Mixing
Proof of second Fundamental theorem of calculus
$begingroup$
Is there any other proof of this?
Second fundamental Theorem of Calculus: If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
My proof
Since $f$ is differentiable on [a,b], then $f'(t)$ exists for all $tin[a,b]$. For each $nin Bbb{N}$, let $P_n$ be an arbitrary partition such that
$$a=x_0<x_1<cdots<x_n=b.$$
Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $(a,b).$ So, by Mean Value Theorem, there exists $t_iin [x_{i-1},x_i]$, for $1leq ileq n$ such that
$$ f(x_i)-f(x_{i-1})=f'(t_i)(x_{i-1}-x_i).$$
Summing these up, we have
$$ f(b)-f(a)=sum^{n}_{i=1}left[f(x_i)-f(x_{i-1})right]=sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i).$$
Integrability of $f'$ on $[a,b]$ implies that
$$ sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i),$$
which implies that $$ L(f',P_n)=sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq f(b)-f(a)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i)=U(f',P_n),$$
As $ntoinfty,$ $$limlimits_{ntoinfty} L(f',P_n)=limlimits_{ntoinfty} U(f',P_n)=int^{b}_a f'(t)dt.$$
Therefore,
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
$endgroup$
add a comment |
$begingroup$
Is there any other proof of this?
Second fundamental Theorem of Calculus: If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
My proof
Since $f$ is differentiable on [a,b], then $f'(t)$ exists for all $tin[a,b]$. For each $nin Bbb{N}$, let $P_n$ be an arbitrary partition such that
$$a=x_0<x_1<cdots<x_n=b.$$
Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $(a,b).$ So, by Mean Value Theorem, there exists $t_iin [x_{i-1},x_i]$, for $1leq ileq n$ such that
$$ f(x_i)-f(x_{i-1})=f'(t_i)(x_{i-1}-x_i).$$
Summing these up, we have
$$ f(b)-f(a)=sum^{n}_{i=1}left[f(x_i)-f(x_{i-1})right]=sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i).$$
Integrability of $f'$ on $[a,b]$ implies that
$$ sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i),$$
which implies that $$ L(f',P_n)=sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq f(b)-f(a)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i)=U(f',P_n),$$
As $ntoinfty,$ $$limlimits_{ntoinfty} L(f',P_n)=limlimits_{ntoinfty} U(f',P_n)=int^{b}_a f'(t)dt.$$
Therefore,
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
$endgroup$
$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
1
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31
add a comment |
$begingroup$
Is there any other proof of this?
Second fundamental Theorem of Calculus: If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
My proof
Since $f$ is differentiable on [a,b], then $f'(t)$ exists for all $tin[a,b]$. For each $nin Bbb{N}$, let $P_n$ be an arbitrary partition such that
$$a=x_0<x_1<cdots<x_n=b.$$
Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $(a,b).$ So, by Mean Value Theorem, there exists $t_iin [x_{i-1},x_i]$, for $1leq ileq n$ such that
$$ f(x_i)-f(x_{i-1})=f'(t_i)(x_{i-1}-x_i).$$
Summing these up, we have
$$ f(b)-f(a)=sum^{n}_{i=1}left[f(x_i)-f(x_{i-1})right]=sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i).$$
Integrability of $f'$ on $[a,b]$ implies that
$$ sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i),$$
which implies that $$ L(f',P_n)=sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq f(b)-f(a)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i)=U(f',P_n),$$
As $ntoinfty,$ $$limlimits_{ntoinfty} L(f',P_n)=limlimits_{ntoinfty} U(f',P_n)=int^{b}_a f'(t)dt.$$
Therefore,
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
$endgroup$
Is there any other proof of this?
Second fundamental Theorem of Calculus: If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
My proof
Since $f$ is differentiable on [a,b], then $f'(t)$ exists for all $tin[a,b]$. For each $nin Bbb{N}$, let $P_n$ be an arbitrary partition such that
$$a=x_0<x_1<cdots<x_n=b.$$
Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $(a,b).$ So, by Mean Value Theorem, there exists $t_iin [x_{i-1},x_i]$, for $1leq ileq n$ such that
$$ f(x_i)-f(x_{i-1})=f'(t_i)(x_{i-1}-x_i).$$
Summing these up, we have
$$ f(b)-f(a)=sum^{n}_{i=1}left[f(x_i)-f(x_{i-1})right]=sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i).$$
Integrability of $f'$ on $[a,b]$ implies that
$$ sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i),$$
which implies that $$ L(f',P_n)=sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)leq f(b)-f(a)leq sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i)=U(f',P_n),$$
As $ntoinfty,$ $$limlimits_{ntoinfty} L(f',P_n)=limlimits_{ntoinfty} U(f',P_n)=int^{b}_a f'(t)dt.$$
Therefore,
$$int^{b}_a f'(t)dt=f(b)-f(a)$$
real-analysis calculus analysis proof-verification riemann-integration
real-analysis calculus analysis proof-verification riemann-integration
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
edited Jan 14 at 16:36
Omojola Micheal
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
asked Jan 14 at 16:09
Omojola MichealOmojola Micheal
1,875324
1,875324
$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
1
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31
add a comment |
$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
1
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31
$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
1
1
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31
add a comment |
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$begingroup$
I suggest that you include the tag "proof-verification" for this kind of question.
$endgroup$
– BigbearZzz
Jan 14 at 16:11
$begingroup$
@BigbearZzz: That's true but I believe this one is correct.
$endgroup$
– Omojola Micheal
Jan 14 at 16:25
$begingroup$
I meant that the tag is used for a question that would involve people checking the proof that you posted.
$endgroup$
– BigbearZzz
Jan 14 at 16:26
1
$begingroup$
@BigbearZzz: You are right! Sorry for my comments. You really wanted the best for me!
$endgroup$
– Omojola Micheal
Jan 14 at 16:31