Changing bounds of definite integral












0












$begingroup$


I am using this baby Rudin definition for Riemann integral:




Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, ldots, x_n$, where
$$ a = x_0 leq x_1 leq cdots leq x_{n-1} leq x_n = b.$$
We write
$$ Delta x_i = x_i - x_{i-1} qquad (i = 1, ldots, n). $$
Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put
$$
begin{align}
M_i &= sup f(x) qquad (x_{i-1} leq x leq x_i), \
m_i &= inf f(x) qquad (x_{i-1} leq x leq x_i), \
U(P, f) &= sum_{i=1}^n M_i Delta x_i, \
L(P, f) &= sum_{i=1}^n m_i Delta x_i,
end{align}
$$
and finally
$$
begin{align}
tag{1} overline{int_a^b} f dx &= inf U(P, f), \
tag{2} underline{int_a^b} f dx &= sup L(P, f),\,
end{align}
$$
where the $inf$ and the $sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.



If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$,
$$ tag{3} int_a^b f dx. $$




How do we get property $displaystyleint_a^b f dx=-displaystyleint_b^a f dx$ from this definition?



(Code borrowed from here)






real-analysis integration definition







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    You don't. That's the convention for defining the integral when $a>b$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:19
















0












$begingroup$


I am using this baby Rudin definition for Riemann integral:




Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, ldots, x_n$, where
$$ a = x_0 leq x_1 leq cdots leq x_{n-1} leq x_n = b.$$
We write
$$ Delta x_i = x_i - x_{i-1} qquad (i = 1, ldots, n). $$
Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put
$$
begin{align}
M_i &= sup f(x) qquad (x_{i-1} leq x leq x_i), \
m_i &= inf f(x) qquad (x_{i-1} leq x leq x_i), \
U(P, f) &= sum_{i=1}^n M_i Delta x_i, \
L(P, f) &= sum_{i=1}^n m_i Delta x_i,
end{align}
$$
and finally
$$
begin{align}
tag{1} overline{int_a^b} f dx &= inf U(P, f), \
tag{2} underline{int_a^b} f dx &= sup L(P, f),\,
end{align}
$$
where the $inf$ and the $sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.



If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$,
$$ tag{3} int_a^b f dx. $$




How do we get property $displaystyleint_a^b f dx=-displaystyleint_b^a f dx$ from this definition?



(Code borrowed from here)






real-analysis integration definition







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    You don't. That's the convention for defining the integral when $a>b$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:19














0












0








0





$begingroup$


I am using this baby Rudin definition for Riemann integral:




Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, ldots, x_n$, where
$$ a = x_0 leq x_1 leq cdots leq x_{n-1} leq x_n = b.$$
We write
$$ Delta x_i = x_i - x_{i-1} qquad (i = 1, ldots, n). $$
Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put
$$
begin{align}
M_i &= sup f(x) qquad (x_{i-1} leq x leq x_i), \
m_i &= inf f(x) qquad (x_{i-1} leq x leq x_i), \
U(P, f) &= sum_{i=1}^n M_i Delta x_i, \
L(P, f) &= sum_{i=1}^n m_i Delta x_i,
end{align}
$$
and finally
$$
begin{align}
tag{1} overline{int_a^b} f dx &= inf U(P, f), \
tag{2} underline{int_a^b} f dx &= sup L(P, f),\,
end{align}
$$
where the $inf$ and the $sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.



If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$,
$$ tag{3} int_a^b f dx. $$




How do we get property $displaystyleint_a^b f dx=-displaystyleint_b^a f dx$ from this definition?



(Code borrowed from here)






real-analysis integration definition







share|cite|improve this question









$endgroup$




I am using this baby Rudin definition for Riemann integral:




Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, ldots, x_n$, where
$$ a = x_0 leq x_1 leq cdots leq x_{n-1} leq x_n = b.$$
We write
$$ Delta x_i = x_i - x_{i-1} qquad (i = 1, ldots, n). $$
Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put
$$
begin{align}
M_i &= sup f(x) qquad (x_{i-1} leq x leq x_i), \
m_i &= inf f(x) qquad (x_{i-1} leq x leq x_i), \
U(P, f) &= sum_{i=1}^n M_i Delta x_i, \
L(P, f) &= sum_{i=1}^n m_i Delta x_i,
end{align}
$$
and finally
$$
begin{align}
tag{1} overline{int_a^b} f dx &= inf U(P, f), \
tag{2} underline{int_a^b} f dx &= sup L(P, f),\,
end{align}
$$
where the $inf$ and the $sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.



If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$,
$$ tag{3} int_a^b f dx. $$




How do we get property $displaystyleint_a^b f dx=-displaystyleint_b^a f dx$ from this definition?



(Code borrowed from here)






real-analysis integration definition




real-analysis integration definition






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 11:18









SilentSilent

2,90232152




2,90232152








  • 3




    $begingroup$
    You don't. That's the convention for defining the integral when $a>b$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:19














  • 3




    $begingroup$
    You don't. That's the convention for defining the integral when $a>b$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 29 '18 at 11:19








3




3




$begingroup$
You don't. That's the convention for defining the integral when $a>b$.
$endgroup$
– Lord Shark the Unknown
Dec 29 '18 at 11:19




$begingroup$
You don't. That's the convention for defining the integral when $a>b$.
$endgroup$
– Lord Shark the Unknown
Dec 29 '18 at 11:19










1 Answer
1






active

oldest

votes


















2












$begingroup$

You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $int_a^afdx=0$ and $int_b^afdx=-int_a^bfdx$.



Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
    $endgroup$
    – Silent
    Dec 29 '18 at 11:34








  • 1




    $begingroup$
    @Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
    $endgroup$
    – SinTan1729
    Dec 29 '18 at 11:49












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1 Answer
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1 Answer
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active

oldest

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2












$begingroup$

You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $int_a^afdx=0$ and $int_b^afdx=-int_a^bfdx$.



Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
    $endgroup$
    – Silent
    Dec 29 '18 at 11:34








  • 1




    $begingroup$
    @Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
    $endgroup$
    – SinTan1729
    Dec 29 '18 at 11:49
















2












$begingroup$

You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $int_a^afdx=0$ and $int_b^afdx=-int_a^bfdx$.



Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
    $endgroup$
    – Silent
    Dec 29 '18 at 11:34








  • 1




    $begingroup$
    @Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
    $endgroup$
    – SinTan1729
    Dec 29 '18 at 11:49














2












2








2





$begingroup$

You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $int_a^afdx=0$ and $int_b^afdx=-int_a^bfdx$.



Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.






share|cite|improve this answer









$endgroup$



You can't, because the definition you quoted deals with integrals for $a>b$. You have to further define that $int_a^afdx=0$ and $int_b^afdx=-int_a^bfdx$.



Why you ask? Because it's logical. Negative sign is often asigned to changing the orientation. Also, it plays nicely with the "area" interpretation of integrals.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 11:29









SinTan1729SinTan1729

2,682723




2,682723








  • 1




    $begingroup$
    Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
    $endgroup$
    – Silent
    Dec 29 '18 at 11:34








  • 1




    $begingroup$
    @Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
    $endgroup$
    – SinTan1729
    Dec 29 '18 at 11:49














  • 1




    $begingroup$
    Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
    $endgroup$
    – Silent
    Dec 29 '18 at 11:34








  • 1




    $begingroup$
    @Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
    $endgroup$
    – SinTan1729
    Dec 29 '18 at 11:49








1




1




$begingroup$
Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
$endgroup$
– Silent
Dec 29 '18 at 11:34






$begingroup$
Yeah. And also, integration by substitution becomes more applicable. Rudin gives theorem only for $varphi$ monotonically increasing, but Wikipedia says that it is true for any differentiable $varphi$.
$endgroup$
– Silent
Dec 29 '18 at 11:34






1




1




$begingroup$
@Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
$endgroup$
– SinTan1729
Dec 29 '18 at 11:49




$begingroup$
@Silent Yup. You can sort of break the interval into pieces where the function is monotonic and change the signs if you need to make it monotonically increasing to use that definition given in Rudin.
$endgroup$
– SinTan1729
Dec 29 '18 at 11:49


















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