Mixing
Clarification on Theorem 1.19, Apostol Calculus
$begingroup$
In Chapter 1 of Apostol Calculus, Apostol provides proof of Theorem 1.19, however only shows the proof for the case of k > 0, saying k < 0 can be shown similarly.
I've tried working out the second half on my own, but I'm having some trouble. I'm hoping somebody can provide a hint or pose a question that will help me along. I've copied the case for $k > 0$ below, and then shown what I have done below that. I'll add that I've gone through every page/exercise up until this point in this book, but I have no experience beyond that (i.e. analysis)
Theorem 1.19: Expansion or Contraction of the Interval of Integration
If $f$ is integrable on $[a, b]$, then for every real $k neq 0$ we have $$int_{a}^{b}{f(x)dx} = frac{1}{k}int_{ka}^{kb}{f(frac{x}{k})dx}$$
Here is the case for $k > 0$, given by Apostol.
Case: $k > 0$
Assume $k > 0$ and define $g$ on the interval $[ka, kb]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. We shall prove that
$$ underline{I}(g) = overline{I}(g) = k cdot int_{a}^{b}{f(x)}dx label{1.17}tag{1.17}$$
Let $s$ be any step function below $g$ on $[ka, kb]$. Then the function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ is a step function below f on $[a,b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. Also by the expansion property for integrals of step functions, we have
$$int_{ka}^{kb}{s(x)dx} = k cdot int_{a}^{b}{s(kx)dx} = k cdot int_{a}^{b}{s_{1}(x)dx}$$
Therefore we have
$$ underline{I}(g) = sup left{ int_{ka}^{kb}{s} :|: s leq gright} = sup left{ k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot int_{a}^{b}{f(x)dx}$$
Similarly, we find $overline{I}(g) = k cdot int_{a}^{b}{f(x)dx}$ which proves (1.17)
Case: $k < 0$
I'm not entirely sure why the proof needs to be split into two cases, however since it was, my thought was to define $g$ on the interval $[kb, ka]$ since $k < 0$ (I'm imagining g is f, reflected over the y-axis). Then define the step function below $g$ and $f$ along the same lines. Later, in line (1), I make use of the reflection property for when lower limit of integration is greater than the upper limit. However at the end of this, I'm left with an extra negative.
Assume $k < 0$ and $b > a$. Define $g$ on the interval $[kb, ka]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. Again we want to show (1.17).
Let $s$ be any step function below $g$ on $[kb, ka]$. Then there is a function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ that is a step function below f on $[a, b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. By the expansion property for integrals of step functions, we have
$$int_{kb}^{ka}{s(x)dx} = k cdot int_{b}^{a}{s(kx)dx} = k cdot int_{b}^{a}{s_{1}(x)dx} = -k cdot int_{a}^{b}{s_{1}(x)dx}label{1}tag{1}$$
$$ underline{I}(g) = sup left{ int_{kb}^{ka}{s} :|: s leq gright} = sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = -k cdot int_{a}^{b}{f(x)dx}label{2}tag{2}$$
It's at this point that I'm stuck. Working it backwards, the only way I see to remove the last negative in (2) is to start with $g$ defined on $[ka, kb]$ but then why was the proof split into two cases if they are identical?
One other thought I had was to explore defining, in (2), $$sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot inf left{ int_{a}^{b}{s_{1}} :|: s_{1} leq fright}$$ but the lower integral of $g$ is not equal to the infimum of the set of the integrals of the step function below $g$.
Thanks for your insight
calculus integration analysis proof-verification proof-explanation
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
$endgroup$
add a comment |
$begingroup$
In Chapter 1 of Apostol Calculus, Apostol provides proof of Theorem 1.19, however only shows the proof for the case of k > 0, saying k < 0 can be shown similarly.
I've tried working out the second half on my own, but I'm having some trouble. I'm hoping somebody can provide a hint or pose a question that will help me along. I've copied the case for $k > 0$ below, and then shown what I have done below that. I'll add that I've gone through every page/exercise up until this point in this book, but I have no experience beyond that (i.e. analysis)
Theorem 1.19: Expansion or Contraction of the Interval of Integration
If $f$ is integrable on $[a, b]$, then for every real $k neq 0$ we have $$int_{a}^{b}{f(x)dx} = frac{1}{k}int_{ka}^{kb}{f(frac{x}{k})dx}$$
Here is the case for $k > 0$, given by Apostol.
Case: $k > 0$
Assume $k > 0$ and define $g$ on the interval $[ka, kb]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. We shall prove that
$$ underline{I}(g) = overline{I}(g) = k cdot int_{a}^{b}{f(x)}dx label{1.17}tag{1.17}$$
Let $s$ be any step function below $g$ on $[ka, kb]$. Then the function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ is a step function below f on $[a,b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. Also by the expansion property for integrals of step functions, we have
$$int_{ka}^{kb}{s(x)dx} = k cdot int_{a}^{b}{s(kx)dx} = k cdot int_{a}^{b}{s_{1}(x)dx}$$
Therefore we have
$$ underline{I}(g) = sup left{ int_{ka}^{kb}{s} :|: s leq gright} = sup left{ k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot int_{a}^{b}{f(x)dx}$$
Similarly, we find $overline{I}(g) = k cdot int_{a}^{b}{f(x)dx}$ which proves (1.17)
Case: $k < 0$
I'm not entirely sure why the proof needs to be split into two cases, however since it was, my thought was to define $g$ on the interval $[kb, ka]$ since $k < 0$ (I'm imagining g is f, reflected over the y-axis). Then define the step function below $g$ and $f$ along the same lines. Later, in line (1), I make use of the reflection property for when lower limit of integration is greater than the upper limit. However at the end of this, I'm left with an extra negative.
Assume $k < 0$ and $b > a$. Define $g$ on the interval $[kb, ka]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. Again we want to show (1.17).
Let $s$ be any step function below $g$ on $[kb, ka]$. Then there is a function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ that is a step function below f on $[a, b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. By the expansion property for integrals of step functions, we have
$$int_{kb}^{ka}{s(x)dx} = k cdot int_{b}^{a}{s(kx)dx} = k cdot int_{b}^{a}{s_{1}(x)dx} = -k cdot int_{a}^{b}{s_{1}(x)dx}label{1}tag{1}$$
$$ underline{I}(g) = sup left{ int_{kb}^{ka}{s} :|: s leq gright} = sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = -k cdot int_{a}^{b}{f(x)dx}label{2}tag{2}$$
It's at this point that I'm stuck. Working it backwards, the only way I see to remove the last negative in (2) is to start with $g$ defined on $[ka, kb]$ but then why was the proof split into two cases if they are identical?
One other thought I had was to explore defining, in (2), $$sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot inf left{ int_{a}^{b}{s_{1}} :|: s_{1} leq fright}$$ but the lower integral of $g$ is not equal to the infimum of the set of the integrals of the step function below $g$.
Thanks for your insight
calculus integration analysis proof-verification proof-explanation
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
$endgroup$
$begingroup$
There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
$endgroup$
– Tsemo Aristide
Feb 3 at 0:09
$begingroup$
@TsemoAristide sorry, can you clarify which part you are referring to?
$endgroup$
– Jake Kirsch
Feb 3 at 11:30
$begingroup$
The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
$endgroup$
– Paramanand Singh
Feb 5 at 7:47
$begingroup$
@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
$endgroup$
– Jake Kirsch
Feb 5 at 14:48
add a comment |
$begingroup$
In Chapter 1 of Apostol Calculus, Apostol provides proof of Theorem 1.19, however only shows the proof for the case of k > 0, saying k < 0 can be shown similarly.
I've tried working out the second half on my own, but I'm having some trouble. I'm hoping somebody can provide a hint or pose a question that will help me along. I've copied the case for $k > 0$ below, and then shown what I have done below that. I'll add that I've gone through every page/exercise up until this point in this book, but I have no experience beyond that (i.e. analysis)
Theorem 1.19: Expansion or Contraction of the Interval of Integration
If $f$ is integrable on $[a, b]$, then for every real $k neq 0$ we have $$int_{a}^{b}{f(x)dx} = frac{1}{k}int_{ka}^{kb}{f(frac{x}{k})dx}$$
Here is the case for $k > 0$, given by Apostol.
Case: $k > 0$
Assume $k > 0$ and define $g$ on the interval $[ka, kb]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. We shall prove that
$$ underline{I}(g) = overline{I}(g) = k cdot int_{a}^{b}{f(x)}dx label{1.17}tag{1.17}$$
Let $s$ be any step function below $g$ on $[ka, kb]$. Then the function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ is a step function below f on $[a,b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. Also by the expansion property for integrals of step functions, we have
$$int_{ka}^{kb}{s(x)dx} = k cdot int_{a}^{b}{s(kx)dx} = k cdot int_{a}^{b}{s_{1}(x)dx}$$
Therefore we have
$$ underline{I}(g) = sup left{ int_{ka}^{kb}{s} :|: s leq gright} = sup left{ k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot int_{a}^{b}{f(x)dx}$$
Similarly, we find $overline{I}(g) = k cdot int_{a}^{b}{f(x)dx}$ which proves (1.17)
Case: $k < 0$
I'm not entirely sure why the proof needs to be split into two cases, however since it was, my thought was to define $g$ on the interval $[kb, ka]$ since $k < 0$ (I'm imagining g is f, reflected over the y-axis). Then define the step function below $g$ and $f$ along the same lines. Later, in line (1), I make use of the reflection property for when lower limit of integration is greater than the upper limit. However at the end of this, I'm left with an extra negative.
Assume $k < 0$ and $b > a$. Define $g$ on the interval $[kb, ka]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. Again we want to show (1.17).
Let $s$ be any step function below $g$ on $[kb, ka]$. Then there is a function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ that is a step function below f on $[a, b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. By the expansion property for integrals of step functions, we have
$$int_{kb}^{ka}{s(x)dx} = k cdot int_{b}^{a}{s(kx)dx} = k cdot int_{b}^{a}{s_{1}(x)dx} = -k cdot int_{a}^{b}{s_{1}(x)dx}label{1}tag{1}$$
$$ underline{I}(g) = sup left{ int_{kb}^{ka}{s} :|: s leq gright} = sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = -k cdot int_{a}^{b}{f(x)dx}label{2}tag{2}$$
It's at this point that I'm stuck. Working it backwards, the only way I see to remove the last negative in (2) is to start with $g$ defined on $[ka, kb]$ but then why was the proof split into two cases if they are identical?
One other thought I had was to explore defining, in (2), $$sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot inf left{ int_{a}^{b}{s_{1}} :|: s_{1} leq fright}$$ but the lower integral of $g$ is not equal to the infimum of the set of the integrals of the step function below $g$.
Thanks for your insight
calculus integration analysis proof-verification proof-explanation
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
$endgroup$
In Chapter 1 of Apostol Calculus, Apostol provides proof of Theorem 1.19, however only shows the proof for the case of k > 0, saying k < 0 can be shown similarly.
I've tried working out the second half on my own, but I'm having some trouble. I'm hoping somebody can provide a hint or pose a question that will help me along. I've copied the case for $k > 0$ below, and then shown what I have done below that. I'll add that I've gone through every page/exercise up until this point in this book, but I have no experience beyond that (i.e. analysis)
Theorem 1.19: Expansion or Contraction of the Interval of Integration
If $f$ is integrable on $[a, b]$, then for every real $k neq 0$ we have $$int_{a}^{b}{f(x)dx} = frac{1}{k}int_{ka}^{kb}{f(frac{x}{k})dx}$$
Here is the case for $k > 0$, given by Apostol.
Case: $k > 0$
Assume $k > 0$ and define $g$ on the interval $[ka, kb]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. We shall prove that
$$ underline{I}(g) = overline{I}(g) = k cdot int_{a}^{b}{f(x)}dx label{1.17}tag{1.17}$$
Let $s$ be any step function below $g$ on $[ka, kb]$. Then the function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ is a step function below f on $[a,b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. Also by the expansion property for integrals of step functions, we have
$$int_{ka}^{kb}{s(x)dx} = k cdot int_{a}^{b}{s(kx)dx} = k cdot int_{a}^{b}{s_{1}(x)dx}$$
Therefore we have
$$ underline{I}(g) = sup left{ int_{ka}^{kb}{s} :|: s leq gright} = sup left{ k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot int_{a}^{b}{f(x)dx}$$
Similarly, we find $overline{I}(g) = k cdot int_{a}^{b}{f(x)dx}$ which proves (1.17)
Case: $k < 0$
I'm not entirely sure why the proof needs to be split into two cases, however since it was, my thought was to define $g$ on the interval $[kb, ka]$ since $k < 0$ (I'm imagining g is f, reflected over the y-axis). Then define the step function below $g$ and $f$ along the same lines. Later, in line (1), I make use of the reflection property for when lower limit of integration is greater than the upper limit. However at the end of this, I'm left with an extra negative.
Assume $k < 0$ and $b > a$. Define $g$ on the interval $[kb, ka]$ by the equation $g(x) = f(x/k)$. Let $underline{I}(g) text{ and } overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. Again we want to show (1.17).
Let $s$ be any step function below $g$ on $[kb, ka]$. Then there is a function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ that is a step function below f on $[a, b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. By the expansion property for integrals of step functions, we have
$$int_{kb}^{ka}{s(x)dx} = k cdot int_{b}^{a}{s(kx)dx} = k cdot int_{b}^{a}{s_{1}(x)dx} = -k cdot int_{a}^{b}{s_{1}(x)dx}label{1}tag{1}$$
$$ underline{I}(g) = sup left{ int_{kb}^{ka}{s} :|: s leq gright} = sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = -k cdot int_{a}^{b}{f(x)dx}label{2}tag{2}$$
It's at this point that I'm stuck. Working it backwards, the only way I see to remove the last negative in (2) is to start with $g$ defined on $[ka, kb]$ but then why was the proof split into two cases if they are identical?
One other thought I had was to explore defining, in (2), $$sup left{ -k cdot int_{a}^{b}{s_{1}} :|: s_{1} leq fright} = k cdot inf left{ int_{a}^{b}{s_{1}} :|: s_{1} leq fright}$$ but the lower integral of $g$ is not equal to the infimum of the set of the integrals of the step function below $g$.
Thanks for your insight
calculus integration analysis proof-verification proof-explanation
calculus integration analysis proof-verification proof-explanation
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
edited Feb 4 at 16:49
Jake Kirsch
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
asked Feb 2 at 23:52
Jake KirschJake Kirsch
687
687
$begingroup$
There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
$endgroup$
– Tsemo Aristide
Feb 3 at 0:09
$begingroup$
@TsemoAristide sorry, can you clarify which part you are referring to?
$endgroup$
– Jake Kirsch
Feb 3 at 11:30
$begingroup$
The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
$endgroup$
– Paramanand Singh
Feb 5 at 7:47
$begingroup$
@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
$endgroup$
– Jake Kirsch
Feb 5 at 14:48
add a comment |
$begingroup$
There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
$endgroup$
– Tsemo Aristide
Feb 3 at 0:09
$begingroup$
@TsemoAristide sorry, can you clarify which part you are referring to?
$endgroup$
– Jake Kirsch
Feb 3 at 11:30
$begingroup$
The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
$endgroup$
– Paramanand Singh
Feb 5 at 7:47
$begingroup$
@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
$endgroup$
– Jake Kirsch
Feb 5 at 14:48
$begingroup$
There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
$endgroup$
– Tsemo Aristide
Feb 3 at 0:09
$begingroup$
There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
$endgroup$
– Tsemo Aristide
Feb 3 at 0:09
$begingroup$
@TsemoAristide sorry, can you clarify which part you are referring to?
$endgroup$
– Jake Kirsch
Feb 3 at 11:30
$begingroup$
@TsemoAristide sorry, can you clarify which part you are referring to?
$endgroup$
– Jake Kirsch
Feb 3 at 11:30
$begingroup$
The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
$endgroup$
– Paramanand Singh
Feb 5 at 7:47
$begingroup$
The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
$endgroup$
– Paramanand Singh
Feb 5 at 7:47
$begingroup$
@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
$endgroup$
– Jake Kirsch
Feb 5 at 14:48
$begingroup$
@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
$endgroup$
– Jake Kirsch
Feb 5 at 14:48
add a comment |
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There is something I dont understand if you make the variable change $u={xover k}$ you dont obtain a proof ?
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– Tsemo Aristide
Feb 3 at 0:09
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@TsemoAristide sorry, can you clarify which part you are referring to?
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– Jake Kirsch
Feb 3 at 11:30
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The case $k<0$ is more properly written as $$int_{a} ^{b} f(x) , dx=-frac{1}{k}int_{kb}^{ka}f(x/k),dx$$ and the proof works best by using Riemann sums instead of Darboux integrals. You can show that each Riemann sum for integral on right is equal to $-k$ times a Riemann sum for the integral on left and vice versa.
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– Paramanand Singh
Feb 5 at 7:47
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@ParamanandSingh unfortunately I'm not familiar with Riemann sums yet (i've only worked through through the introduction and ch 1 of apostol vol 1). However your point helps me see where I was getting tripped up. I showed the upper and lower integral of $g$ is equal to $-k cdot int_{a}^{b}{f}$ which is what I was supposed to do for the case of $k < 0$
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– Jake Kirsch
Feb 5 at 14:48