Mixing
Riemann Integral of discontinuous function
$begingroup$
So I know you can integrate some discontinuous functions when the function is discontinuous at a finite number of points. So you can integrate for example this function on a interval [-1,3]:
$$ f(x):=begin{cases}
x & x< frac{1}{4} \
frac{1}{x} & xgeq frac{1}{4}
end{cases}
$$
But can you integrate this function on a interval the discontinuous point as an end point? Like [-1,1/4]?
integration riemann-integration discontinuous-functions
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
$endgroup$
add a comment |
$begingroup$
So I know you can integrate some discontinuous functions when the function is discontinuous at a finite number of points. So you can integrate for example this function on a interval [-1,3]:
$$ f(x):=begin{cases}
x & x< frac{1}{4} \
frac{1}{x} & xgeq frac{1}{4}
end{cases}
$$
But can you integrate this function on a interval the discontinuous point as an end point? Like [-1,1/4]?
integration riemann-integration discontinuous-functions
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
$endgroup$
$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
1
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36
add a comment |
$begingroup$
So I know you can integrate some discontinuous functions when the function is discontinuous at a finite number of points. So you can integrate for example this function on a interval [-1,3]:
$$ f(x):=begin{cases}
x & x< frac{1}{4} \
frac{1}{x} & xgeq frac{1}{4}
end{cases}
$$
But can you integrate this function on a interval the discontinuous point as an end point? Like [-1,1/4]?
integration riemann-integration discontinuous-functions
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
$endgroup$
So I know you can integrate some discontinuous functions when the function is discontinuous at a finite number of points. So you can integrate for example this function on a interval [-1,3]:
$$ f(x):=begin{cases}
x & x< frac{1}{4} \
frac{1}{x} & xgeq frac{1}{4}
end{cases}
$$
But can you integrate this function on a interval the discontinuous point as an end point? Like [-1,1/4]?
integration riemann-integration discontinuous-functions
integration riemann-integration discontinuous-functions
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
edited Jan 19 at 11:27
Dries De Witte
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
asked Jan 19 at 11:17
Dries De WitteDries De Witte
103
103
$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
1
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36
add a comment |
$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
1
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36
$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
1
1
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36
add a comment |
0
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$begingroup$
but this function $f$ is continuous?
$endgroup$
– BigbearZzz
Jan 19 at 11:22
$begingroup$
my mistake, got the wrong function
$endgroup$
– Dries De Witte
Jan 19 at 11:24
1
$begingroup$
It is right-continuous, hence integrable, if you redefine at 1/4
$endgroup$
– Tito Eliatron
Jan 19 at 11:36