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.
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
\(\int f+c ).\)
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 \(I. \quad \square\)
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?)
(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.
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*). \(\quad \square\)
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,
\[|F(b)-F(a)| \leq G(b)-G(a)=\int_{a}^{b}|f|. \quad \square\]
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, \(F=\int f\)
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. \(\quad \square\)
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.
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) \(\int_{b}^{a} f=-\int_{a}^{b} f\).
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).
(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) \(\int_{0}^{\pi} e^{i x} d x=\int_{0}^{\pi}(\cos x+i \cdot \sin x) d x=\left.(\sin x-i \cdot \cos x)\right|_{0} ^{\pi}=2i.\)
If \(f=0\) on \(I-Q,\) then \(\int f\) exists on \(I,\) and
\[\left|\int_{a}^{b} f\right|=\int_{a}^{b}|f|=0 \quad \text { for } a, b \in I.\]
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.)
(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.
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).
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 \([-2,2].\)