Mixing
Proof of first Fundamental theorem of calculus
$begingroup$
Can you please, check if my proof is correct?
Suppose that $f:[a,b]to Bbb{R}$ is continuous and $F(x)=int^{x}_{a}f(t)dt$, then $Fin C^{1}[a,b]$ and
$$dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
MY PROOF: Credits to Aweygan for the correction
Let $x_0in[a,b]$ and $epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $delta>0$ such that $|t-x_0|<delta$ implies $$|f(t)-f(x_0)|<epsilon.$$
Thus, $$f(x_0)=dfrac{1}{x-x_0}int^{x}_{x_0}f(x_0)dt,;;text{where};;xneq x_0.$$
For any $xin (a,b),$ with $0<|x-x_0|<delta,$ such that $x_1=min{x,x_0}$ and $x_2=max{x,x_0}$. So, we have
begin{align}left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \&leq dfrac{1}{|x-x_0|}int^{x}_{x_0} left|f(t)-f(x_0) right|dt\&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt\&< dfrac{1}{|x-x_0|}epsilon|x_1-x_2| \&leq dfrac{1}{|x-x_0|}epsilon|x-x_0| =epsilon end{align}
Hence,
$$Fin C^{1}[a,b];;text{and};;dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
$endgroup$
|
show 6 more comments
$begingroup$
Can you please, check if my proof is correct?
Suppose that $f:[a,b]to Bbb{R}$ is continuous and $F(x)=int^{x}_{a}f(t)dt$, then $Fin C^{1}[a,b]$ and
$$dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
MY PROOF: Credits to Aweygan for the correction
Let $x_0in[a,b]$ and $epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $delta>0$ such that $|t-x_0|<delta$ implies $$|f(t)-f(x_0)|<epsilon.$$
Thus, $$f(x_0)=dfrac{1}{x-x_0}int^{x}_{x_0}f(x_0)dt,;;text{where};;xneq x_0.$$
For any $xin (a,b),$ with $0<|x-x_0|<delta,$ such that $x_1=min{x,x_0}$ and $x_2=max{x,x_0}$. So, we have
begin{align}left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \&leq dfrac{1}{|x-x_0|}int^{x}_{x_0} left|f(t)-f(x_0) right|dt\&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt\&< dfrac{1}{|x-x_0|}epsilon|x_1-x_2| \&leq dfrac{1}{|x-x_0|}epsilon|x-x_0| =epsilon end{align}
Hence,
$$Fin C^{1}[a,b];;text{and};;dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
$endgroup$
$begingroup$
It's not sufficient to prove the limit is $f(x)$ only from one side.
$endgroup$
– egreg
Jan 14 at 13:33
$begingroup$
You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
$endgroup$
– Aweygan
Jan 14 at 13:47
$begingroup$
You're still computing a one-sided limit.
$endgroup$
– egreg
Jan 14 at 13:49
$begingroup$
It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
$endgroup$
– Aweygan
Jan 14 at 13:55
$begingroup$
@Aweygan: I want to rewrite it, now. Some moments, please!
$endgroup$
– Omojola Micheal
Jan 14 at 13:56
|
show 6 more comments
$begingroup$
Can you please, check if my proof is correct?
Suppose that $f:[a,b]to Bbb{R}$ is continuous and $F(x)=int^{x}_{a}f(t)dt$, then $Fin C^{1}[a,b]$ and
$$dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
MY PROOF: Credits to Aweygan for the correction
Let $x_0in[a,b]$ and $epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $delta>0$ such that $|t-x_0|<delta$ implies $$|f(t)-f(x_0)|<epsilon.$$
Thus, $$f(x_0)=dfrac{1}{x-x_0}int^{x}_{x_0}f(x_0)dt,;;text{where};;xneq x_0.$$
For any $xin (a,b),$ with $0<|x-x_0|<delta,$ such that $x_1=min{x,x_0}$ and $x_2=max{x,x_0}$. So, we have
begin{align}left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \&leq dfrac{1}{|x-x_0|}int^{x}_{x_0} left|f(t)-f(x_0) right|dt\&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt\&< dfrac{1}{|x-x_0|}epsilon|x_1-x_2| \&leq dfrac{1}{|x-x_0|}epsilon|x-x_0| =epsilon end{align}
Hence,
$$Fin C^{1}[a,b];;text{and};;dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
$endgroup$
Can you please, check if my proof is correct?
Suppose that $f:[a,b]to Bbb{R}$ is continuous and $F(x)=int^{x}_{a}f(t)dt$, then $Fin C^{1}[a,b]$ and
$$dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
MY PROOF: Credits to Aweygan for the correction
Let $x_0in[a,b]$ and $epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $delta>0$ such that $|t-x_0|<delta$ implies $$|f(t)-f(x_0)|<epsilon.$$
Thus, $$f(x_0)=dfrac{1}{x-x_0}int^{x}_{x_0}f(x_0)dt,;;text{where};;xneq x_0.$$
For any $xin (a,b),$ with $0<|x-x_0|<delta,$ such that $x_1=min{x,x_0}$ and $x_2=max{x,x_0}$. So, we have
begin{align}left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \&leq dfrac{1}{|x-x_0|}int^{x}_{x_0} left|f(t)-f(x_0) right|dt\&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt\&< dfrac{1}{|x-x_0|}epsilon|x_1-x_2| \&leq dfrac{1}{|x-x_0|}epsilon|x-x_0| =epsilon end{align}
Hence,
$$Fin C^{1}[a,b];;text{and};;dfrac{d}{dx}int^{x}_{a}f(t)dt:=F'(x)=f(x)$$
real-analysis calculus analysis proof-verification riemann-integration
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
edited Jan 14 at 15:15
Omojola Micheal
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 13:26
Omojola MichealOmojola Micheal
1,875324
1,875324
$begingroup$
It's not sufficient to prove the limit is $f(x)$ only from one side.
$endgroup$
– egreg
Jan 14 at 13:33
$begingroup$
You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
$endgroup$
– Aweygan
Jan 14 at 13:47
$begingroup$
You're still computing a one-sided limit.
$endgroup$
– egreg
Jan 14 at 13:49
$begingroup$
It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
$endgroup$
– Aweygan
Jan 14 at 13:55
$begingroup$
@Aweygan: I want to rewrite it, now. Some moments, please!
$endgroup$
– Omojola Micheal
Jan 14 at 13:56
|
show 6 more comments
$begingroup$
It's not sufficient to prove the limit is $f(x)$ only from one side.
$endgroup$
– egreg
Jan 14 at 13:33
$begingroup$
You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
$endgroup$
– Aweygan
Jan 14 at 13:47
$begingroup$
You're still computing a one-sided limit.
$endgroup$
– egreg
Jan 14 at 13:49
$begingroup$
It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
$endgroup$
– Aweygan
Jan 14 at 13:55
$begingroup$
@Aweygan: I want to rewrite it, now. Some moments, please!
$endgroup$
– Omojola Micheal
Jan 14 at 13:56
$begingroup$
It's not sufficient to prove the limit is $f(x)$ only from one side.
$endgroup$
– egreg
Jan 14 at 13:33
$begingroup$
It's not sufficient to prove the limit is $f(x)$ only from one side.
$endgroup$
– egreg
Jan 14 at 13:33
$begingroup$
You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
$endgroup$
– Aweygan
Jan 14 at 13:47
$begingroup$
You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
$endgroup$
– Aweygan
Jan 14 at 13:47
$begingroup$
You're still computing a one-sided limit.
$endgroup$
– egreg
Jan 14 at 13:49
$begingroup$
You're still computing a one-sided limit.
$endgroup$
– egreg
Jan 14 at 13:49
$begingroup$
It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
$endgroup$
– Aweygan
Jan 14 at 13:55
$begingroup$
It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
$endgroup$
– Aweygan
Jan 14 at 13:55
$begingroup$
@Aweygan: I want to rewrite it, now. Some moments, please!
$endgroup$
– Omojola Micheal
Jan 14 at 13:56
$begingroup$
@Aweygan: I want to rewrite it, now. Some moments, please!
$endgroup$
– Omojola Micheal
Jan 14 at 13:56
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=min{x,x_0}$, $x_2=max{x,x_0}$ and do the following:
begin{align}
left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \
&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt
\&< dfrac{1}{|x-x_0|}epsilon|x-x_0| \
&=epsilon.
end{align}
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
add a comment |
$begingroup$
You can also directly use the mean value theorem for integrals:
$$
lim_{hdownarrow0}frac{int_{a}^{x}f(t)dt-int_{a}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{int_{x}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{f(c_{h})h}{h}
$$
where $c_{h}$ is between $x$ and $x+h$. Therefore, $lim_{hdownarrow0}c_{h}=x$.
By continuity, $lim_{hdownarrow0}f(c_{h})=f(x)$.
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=min{x,x_0}$, $x_2=max{x,x_0}$ and do the following:
begin{align}
left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \
&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt
\&< dfrac{1}{|x-x_0|}epsilon|x-x_0| \
&=epsilon.
end{align}
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
add a comment |
$begingroup$
It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=min{x,x_0}$, $x_2=max{x,x_0}$ and do the following:
begin{align}
left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \
&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt
\&< dfrac{1}{|x-x_0|}epsilon|x-x_0| \
&=epsilon.
end{align}
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
add a comment |
$begingroup$
It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=min{x,x_0}$, $x_2=max{x,x_0}$ and do the following:
begin{align}
left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \
&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt
\&< dfrac{1}{|x-x_0|}epsilon|x-x_0| \
&=epsilon.
end{align}
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
$endgroup$
It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=min{x,x_0}$, $x_2=max{x,x_0}$ and do the following:
begin{align}
left| dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) right|&= left| dfrac{1}{x-x_0}int^{x}_{x_0}(f(t)-f(x_0))dt right| \
&leq dfrac{1}{|x-x_0|}int^{x_2}_{x_1} left|f(t)-f(x_0) right|dt
\&< dfrac{1}{|x-x_0|}epsilon|x-x_0| \
&=epsilon.
end{align}
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
edited Jan 14 at 15:08
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
answered Jan 14 at 15:05
AweyganAweygan
14.2k21441
14.2k21441
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
add a comment |
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
Thanks a lot for the explanation (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 15:07
add a comment |
$begingroup$
You can also directly use the mean value theorem for integrals:
$$
lim_{hdownarrow0}frac{int_{a}^{x}f(t)dt-int_{a}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{int_{x}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{f(c_{h})h}{h}
$$
where $c_{h}$ is between $x$ and $x+h$. Therefore, $lim_{hdownarrow0}c_{h}=x$.
By continuity, $lim_{hdownarrow0}f(c_{h})=f(x)$.
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
$endgroup$
add a comment |
$begingroup$
You can also directly use the mean value theorem for integrals:
$$
lim_{hdownarrow0}frac{int_{a}^{x}f(t)dt-int_{a}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{int_{x}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{f(c_{h})h}{h}
$$
where $c_{h}$ is between $x$ and $x+h$. Therefore, $lim_{hdownarrow0}c_{h}=x$.
By continuity, $lim_{hdownarrow0}f(c_{h})=f(x)$.
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
$endgroup$
add a comment |
$begingroup$
You can also directly use the mean value theorem for integrals:
$$
lim_{hdownarrow0}frac{int_{a}^{x}f(t)dt-int_{a}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{int_{x}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{f(c_{h})h}{h}
$$
where $c_{h}$ is between $x$ and $x+h$. Therefore, $lim_{hdownarrow0}c_{h}=x$.
By continuity, $lim_{hdownarrow0}f(c_{h})=f(x)$.
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
$endgroup$
You can also directly use the mean value theorem for integrals:
$$
lim_{hdownarrow0}frac{int_{a}^{x}f(t)dt-int_{a}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{int_{x}^{x+h}f(t)dt}{h}=lim_{hdownarrow0}frac{f(c_{h})h}{h}
$$
where $c_{h}$ is between $x$ and $x+h$. Therefore, $lim_{hdownarrow0}c_{h}=x$.
By continuity, $lim_{hdownarrow0}f(c_{h})=f(x)$.
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
answered Jan 14 at 15:12
parsiadparsiad
17.7k32353
17.7k32353
add a comment |
add a comment |
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It's not sufficient to prove the limit is $f(x)$ only from one side.
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– egreg
Jan 14 at 13:33
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You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-delta<h<0$).
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– Aweygan
Jan 14 at 13:47
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You're still computing a one-sided limit.
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– egreg
Jan 14 at 13:49
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It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this.
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– Aweygan
Jan 14 at 13:55
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@Aweygan: I want to rewrite it, now. Some moments, please!
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– Omojola Micheal
Jan 14 at 13:56