What happened in the integral in this question?












0












$begingroup$


I'm working through an integral problem and it seems that there is a step missing, either in the solution or in my understanding (probably the latter).



The question is: Let $f(x)$ be continuous on $mathbb{R}$. Then:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_0^af(-x)dx$$



Here is what happens:



We begin with basic breaking up of the bounds:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_{-a}^0f(x)dx$$



We have $u = -x$, $f(x)$ is known to be continuous on the bounds of the intgral, and



$$int_0^a f(-u)du$$



$$= int_0^af(-x)dx$$



I'm not clear about the last step. They substituted $x = -u$, I get that, but there appears to be an extra negative sign in there. What is the missing step?



So far I've been able to figure out that it involves the substitution: if $x = -u$ then the upper bound of integration becomes $-a$. The intermediate integral should be:



$$-int_0^{-a}f(x)dx = int_{-a}^0f(x)dx$$



Neither of these seem to agree with the last step.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just select another alphabet. Nothing more. Or to say $x = u$.
    $endgroup$
    – xbh
    Dec 5 '18 at 17:59












  • $begingroup$
    You keep repeating that they substitute $x=-u$. That's not what they do
    $endgroup$
    – Federico
    Dec 5 '18 at 18:01










  • $begingroup$
    You can call the variable of integration whatever you like, it’s a dummy variable
    $endgroup$
    – qbert
    Dec 5 '18 at 18:10
















0












$begingroup$


I'm working through an integral problem and it seems that there is a step missing, either in the solution or in my understanding (probably the latter).



The question is: Let $f(x)$ be continuous on $mathbb{R}$. Then:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_0^af(-x)dx$$



Here is what happens:



We begin with basic breaking up of the bounds:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_{-a}^0f(x)dx$$



We have $u = -x$, $f(x)$ is known to be continuous on the bounds of the intgral, and



$$int_0^a f(-u)du$$



$$= int_0^af(-x)dx$$



I'm not clear about the last step. They substituted $x = -u$, I get that, but there appears to be an extra negative sign in there. What is the missing step?



So far I've been able to figure out that it involves the substitution: if $x = -u$ then the upper bound of integration becomes $-a$. The intermediate integral should be:



$$-int_0^{-a}f(x)dx = int_{-a}^0f(x)dx$$



Neither of these seem to agree with the last step.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just select another alphabet. Nothing more. Or to say $x = u$.
    $endgroup$
    – xbh
    Dec 5 '18 at 17:59












  • $begingroup$
    You keep repeating that they substitute $x=-u$. That's not what they do
    $endgroup$
    – Federico
    Dec 5 '18 at 18:01










  • $begingroup$
    You can call the variable of integration whatever you like, it’s a dummy variable
    $endgroup$
    – qbert
    Dec 5 '18 at 18:10














0












0








0





$begingroup$


I'm working through an integral problem and it seems that there is a step missing, either in the solution or in my understanding (probably the latter).



The question is: Let $f(x)$ be continuous on $mathbb{R}$. Then:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_0^af(-x)dx$$



Here is what happens:



We begin with basic breaking up of the bounds:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_{-a}^0f(x)dx$$



We have $u = -x$, $f(x)$ is known to be continuous on the bounds of the intgral, and



$$int_0^a f(-u)du$$



$$= int_0^af(-x)dx$$



I'm not clear about the last step. They substituted $x = -u$, I get that, but there appears to be an extra negative sign in there. What is the missing step?



So far I've been able to figure out that it involves the substitution: if $x = -u$ then the upper bound of integration becomes $-a$. The intermediate integral should be:



$$-int_0^{-a}f(x)dx = int_{-a}^0f(x)dx$$



Neither of these seem to agree with the last step.






calculus integration







share|cite|improve this question











$endgroup$




I'm working through an integral problem and it seems that there is a step missing, either in the solution or in my understanding (probably the latter).



The question is: Let $f(x)$ be continuous on $mathbb{R}$. Then:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_0^af(-x)dx$$



Here is what happens:



We begin with basic breaking up of the bounds:



$$int_{-a}^af(x)dx = int_0^af(x)dx + int_{-a}^0f(x)dx$$



We have $u = -x$, $f(x)$ is known to be continuous on the bounds of the intgral, and



$$int_0^a f(-u)du$$



$$= int_0^af(-x)dx$$



I'm not clear about the last step. They substituted $x = -u$, I get that, but there appears to be an extra negative sign in there. What is the missing step?



So far I've been able to figure out that it involves the substitution: if $x = -u$ then the upper bound of integration becomes $-a$. The intermediate integral should be:



$$-int_0^{-a}f(x)dx = int_{-a}^0f(x)dx$$



Neither of these seem to agree with the last step.






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 18:04







Michael Stachowsky

















asked Dec 5 '18 at 17:58









Michael StachowskyMichael Stachowsky

1,252417




1,252417












  • $begingroup$
    Just select another alphabet. Nothing more. Or to say $x = u$.
    $endgroup$
    – xbh
    Dec 5 '18 at 17:59












  • $begingroup$
    You keep repeating that they substitute $x=-u$. That's not what they do
    $endgroup$
    – Federico
    Dec 5 '18 at 18:01










  • $begingroup$
    You can call the variable of integration whatever you like, it’s a dummy variable
    $endgroup$
    – qbert
    Dec 5 '18 at 18:10


















  • $begingroup$
    Just select another alphabet. Nothing more. Or to say $x = u$.
    $endgroup$
    – xbh
    Dec 5 '18 at 17:59












  • $begingroup$
    You keep repeating that they substitute $x=-u$. That's not what they do
    $endgroup$
    – Federico
    Dec 5 '18 at 18:01










  • $begingroup$
    You can call the variable of integration whatever you like, it’s a dummy variable
    $endgroup$
    – qbert
    Dec 5 '18 at 18:10
















$begingroup$
Just select another alphabet. Nothing more. Or to say $x = u$.
$endgroup$
– xbh
Dec 5 '18 at 17:59






$begingroup$
Just select another alphabet. Nothing more. Or to say $x = u$.
$endgroup$
– xbh
Dec 5 '18 at 17:59














$begingroup$
You keep repeating that they substitute $x=-u$. That's not what they do
$endgroup$
– Federico
Dec 5 '18 at 18:01




$begingroup$
You keep repeating that they substitute $x=-u$. That's not what they do
$endgroup$
– Federico
Dec 5 '18 at 18:01












$begingroup$
You can call the variable of integration whatever you like, it’s a dummy variable
$endgroup$
– qbert
Dec 5 '18 at 18:10




$begingroup$
You can call the variable of integration whatever you like, it’s a dummy variable
$endgroup$
– qbert
Dec 5 '18 at 18:10










1 Answer
1






active

oldest

votes


















1












$begingroup$

They substituted $x=u$, which admittedly is a pretty dumb thing to do.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:00










  • $begingroup$
    They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
    $endgroup$
    – Federico
    Dec 5 '18 at 18:02










  • $begingroup$
    I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:05










  • $begingroup$
    What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
    $endgroup$
    – Federico
    Dec 5 '18 at 18:11










  • $begingroup$
    My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
    $endgroup$
    – Federico
    Dec 5 '18 at 18:12













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

They substituted $x=u$, which admittedly is a pretty dumb thing to do.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:00










  • $begingroup$
    They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
    $endgroup$
    – Federico
    Dec 5 '18 at 18:02










  • $begingroup$
    I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:05










  • $begingroup$
    What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
    $endgroup$
    – Federico
    Dec 5 '18 at 18:11










  • $begingroup$
    My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
    $endgroup$
    – Federico
    Dec 5 '18 at 18:12


















1












$begingroup$

They substituted $x=u$, which admittedly is a pretty dumb thing to do.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:00










  • $begingroup$
    They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
    $endgroup$
    – Federico
    Dec 5 '18 at 18:02










  • $begingroup$
    I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:05










  • $begingroup$
    What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
    $endgroup$
    – Federico
    Dec 5 '18 at 18:11










  • $begingroup$
    My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
    $endgroup$
    – Federico
    Dec 5 '18 at 18:12
















1












1








1





$begingroup$

They substituted $x=u$, which admittedly is a pretty dumb thing to do.






share|cite|improve this answer









$endgroup$



They substituted $x=u$, which admittedly is a pretty dumb thing to do.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 18:00









FedericoFederico

5,024514




5,024514












  • $begingroup$
    Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:00










  • $begingroup$
    They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
    $endgroup$
    – Federico
    Dec 5 '18 at 18:02










  • $begingroup$
    I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:05










  • $begingroup$
    What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
    $endgroup$
    – Federico
    Dec 5 '18 at 18:11










  • $begingroup$
    My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
    $endgroup$
    – Federico
    Dec 5 '18 at 18:12




















  • $begingroup$
    Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:00










  • $begingroup$
    They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
    $endgroup$
    – Federico
    Dec 5 '18 at 18:02










  • $begingroup$
    I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
    $endgroup$
    – Michael Stachowsky
    Dec 5 '18 at 18:05










  • $begingroup$
    What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
    $endgroup$
    – Federico
    Dec 5 '18 at 18:11










  • $begingroup$
    My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
    $endgroup$
    – Federico
    Dec 5 '18 at 18:12


















$begingroup$
Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
$endgroup$
– Michael Stachowsky
Dec 5 '18 at 18:00




$begingroup$
Do you mean they did another substitution at the end? If so, doesn't that mean that original variable $x$ is actually $-x$?
$endgroup$
– Michael Stachowsky
Dec 5 '18 at 18:00












$begingroup$
They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
$endgroup$
– Federico
Dec 5 '18 at 18:02




$begingroup$
They substitute $x=u$. $0$ remains $0$, $a$ remains $a$, $-u$ becomes $-x$, $du$ becomes $dx$. Where is the difficulty?
$endgroup$
– Federico
Dec 5 '18 at 18:02












$begingroup$
I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
$endgroup$
– Michael Stachowsky
Dec 5 '18 at 18:05




$begingroup$
I've edited the question to explain a bit better. The whole question is about breaking an integral apart into two pieces. My issue is that in the expression, if all they did was re-name $u=x$, then the two variables $x$ are different, are they not?
$endgroup$
– Michael Stachowsky
Dec 5 '18 at 18:05












$begingroup$
What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
$endgroup$
– Federico
Dec 5 '18 at 18:11




$begingroup$
What does "the two variables $x$ are different" mean? That is the variable of integration. It is local to that integral
$endgroup$
– Federico
Dec 5 '18 at 18:11












$begingroup$
My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
$endgroup$
– Federico
Dec 5 '18 at 18:12






$begingroup$
My answer still addresses the step that you are not understanding, regarding of whatever context this fits in
$endgroup$
– Federico
Dec 5 '18 at 18:12




















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