$int_{-h}^{h}f(x)g(x)dx=f(C_h)int_{-h}^{h}g(x)dx$
$begingroup$
$f:[-1,1]rightarrow mathbb{R}$ is continuous and $g:[-1,1]rightarrow mathbb{R}$ is nonnegative integrable. Show that there exist $0<C_h<h$ with $0<h<1$ such that
$$int_{-h}^{h}f(x)g(x)dx=f(C_h)int_{-h}^{h}g(x)dx$$
I think this question can be solved by mean value theorem for integral, but still don't know how to solve it. Could anyone give me some advice?
calculus
asked Jan 9 at 16:00
ReinherdReinherd
764
$endgroup$
add a comment |
$begingroup$
$f:[-1,1]rightarrow mathbb{R}$ is continuous and $g:[-1,1]rightarrow mathbb{R}$ is nonnegative integrable. Show that there exist $0<C_h<h$ with $0<h<1$ such that
$$int_{-h}^{h}f(x)g(x)dx=f(C_h)int_{-h}^{h}g(x)dx$$
I think this question can be solved by mean value theorem for integral, but still don't know how to solve it. Could anyone give me some advice?
calculus
asked Jan 9 at 16:00
ReinherdReinherd
764
$endgroup$
$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20
add a comment |
$begingroup$
$f:[-1,1]rightarrow mathbb{R}$ is continuous and $g:[-1,1]rightarrow mathbb{R}$ is nonnegative integrable. Show that there exist $0<C_h<h$ with $0<h<1$ such that
$$int_{-h}^{h}f(x)g(x)dx=f(C_h)int_{-h}^{h}g(x)dx$$
I think this question can be solved by mean value theorem for integral, but still don't know how to solve it. Could anyone give me some advice?
calculus
asked Jan 9 at 16:00
ReinherdReinherd
764
$endgroup$
$f:[-1,1]rightarrow mathbb{R}$ is continuous and $g:[-1,1]rightarrow mathbb{R}$ is nonnegative integrable. Show that there exist $0<C_h<h$ with $0<h<1$ such that
$$int_{-h}^{h}f(x)g(x)dx=f(C_h)int_{-h}^{h}g(x)dx$$
I think this question can be solved by mean value theorem for integral, but still don't know how to solve it. Could anyone give me some advice?
calculus
calculus
asked Jan 9 at 16:00
ReinherdReinherd
764
asked Jan 9 at 16:00
ReinherdReinherd
764
asked Jan 9 at 16:00
ReinherdReinherd
764
asked Jan 9 at 16:00
ReinherdReinherd
764
asked Jan 9 at 16:00
ReinherdReinherd
764
764
$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20
add a comment |
$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20
$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't think this is correct.
Let $f(x)=x, g(x)=x^2$, then
$$
int_{-h}^h fg = 0, int_{-h}^h g >0,
$$
then if the equation holds, then $f(C_h)=0$, then $C_h = 0$, contradiction.
answered Jan 9 at 16:08
xbhxbh
6,2151522
$endgroup$
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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votes
$begingroup$
I don't think this is correct.
Let $f(x)=x, g(x)=x^2$, then
$$
int_{-h}^h fg = 0, int_{-h}^h g >0,
$$
then if the equation holds, then $f(C_h)=0$, then $C_h = 0$, contradiction.
answered Jan 9 at 16:08
xbhxbh
6,2151522
$endgroup$
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
add a comment |
$begingroup$
I don't think this is correct.
Let $f(x)=x, g(x)=x^2$, then
$$
int_{-h}^h fg = 0, int_{-h}^h g >0,
$$
then if the equation holds, then $f(C_h)=0$, then $C_h = 0$, contradiction.
answered Jan 9 at 16:08
xbhxbh
6,2151522
$endgroup$
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
add a comment |
$begingroup$
I don't think this is correct.
Let $f(x)=x, g(x)=x^2$, then
$$
int_{-h}^h fg = 0, int_{-h}^h g >0,
$$
then if the equation holds, then $f(C_h)=0$, then $C_h = 0$, contradiction.
answered Jan 9 at 16:08
xbhxbh
6,2151522
$endgroup$
I don't think this is correct.
Let $f(x)=x, g(x)=x^2$, then
$$
int_{-h}^h fg = 0, int_{-h}^h g >0,
$$
then if the equation holds, then $f(C_h)=0$, then $C_h = 0$, contradiction.
answered Jan 9 at 16:08
xbhxbh
6,2151522
answered Jan 9 at 16:08
xbhxbh
6,2151522
answered Jan 9 at 16:08
xbhxbh
6,2151522
answered Jan 9 at 16:08
xbhxbh
6,2151522
6,2151522
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
add a comment |
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
thank you. maybe I should modify the given condition to $0leq C_h <h$?
$endgroup$
– Reinherd
Jan 9 at 16:25
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
$begingroup$
I don't think you could prove this…… The 1st MVT of definite integral only guarantees $C_h in [-1,1]$, so maybe you might find some other counterexamples.
$endgroup$
– xbh
Jan 9 at 16:27
add a comment |
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$begingroup$
That is the First mean value theorem for definite integrals.
$endgroup$
– Martin R
Jan 9 at 16:06
$begingroup$
thank you, this is indeed almost the same question.
$endgroup$
– Reinherd
Jan 9 at 16:20