F(a) \leq F(b),$ and so

$$\int_{a}^{b} f=F(b)-F(a) \geq 0.$$

One proves (i') similarly.

(ii) If $f-g \geq 0,$ then by (i),

$$\int_{a}^{b}(f-g)=\int_{a}^{b} f-\int_{a}^{b} g \geq 0,$$

so $\int_{a}^{b} f \geq \int_{a}^{b} g,$ as claimed.

(iii) Let $f \geq 0$ on $I$ and $a \leq c \leq d \leq b.$ Then by (i),

$$\int_{a}^{c} f \geq 0 \text { and } \int_{d}^{b} f \geq 0.$$

Thus by Corollary 6,

$$\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{d} f+\int_{d}^{b} f \geq \int_{c}^{d} f,$$

as asserted.

(iv) Seeking a contradiction, suppose $\int_{a}^{b} f=0, f \geq 0$ on $I,$ yet $f(p)>0$ for some $p \in I-P$ ($P$ as above), so $F^{\prime}(p)=f(p)>0$.

Now if $a \leq p<b,$ Lemma 1 of §2 yields $F(c)>F(p)$ for some $c \in(p, b].$ Then by (iii),

$$\int_{a}^{b} f \geq \int_{p}^{c} f=F(c)-F(p)>0,$$

contrary to $\int_{a}^{b} f=0;$ similarly in case $a<p \leq b. \quad \square$

Note 4. Hence

$$\int_{a}^{b}|f|=0 \text { implies } f=0 \text { on }[a, b]-P$$

($P$ countable), even for vector-valued functions (for $|f|$ is always real, and so Theorem 3 applies).

However, $\int_{a}^{b} f=0$ does not suffice, even for real functions (unless $f$ is signconstant). For example,

$$\int_{0}^{2 \pi} \sin x d x=0, \text { yet } \sin x \not \equiv 0 \text { on any } I-P.$$

See also Example (b).

Corollary $\PageIndex{9}$ (first law of the mean)

If $f$ is real and $\int f$ exists on $[a, b],$ exact on $(a, b),$ then

$$\int_{a}^{b} f=f(q)(b-a) \text { for some } q \in(a, b).$$

Proof

Apply Corollary 3 in §2 to the function $F=\int f. \quad \square$

Caution: Corollary 9 may fail if $\int f$ is inexact at some $p \in(a, b).$ (Exactness on $[a, b]-Q$ does not suffice, as it does not in Corollary 3 of §2, used here.) Thus in Example (b) above, $\int_{-2}^{2} f=0.$ Yet for no $q$ is $f(q)(2+2)=0,$ since $f(q)=\pm 1.$ The reason is that $\int f$ is inexact just at $0,$ an interior point of $[-2,2].$