5.5.E: Problems on Antiderivatives
- Page ID
- 23756
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Prove in detail Corollaries \(3,4,6,7,8,\) and 9 and Theorem 3\(\left(\mathrm{i}^{\prime}\right)\) and \((\text { iv }) .\)
In Examples (a) and (b) discuss continuity and differentiability of \(f\) and \(F\) at \(0 .\) In (a) show that \(\int f\) does not exist on any interval \((-a, a) .\)
[Hint: Use Theorem 1.]
Show that Theorem 2 holds also if \(g\) is relatively continuous on \(I\) and differentiable on \(I-Q\).
Under the assumptions of Theorem \(2,\) show that if \(g\) is one to one on \(I,\) then automatically \(\int f\) is exact on \(g[I-Q](Q \text { countable). }\)
[Hint: If \(F=\int f\) on \(g[I],\) then
\[
F^{\prime}=f \text { on } g[I]-P, P \text { countable. }
\]
Let \(Q=g^{-1}[P] .\) Use Problem 6 of Chapter \(1,§§4-7\) and Problem 2 of Chapter 1 §9 to show that \(Q\) is countable and \(g[I]-P=g[I-Q] .\)
Prove Corollary 5 for dot products \(f \cdot g\) of vector-valued functions.
Prove that if \(\int f\) exists on \([a, p]\) and \([p, b],\) then it exists on \([a, b] .\) By induction, extend this to unions of \(n\) adjacent intervals.
[ Hint: Choose \(F=\int f\) on \([a, p]\) and \(G=\int f\) on \([p, b]\) such that \(F(p)=G(p)\). (Why do such \(F, G\) exist?) Then construct a primitive \(H=\int f\) that is relatively \(\text { continuous on } a l l \text { of }[a, b] .]\)
Prove the weighted law of the mean: If \(g\) is real and nonnegative on \(I=[a, b],\) and if \(\int g\) and \(\int g f\) exist on \(I\) for some \(f : E^{1} \rightarrow E,\) then there is a finite \(c \in E\) with
\[
\int_{a}^{b} g f=c \int_{a}^{b} g .
\]
(The value \(c\) is called a \(g\)-weighted mean of \(f\).)
[Hint: If \(\int_{a}^{b} g>0,\) put
\[
c=\int_{a}^{b} g f / \int_{a}^{b} g .
\]
\(\left.\text { If } \int_{a}^{b} g=0, \text { use Theorem } 3(\mathrm{i}) \text { and (iv) to show that also } \int_{a}^{b} g f=0, \text { so any } c \text { will do. }\right]\)
In Problem \(7,\) prove that if, in addition, \(f\) is real and has the Darboux property on \(I,\) then \(c=f(q)\) for some \(q \in I\) (the second law of the mean).
[Hint: Choose \(c\) as in Problem 7. If \(\int_{a}^{b} g>0,\) put
\[
m=\inf f[I] \text { and } M=\sup f[I], \text { in } E^{*} ,
\]
so \(m \leq f \leq M\) on \(I .\) Deduce that
\[
m \int_{a}^{b} g \leq \int_{a}^{b} g f \leq M \int_{a}^{b} g ,
\]
whence \(m \leq c \leq M\).
If \(m<c<M,\) then \(f(x)<c<f(y)\) for some \(x, y \in I\) (why?), so the Darboux property applies.
If \(c=m,\) then \(g \cdot(f-c) \geq 0\) and Theorem 3\((\text { iv })\) yields \(g f=g c\) on \(I-P .\) (Why?) Deduce that \(f(q)=c\) if \(g(q) \neq 0\) and \(q \in I-P .\) (Why does such a \(q\) exist?)
\(\text { What if } c=M ?]\)
Taking \(g(x) \equiv 1\) in Problem \(8,\) obtain a new version of Corollary \(9 .\) State it precisely!
\(\Rightarrow 10.\) Prove that if \(F=\int f\) on \(I=(a, b)\) and \(f\) is right (left) continuous and finite at \(p \in I,\) then
\[
\left.f(p)=F_{+}^{\prime}(p) \text { (respectively, } F_{-}^{\prime}(p)\right) .
\]
Deduce that if \(f\) is continuous and finite on \(I,\) all its primitives on \(I\) are exact on \(I .\)
[Hint: Fix \(\varepsilon>0 .\) If \(f\) is right continuous at \(p,\) there is \(c \in I(c>p),\) with
\[
|f(x)-f(p)|<\varepsilon \text { for } x \in[p, c) .
\]
Fix such an \(x .\) Let
\[
G(t)=F(t)-t f(p), \quad t \in E^{1} .
\]
Deduce that \(G^{\prime}(t)=f(t)-f(p)\) for \(t \in I-Q\).
By Corollary 1 of §4,
\[
|G(x)-G(p)|=|F(x)-F(p)-(x-p) f(p)| \leq M(x-p) ,
\]
with \(M \leq \varepsilon .\) (Why?) Hence
\[
\left|\frac{\Delta F}{\Delta x}-f(p)\right| \leq \varepsilon \text { for } x \in[p, c) ,
\]
and so
\[
\lim _{x \rightarrow p^{+}} \frac{\Delta F}{\Delta x}=f(p) \quad(\text { why? }) ;
\]
\(\text { similarly for a left-continuous } f .]\)
State and solve Problem 10 for the case \(I=[a, b]\).
(i) Prove that if \(f\) is constant \((f=c \neq \pm \infty)\) on \(I-Q,\) then
\[
\int_{a}^{b} f=(b-a) c \quad \text { for } a, b \in I .
\]
(ii) Hence prove that if \(f=c_{k} \neq \pm \infty\) on
\[
I_{k}=\left[a_{k}, a_{k+1}\right), \quad a=a_{0}<a_{1}<\cdots<a_{n}=b ,
\]
then \(\int f\) exists on \([a, b],\) and
\[
\int_{a}^{b} f=\sum_{k=0}^{n-1}\left(a_{k+1}-a_{k}\right) c_{k} .
\]
Show that this is true also if \(f=c_{k} \neq \pm \infty\) on \(I_{k}-Q_{k}\).
\(\text { [Hint: Use Problem } 6 .]\)
Prove that if \(\int f\) exists on each \(I_{n}=\left[a_{n}, b_{n}\right],\) where
\[
a_{n+1} \leq a_{n} \leq b_{n} \leq b_{n+1}, \quad n=1,2, \ldots ,
\]
then \(\int f\) exists on
\[
I=\bigcup_{n=1}^{\infty}\left[a_{n}, b_{n}\right] ,
\]
itself an interval with endpoints \(a=\inf a_{n}\) and \(b=\sup b_{n}, a, b \in E^{*}\).
[Hint: Fix some \(c \in I_{1}\). Define
\[
H_{n}(t)=\int_{c}^{t} f \text { on } I_{n}, n=1,2, \ldots .
\]
Prove that
\[
(\forall n \leq m) \quad H_{n}=H_{m} \text { on } I_{n}\left(\text { since }\left\{I_{n}\right\} \uparrow\right) .
\]
Thus \(H_{n}(t)\) is the same for all \(n\) such that \(t \in I_{n},\) so we may simply write \(H\) for \(H_{n}\) on \(I=\bigcup_{n=1}^{\infty} I_{n} .\) Show that \(H=\int f\) on all of \(I ;\) verify that \(I\) is, indeed, an interval.]
Continuing Problem \(13,\) prove that \(\int f\) exists on an interval \(I\) iff it exists on each closed subinterval \([a, b] \subseteq I .\)
[Hint: Show that each \(I\) is the union of an expanding sequence \(I_{n}=\left[a_{n}, b_{n}\right] .\) For example, if \(I=(a, b), a, b \in E^{1},\) put
\[
a_{n}=a+\frac{1}{n} \text { and } b_{n}=b-\frac{1}{n} \text { for large } n \text { (how large?) } ,
\]
and show that
\[
\left.I=\bigcup_{n}\left[a_{n}, b_{n}\right] \text { over such } n .\right]
\]