Mixing
Let $I=int_{0}^{1}f(x)x^2dx$
$begingroup$
I came across the following problem that says:
Suppose $f$ is a continuous real-valued function. Let $I=int_0^1 f(x)x^2 , dx$.Then it is necessarily true that $I$ equals :
(A)$frac{f(1)}{3}-frac{f(0)}{3}$,
(B)$frac{f(c)}{3}$ for some $c in [0,1]$,
(C)$f(1/3)-f(0)$,
(D)$f(c)$ for some $c in [0,1]$.
I was thinking about it but do not know how to tackle it.Can someone point me in the right direction? Thanks in advance for your time.
real-analysis analysis
edited Jan 19 '13 at 19:38
Michael Hardy
1
$endgroup$
add a comment |
$begingroup$
I came across the following problem that says:
Suppose $f$ is a continuous real-valued function. Let $I=int_0^1 f(x)x^2 , dx$.Then it is necessarily true that $I$ equals :
(A)$frac{f(1)}{3}-frac{f(0)}{3}$,
(B)$frac{f(c)}{3}$ for some $c in [0,1]$,
(C)$f(1/3)-f(0)$,
(D)$f(c)$ for some $c in [0,1]$.
I was thinking about it but do not know how to tackle it.Can someone point me in the right direction? Thanks in advance for your time.
real-analysis analysis
edited Jan 19 '13 at 19:38
Michael Hardy
1
$endgroup$
2
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42
add a comment |
$begingroup$
I came across the following problem that says:
Suppose $f$ is a continuous real-valued function. Let $I=int_0^1 f(x)x^2 , dx$.Then it is necessarily true that $I$ equals :
(A)$frac{f(1)}{3}-frac{f(0)}{3}$,
(B)$frac{f(c)}{3}$ for some $c in [0,1]$,
(C)$f(1/3)-f(0)$,
(D)$f(c)$ for some $c in [0,1]$.
I was thinking about it but do not know how to tackle it.Can someone point me in the right direction? Thanks in advance for your time.
real-analysis analysis
edited Jan 19 '13 at 19:38
Michael Hardy
1
$endgroup$
I came across the following problem that says:
Suppose $f$ is a continuous real-valued function. Let $I=int_0^1 f(x)x^2 , dx$.Then it is necessarily true that $I$ equals :
(A)$frac{f(1)}{3}-frac{f(0)}{3}$,
(B)$frac{f(c)}{3}$ for some $c in [0,1]$,
(C)$f(1/3)-f(0)$,
(D)$f(c)$ for some $c in [0,1]$.
I was thinking about it but do not know how to tackle it.Can someone point me in the right direction? Thanks in advance for your time.
real-analysis analysis
real-analysis analysis
edited Jan 19 '13 at 19:38
Michael Hardy
1
edited Jan 19 '13 at 19:38
Michael Hardy
1
edited Jan 19 '13 at 19:38
Michael Hardy
1
edited Jan 19 '13 at 19:38
Michael Hardy
1
edited Jan 19 '13 at 19:38
Michael Hardy
1
1
asked Jan 19 '13 at 19:22
user53386
2
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42
add a comment |
2
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42
2
2
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use the mean value theorem for integration to deduce that (B) is true.
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
$endgroup$
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
add a comment |
$begingroup$
A continuous function takes all values between minimum and maximum. It is evident that:
$frac{f_{min}}{3} le I le frac{f_{max}}{3}$
i.e. :
$f_{min} le 3 I le f_{max}$
so there is $c$ so that $3I=f(c)$, i.e. $I=frac{f(c)}{3}$
answered Feb 1 at 15:42
ThomasThomas
17819
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the mean value theorem for integration to deduce that (B) is true.
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
$endgroup$
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
add a comment |
$begingroup$
Use the mean value theorem for integration to deduce that (B) is true.
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
$endgroup$
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
add a comment |
$begingroup$
Use the mean value theorem for integration to deduce that (B) is true.
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
$endgroup$
Use the mean value theorem for integration to deduce that (B) is true.
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
edited Jan 19 '13 at 19:32
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
answered Jan 19 '13 at 19:24
AmrAmr
14.4k43296
14.4k43296
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
add a comment |
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
$begingroup$
Thanks a lot.I have got it.
$endgroup$
– user53386
Jan 19 '13 at 19:30
add a comment |
$begingroup$
A continuous function takes all values between minimum and maximum. It is evident that:
$frac{f_{min}}{3} le I le frac{f_{max}}{3}$
i.e. :
$f_{min} le 3 I le f_{max}$
so there is $c$ so that $3I=f(c)$, i.e. $I=frac{f(c)}{3}$
answered Feb 1 at 15:42
ThomasThomas
17819
$endgroup$
add a comment |
$begingroup$
A continuous function takes all values between minimum and maximum. It is evident that:
$frac{f_{min}}{3} le I le frac{f_{max}}{3}$
i.e. :
$f_{min} le 3 I le f_{max}$
so there is $c$ so that $3I=f(c)$, i.e. $I=frac{f(c)}{3}$
answered Feb 1 at 15:42
ThomasThomas
17819
$endgroup$
add a comment |
$begingroup$
A continuous function takes all values between minimum and maximum. It is evident that:
$frac{f_{min}}{3} le I le frac{f_{max}}{3}$
i.e. :
$f_{min} le 3 I le f_{max}$
so there is $c$ so that $3I=f(c)$, i.e. $I=frac{f(c)}{3}$
answered Feb 1 at 15:42
ThomasThomas
17819
$endgroup$
A continuous function takes all values between minimum and maximum. It is evident that:
$frac{f_{min}}{3} le I le frac{f_{max}}{3}$
i.e. :
$f_{min} le 3 I le f_{max}$
so there is $c$ so that $3I=f(c)$, i.e. $I=frac{f(c)}{3}$
answered Feb 1 at 15:42
ThomasThomas
17819
answered Feb 1 at 15:42
ThomasThomas
17819
answered Feb 1 at 15:42
ThomasThomas
17819
answered Feb 1 at 15:42
ThomasThomas
17819
17819
add a comment |
add a comment |
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2
$begingroup$
If you use the function $f(x)=1$ you eliminate $A,C,D$.
$endgroup$
– P..
Jan 19 '13 at 19:42