Additivity of integral
$begingroup$
Let $a<c<b$. The function $f$ is integrable over $]a,b[ iff f$ is integrable over $]a,c[$ and $]c,b[$. We get: $int_a^bf=int_a^cf+int_c^bf $
In the proof of this theorem ($Leftarrow$), we take an arbitrary partition $pi_L$ of $]a,c[$ and a partition $pi_R$ of $]c,b[$. Then we consider the partition $pi := pi_L cup pi_R$ of $]a,b[$.
We get that $s_{pi_L}+s_{pi_R} = s_pi le underline{int_a^b} f$. Why does this equality hold here?
calculus
asked Jan 12 at 8:09
ZacharyZachary
3229
$endgroup$
add a comment |
$begingroup$
Let $a<c<b$. The function $f$ is integrable over $]a,b[ iff f$ is integrable over $]a,c[$ and $]c,b[$. We get: $int_a^bf=int_a^cf+int_c^bf $
In the proof of this theorem ($Leftarrow$), we take an arbitrary partition $pi_L$ of $]a,c[$ and a partition $pi_R$ of $]c,b[$. Then we consider the partition $pi := pi_L cup pi_R$ of $]a,b[$.
We get that $s_{pi_L}+s_{pi_R} = s_pi le underline{int_a^b} f$. Why does this equality hold here?
calculus
asked Jan 12 at 8:09
ZacharyZachary
3229
$endgroup$
add a comment |
$begingroup$
Let $a<c<b$. The function $f$ is integrable over $]a,b[ iff f$ is integrable over $]a,c[$ and $]c,b[$. We get: $int_a^bf=int_a^cf+int_c^bf $
In the proof of this theorem ($Leftarrow$), we take an arbitrary partition $pi_L$ of $]a,c[$ and a partition $pi_R$ of $]c,b[$. Then we consider the partition $pi := pi_L cup pi_R$ of $]a,b[$.
We get that $s_{pi_L}+s_{pi_R} = s_pi le underline{int_a^b} f$. Why does this equality hold here?
calculus
asked Jan 12 at 8:09
ZacharyZachary
3229
$endgroup$
Let $a<c<b$. The function $f$ is integrable over $]a,b[ iff f$ is integrable over $]a,c[$ and $]c,b[$. We get: $int_a^bf=int_a^cf+int_c^bf $
In the proof of this theorem ($Leftarrow$), we take an arbitrary partition $pi_L$ of $]a,c[$ and a partition $pi_R$ of $]c,b[$. Then we consider the partition $pi := pi_L cup pi_R$ of $]a,b[$.
We get that $s_{pi_L}+s_{pi_R} = s_pi le underline{int_a^b} f$. Why does this equality hold here?
calculus
calculus
asked Jan 12 at 8:09
ZacharyZachary
3229
asked Jan 12 at 8:09
ZacharyZachary
3229
asked Jan 12 at 8:09
ZacharyZachary
3229
asked Jan 12 at 8:09
ZacharyZachary
3229
asked Jan 12 at 8:09
ZacharyZachary
3229
3229
add a comment |
add a comment |
1 Answer
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The term $underline{int_a^b} f(x),dx$ is defined as the supremum of the set $s_{pi}$ where $pi$ is a generic partition of the interval $[a,b]$. In particular $pi_L cup pi_R$ is a partition of $[a,b]$ and
$$s_{pi_L cup pi_R}leq underline{int_a^b} f(x),dx.$$
Moreover, if $pi_L$ is $a=x_0<x_1<dots<x_n=c$ and $pi_R$ is $c=x_{n}<x_{n+1}<dots<x_{n+m}=b$ then $pi_L cup pi_R$ is $a=x_0<x_1<dots<x_{n+m}=b$ and
$$s_{pi_L}+s_{pi_R}=sum_{k=1}^n m_k(x_k-x_{k-1})+sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{pi_L cup pi_R}$$
where $m_k=inf_{tin [x_{k-1},x_k]} f(t).$
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The term $underline{int_a^b} f(x),dx$ is defined as the supremum of the set $s_{pi}$ where $pi$ is a generic partition of the interval $[a,b]$. In particular $pi_L cup pi_R$ is a partition of $[a,b]$ and
$$s_{pi_L cup pi_R}leq underline{int_a^b} f(x),dx.$$
Moreover, if $pi_L$ is $a=x_0<x_1<dots<x_n=c$ and $pi_R$ is $c=x_{n}<x_{n+1}<dots<x_{n+m}=b$ then $pi_L cup pi_R$ is $a=x_0<x_1<dots<x_{n+m}=b$ and
$$s_{pi_L}+s_{pi_R}=sum_{k=1}^n m_k(x_k-x_{k-1})+sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{pi_L cup pi_R}$$
where $m_k=inf_{tin [x_{k-1},x_k]} f(t).$
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
add a comment |
$begingroup$
The term $underline{int_a^b} f(x),dx$ is defined as the supremum of the set $s_{pi}$ where $pi$ is a generic partition of the interval $[a,b]$. In particular $pi_L cup pi_R$ is a partition of $[a,b]$ and
$$s_{pi_L cup pi_R}leq underline{int_a^b} f(x),dx.$$
Moreover, if $pi_L$ is $a=x_0<x_1<dots<x_n=c$ and $pi_R$ is $c=x_{n}<x_{n+1}<dots<x_{n+m}=b$ then $pi_L cup pi_R$ is $a=x_0<x_1<dots<x_{n+m}=b$ and
$$s_{pi_L}+s_{pi_R}=sum_{k=1}^n m_k(x_k-x_{k-1})+sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{pi_L cup pi_R}$$
where $m_k=inf_{tin [x_{k-1},x_k]} f(t).$
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
$endgroup$
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
add a comment |
$begingroup$
The term $underline{int_a^b} f(x),dx$ is defined as the supremum of the set $s_{pi}$ where $pi$ is a generic partition of the interval $[a,b]$. In particular $pi_L cup pi_R$ is a partition of $[a,b]$ and
$$s_{pi_L cup pi_R}leq underline{int_a^b} f(x),dx.$$
Moreover, if $pi_L$ is $a=x_0<x_1<dots<x_n=c$ and $pi_R$ is $c=x_{n}<x_{n+1}<dots<x_{n+m}=b$ then $pi_L cup pi_R$ is $a=x_0<x_1<dots<x_{n+m}=b$ and
$$s_{pi_L}+s_{pi_R}=sum_{k=1}^n m_k(x_k-x_{k-1})+sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{pi_L cup pi_R}$$
where $m_k=inf_{tin [x_{k-1},x_k]} f(t).$
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
$endgroup$
The term $underline{int_a^b} f(x),dx$ is defined as the supremum of the set $s_{pi}$ where $pi$ is a generic partition of the interval $[a,b]$. In particular $pi_L cup pi_R$ is a partition of $[a,b]$ and
$$s_{pi_L cup pi_R}leq underline{int_a^b} f(x),dx.$$
Moreover, if $pi_L$ is $a=x_0<x_1<dots<x_n=c$ and $pi_R$ is $c=x_{n}<x_{n+1}<dots<x_{n+m}=b$ then $pi_L cup pi_R$ is $a=x_0<x_1<dots<x_{n+m}=b$ and
$$s_{pi_L}+s_{pi_R}=sum_{k=1}^n m_k(x_k-x_{k-1})+sum_{k=n+1}^{n+m} m_k(x_k-x_{k-1})=sum_{k=1}^{n+m}m_k(x_k-x_{k-1}) =s_{pi_L cup pi_R}$$
where $m_k=inf_{tin [x_{k-1},x_k]} f(t).$
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
edited Jan 12 at 8:29
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
answered Jan 12 at 8:20
Robert ZRobert Z
102k1072145
102k1072145
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
add a comment |
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
@Zachary Any further doubts?
$endgroup$
– Robert Z
Jan 12 at 8:40
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
$begingroup$
No, definitely not! Very clear explanation. Thank you!
$endgroup$
– Zachary
Jan 12 at 8:44
add a comment |
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