Mixing
Antiderivative Supremum inequality $F(b)-F(a) le (b-a)sup{f(x):x in [a,b]}$
$begingroup$
Hello everyone I am suppose to show the following:
Let F be the antiderivative of f, show that
$$ F(b)-F(a) le sup{f(x):x in [a,b]} $$
I figured I could use the Mean-Value-Theorem since $F'(x)=f(x)$ so
$frac{F(b)-F(a)}{b-a}=f(x)$ for some $x in [a,b]$ but then the $(b-a)$ part gets in the way. Maybe some trick with the more general form $frac{F(b)-F(a)}{g(b)-g(a)}=frac{f(x)}{g'(x)}$?
I think one could solve it using some facts about the Riemann Integral and upper Darboux sums like:
$$F(b)-F(a) = int_a^b f(x) dx le sum^{n-1}_{i=0} sup_{tin[x_i,x_{i+1}]}f(t)(x_{i+1}-x_i)$$
since it has all the parts but we are not allowed to use this yet.
Thank you in advance!
Edit
There was an error in the assignment. It was to show that:
$$ F(b)-F(a) le (b-a)sup{f(x):x in [a,b]} $$
Which is just the Mean-Value-Theorem:
$$frac{F(b)-F(a)}{b-a}=f(x)lesup{f(x):x in [a,b]}$$
and done.
real-analysis integration analysis inequality supremum-and-infimum
edited Jan 19 at 20:35
clathratus
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
$endgroup$
add a comment |
$begingroup$
Hello everyone I am suppose to show the following:
Let F be the antiderivative of f, show that
$$ F(b)-F(a) le sup{f(x):x in [a,b]} $$
I figured I could use the Mean-Value-Theorem since $F'(x)=f(x)$ so
$frac{F(b)-F(a)}{b-a}=f(x)$ for some $x in [a,b]$ but then the $(b-a)$ part gets in the way. Maybe some trick with the more general form $frac{F(b)-F(a)}{g(b)-g(a)}=frac{f(x)}{g'(x)}$?
I think one could solve it using some facts about the Riemann Integral and upper Darboux sums like:
$$F(b)-F(a) = int_a^b f(x) dx le sum^{n-1}_{i=0} sup_{tin[x_i,x_{i+1}]}f(t)(x_{i+1}-x_i)$$
since it has all the parts but we are not allowed to use this yet.
Thank you in advance!
Edit
There was an error in the assignment. It was to show that:
$$ F(b)-F(a) le (b-a)sup{f(x):x in [a,b]} $$
Which is just the Mean-Value-Theorem:
$$frac{F(b)-F(a)}{b-a}=f(x)lesup{f(x):x in [a,b]}$$
and done.
real-analysis integration analysis inequality supremum-and-infimum
edited Jan 19 at 20:35
clathratus
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
$endgroup$
add a comment |
$begingroup$
Hello everyone I am suppose to show the following:
Let F be the antiderivative of f, show that
$$ F(b)-F(a) le sup{f(x):x in [a,b]} $$
I figured I could use the Mean-Value-Theorem since $F'(x)=f(x)$ so
$frac{F(b)-F(a)}{b-a}=f(x)$ for some $x in [a,b]$ but then the $(b-a)$ part gets in the way. Maybe some trick with the more general form $frac{F(b)-F(a)}{g(b)-g(a)}=frac{f(x)}{g'(x)}$?
I think one could solve it using some facts about the Riemann Integral and upper Darboux sums like:
$$F(b)-F(a) = int_a^b f(x) dx le sum^{n-1}_{i=0} sup_{tin[x_i,x_{i+1}]}f(t)(x_{i+1}-x_i)$$
since it has all the parts but we are not allowed to use this yet.
Thank you in advance!
Edit
There was an error in the assignment. It was to show that:
$$ F(b)-F(a) le (b-a)sup{f(x):x in [a,b]} $$
Which is just the Mean-Value-Theorem:
$$frac{F(b)-F(a)}{b-a}=f(x)lesup{f(x):x in [a,b]}$$
and done.
real-analysis integration analysis inequality supremum-and-infimum
edited Jan 19 at 20:35
clathratus
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
$endgroup$
Hello everyone I am suppose to show the following:
Let F be the antiderivative of f, show that
$$ F(b)-F(a) le sup{f(x):x in [a,b]} $$
I figured I could use the Mean-Value-Theorem since $F'(x)=f(x)$ so
$frac{F(b)-F(a)}{b-a}=f(x)$ for some $x in [a,b]$ but then the $(b-a)$ part gets in the way. Maybe some trick with the more general form $frac{F(b)-F(a)}{g(b)-g(a)}=frac{f(x)}{g'(x)}$?
I think one could solve it using some facts about the Riemann Integral and upper Darboux sums like:
$$F(b)-F(a) = int_a^b f(x) dx le sum^{n-1}_{i=0} sup_{tin[x_i,x_{i+1}]}f(t)(x_{i+1}-x_i)$$
since it has all the parts but we are not allowed to use this yet.
Thank you in advance!
Edit
There was an error in the assignment. It was to show that:
$$ F(b)-F(a) le (b-a)sup{f(x):x in [a,b]} $$
Which is just the Mean-Value-Theorem:
$$frac{F(b)-F(a)}{b-a}=f(x)lesup{f(x):x in [a,b]}$$
and done.
real-analysis integration analysis inequality supremum-and-infimum
real-analysis integration analysis inequality supremum-and-infimum
edited Jan 19 at 20:35
clathratus
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
edited Jan 19 at 20:35
clathratus
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
edited Jan 19 at 20:35
clathratus
4,725337
edited Jan 19 at 20:35
clathratus
4,725337
edited Jan 19 at 20:35
clathratus
4,725337
4,725337
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
asked Jan 19 at 0:00
MatthiasPichlerMatthiasPichler
424
424
add a comment |
add a comment |
1 Answer
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$begingroup$
That's not true, a couterexample is $f(x)=1$.
So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=sup_{[a,b]}f(x)$.
answered Jan 19 at 0:14
ecrinecrin
3477
$endgroup$
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's not true, a couterexample is $f(x)=1$.
So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=sup_{[a,b]}f(x)$.
answered Jan 19 at 0:14
ecrinecrin
3477
$endgroup$
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
add a comment |
$begingroup$
That's not true, a couterexample is $f(x)=1$.
So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=sup_{[a,b]}f(x)$.
answered Jan 19 at 0:14
ecrinecrin
3477
$endgroup$
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
add a comment |
$begingroup$
That's not true, a couterexample is $f(x)=1$.
So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=sup_{[a,b]}f(x)$.
answered Jan 19 at 0:14
ecrinecrin
3477
$endgroup$
That's not true, a couterexample is $f(x)=1$.
So that $F(x)=x+c$, then if $b>a+1$ $F(b)-F(a)=b-a>1=sup_{[a,b]}f(x)$.
answered Jan 19 at 0:14
ecrinecrin
3477
answered Jan 19 at 0:14
ecrinecrin
3477
answered Jan 19 at 0:14
ecrinecrin
3477
answered Jan 19 at 0:14
ecrinecrin
3477
3477
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
add a comment |
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
$begingroup$
Yes indeed, the Professor just posted an update to the assignment. Thank you
$endgroup$
– MatthiasPichler
Jan 19 at 0:27
add a comment |
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