Riemann integral and fundamental theorem of calculus
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1
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My statement in textbook is Suppose $H(x) = int_{a}^{x} h(t) dt$ is differentiable at $c in [a,b]$, then $h$ is continuous at c. I tried to provide a counter-example by considering the piecewise function:
$h : [-1,1] to mathbb{R}$
$$ h(x) = begin{cases}
1 & , x <0 \
-1 &, x geq 0
end{cases}$$
Is there something wrong with this?
real-analysis
asked Nov 15 at 1:39
Dong Le
516
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up vote
1
down vote
favorite
My statement in textbook is Suppose $H(x) = int_{a}^{x} h(t) dt$ is differentiable at $c in [a,b]$, then $h$ is continuous at c. I tried to provide a counter-example by considering the piecewise function:
$h : [-1,1] to mathbb{R}$
$$ h(x) = begin{cases}
1 & , x <0 \
-1 &, x geq 0
end{cases}$$
Is there something wrong with this?
real-analysis
asked Nov 15 at 1:39
Dong Le
516
Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My statement in textbook is Suppose $H(x) = int_{a}^{x} h(t) dt$ is differentiable at $c in [a,b]$, then $h$ is continuous at c. I tried to provide a counter-example by considering the piecewise function:
$h : [-1,1] to mathbb{R}$
$$ h(x) = begin{cases}
1 & , x <0 \
-1 &, x geq 0
end{cases}$$
Is there something wrong with this?
real-analysis
asked Nov 15 at 1:39
Dong Le
516
My statement in textbook is Suppose $H(x) = int_{a}^{x} h(t) dt$ is differentiable at $c in [a,b]$, then $h$ is continuous at c. I tried to provide a counter-example by considering the piecewise function:
$h : [-1,1] to mathbb{R}$
$$ h(x) = begin{cases}
1 & , x <0 \
-1 &, x geq 0
end{cases}$$
Is there something wrong with this?
real-analysis
real-analysis
asked Nov 15 at 1:39
Dong Le
516
asked Nov 15 at 1:39
Dong Le
516
asked Nov 15 at 1:39
Dong Le
516
asked Nov 15 at 1:39
Dong Le
516
asked Nov 15 at 1:39
Dong Le
516
516
Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12
add a comment |
Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12
Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12
Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12
add a comment |
1 Answer
1
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votes
up vote
3
down vote
We get $$H=begin{cases}1+x&x<0\1-x&xgeq0.end{cases}$$
This function is not differentiable at $0$.
answered Nov 15 at 1:52
Melody
39210
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We get $$H=begin{cases}1+x&x<0\1-x&xgeq0.end{cases}$$
This function is not differentiable at $0$.
answered Nov 15 at 1:52
Melody
39210
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
add a comment |
up vote
3
down vote
We get $$H=begin{cases}1+x&x<0\1-x&xgeq0.end{cases}$$
This function is not differentiable at $0$.
answered Nov 15 at 1:52
Melody
39210
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
add a comment |
up vote
3
down vote
up vote
3
down vote
We get $$H=begin{cases}1+x&x<0\1-x&xgeq0.end{cases}$$
This function is not differentiable at $0$.
answered Nov 15 at 1:52
Melody
39210
We get $$H=begin{cases}1+x&x<0\1-x&xgeq0.end{cases}$$
This function is not differentiable at $0$.
answered Nov 15 at 1:52
Melody
39210
answered Nov 15 at 1:52
Melody
39210
answered Nov 15 at 1:52
Melody
39210
answered Nov 15 at 1:52
Melody
39210
39210
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
add a comment |
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
1
1
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
More simply, that is $1-|x|$.
– mr_e_man
Nov 15 at 2:07
add a comment |
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Your book has a serious mistake. The theorem is if $h$ is continuous at $c$ then $H$ is differentiable at $c$ and not the other way around.
– Paramanand Singh
Nov 15 at 3:12