Relaxing hypotheses for the mean value theorem for integrals
If a function $f$ is continuous in $[0,Delta]$ it is pretty easy to prove that
$$
exists cin(0,Delta):frac{1}{Delta},int_{0}^{Delta}f(t)dt=f(c)
$$
It is enough to apply Lagrange's to the function $F(t)=int_0^tf(s)ds$. Is it possible to derive the same result assuming only that $f$ is Riemann-integrable in $[0,Delta]$ ?
riemann-integration
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
add a comment |
If a function $f$ is continuous in $[0,Delta]$ it is pretty easy to prove that
$$
exists cin(0,Delta):frac{1}{Delta},int_{0}^{Delta}f(t)dt=f(c)
$$
It is enough to apply Lagrange's to the function $F(t)=int_0^tf(s)ds$. Is it possible to derive the same result assuming only that $f$ is Riemann-integrable in $[0,Delta]$ ?
riemann-integration
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
add a comment |
If a function $f$ is continuous in $[0,Delta]$ it is pretty easy to prove that
$$
exists cin(0,Delta):frac{1}{Delta},int_{0}^{Delta}f(t)dt=f(c)
$$
It is enough to apply Lagrange's to the function $F(t)=int_0^tf(s)ds$. Is it possible to derive the same result assuming only that $f$ is Riemann-integrable in $[0,Delta]$ ?
riemann-integration
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
If a function $f$ is continuous in $[0,Delta]$ it is pretty easy to prove that
$$
exists cin(0,Delta):frac{1}{Delta},int_{0}^{Delta}f(t)dt=f(c)
$$
It is enough to apply Lagrange's to the function $F(t)=int_0^tf(s)ds$. Is it possible to derive the same result assuming only that $f$ is Riemann-integrable in $[0,Delta]$ ?
riemann-integration
riemann-integration
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
asked Nov 21 '18 at 16:33
AlmostSureUser
306417
306417
add a comment |
add a comment |
1 Answer
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No. For example consider the Riemann integrable function $f:[0,1]rightarrow {0,1}$ defined by
$$f(x)=
begin{cases}
0, & 0leq xleq frac{1}{2}\
1, & frac{1}{2}<xleq 1.
end{cases}$$
Then $int_{0}^{1}f(x)dx=frac{1}{2}$(For this example $Delta=1$) which is not equals to $f(c)$ for any $cin[0,1].$
answered Nov 21 '18 at 16:51
Surajit
5689
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
No. For example consider the Riemann integrable function $f:[0,1]rightarrow {0,1}$ defined by
$$f(x)=
begin{cases}
0, & 0leq xleq frac{1}{2}\
1, & frac{1}{2}<xleq 1.
end{cases}$$
Then $int_{0}^{1}f(x)dx=frac{1}{2}$(For this example $Delta=1$) which is not equals to $f(c)$ for any $cin[0,1].$
answered Nov 21 '18 at 16:51
Surajit
5689
add a comment |
No. For example consider the Riemann integrable function $f:[0,1]rightarrow {0,1}$ defined by
$$f(x)=
begin{cases}
0, & 0leq xleq frac{1}{2}\
1, & frac{1}{2}<xleq 1.
end{cases}$$
Then $int_{0}^{1}f(x)dx=frac{1}{2}$(For this example $Delta=1$) which is not equals to $f(c)$ for any $cin[0,1].$
answered Nov 21 '18 at 16:51
Surajit
5689
add a comment |
No. For example consider the Riemann integrable function $f:[0,1]rightarrow {0,1}$ defined by
$$f(x)=
begin{cases}
0, & 0leq xleq frac{1}{2}\
1, & frac{1}{2}<xleq 1.
end{cases}$$
Then $int_{0}^{1}f(x)dx=frac{1}{2}$(For this example $Delta=1$) which is not equals to $f(c)$ for any $cin[0,1].$
answered Nov 21 '18 at 16:51
Surajit
5689
No. For example consider the Riemann integrable function $f:[0,1]rightarrow {0,1}$ defined by
$$f(x)=
begin{cases}
0, & 0leq xleq frac{1}{2}\
1, & frac{1}{2}<xleq 1.
end{cases}$$
Then $int_{0}^{1}f(x)dx=frac{1}{2}$(For this example $Delta=1$) which is not equals to $f(c)$ for any $cin[0,1].$
answered Nov 21 '18 at 16:51
Surajit
5689
answered Nov 21 '18 at 16:51
Surajit
5689
answered Nov 21 '18 at 16:51
Surajit
5689
answered Nov 21 '18 at 16:51
Surajit
5689
5689
add a comment |
add a comment |
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