Solution verification for evaluating $limlimits_{x to +infty}dfrac{int_0^{x}|sin t|{rm d}t}{x}$
My Solution
Since $x to +infty$，we may assume $x>pi$. Then
$$ exists n in mathbb{N}:npileq x< (n+1)pi.$$
Thus
$$2n=int_0^{npi}|sin t|{rm d}tleq int_0^x|sin t|{rm d}t<int_0^{(n+1)pi}|sin t|{rm d}t=2(n+1).$$
Further
$$frac{2n}{(n+1)pi}<dfrac{int_0^{x}|sin t|{rm d}t}{x}<frac{2(n+1)}{npi}.$$
When $x to +infty$，$n to infty$. We may obatin
$$frac{2n}{(n+1)pi},frac{2(n+1)}{npi} to frac{2}{pi}(n to infty).$$
By the squeeze theorem, it follows that
$$lim_{x to +infty}dfrac{int_0^{x}|sin t|{rm d}t}{x}=frac{2}{pi}.$$
limits proof-verification definite-integrals
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
add a comment |
My Solution
Since $x to +infty$，we may assume $x>pi$. Then
$$ exists n in mathbb{N}:npileq x< (n+1)pi.$$
Thus
$$2n=int_0^{npi}|sin t|{rm d}tleq int_0^x|sin t|{rm d}t<int_0^{(n+1)pi}|sin t|{rm d}t=2(n+1).$$
Further
$$frac{2n}{(n+1)pi}<dfrac{int_0^{x}|sin t|{rm d}t}{x}<frac{2(n+1)}{npi}.$$
When $x to +infty$，$n to infty$. We may obatin
$$frac{2n}{(n+1)pi},frac{2(n+1)}{npi} to frac{2}{pi}(n to infty).$$
By the squeeze theorem, it follows that
$$lim_{x to +infty}dfrac{int_0^{x}|sin t|{rm d}t}{x}=frac{2}{pi}.$$
limits proof-verification definite-integrals
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
1
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48
add a comment |
My Solution
Since $x to +infty$，we may assume $x>pi$. Then
$$ exists n in mathbb{N}:npileq x< (n+1)pi.$$
Thus
$$2n=int_0^{npi}|sin t|{rm d}tleq int_0^x|sin t|{rm d}t<int_0^{(n+1)pi}|sin t|{rm d}t=2(n+1).$$
Further
$$frac{2n}{(n+1)pi}<dfrac{int_0^{x}|sin t|{rm d}t}{x}<frac{2(n+1)}{npi}.$$
When $x to +infty$，$n to infty$. We may obatin
$$frac{2n}{(n+1)pi},frac{2(n+1)}{npi} to frac{2}{pi}(n to infty).$$
By the squeeze theorem, it follows that
$$lim_{x to +infty}dfrac{int_0^{x}|sin t|{rm d}t}{x}=frac{2}{pi}.$$
limits proof-verification definite-integrals
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
My Solution
Since $x to +infty$，we may assume $x>pi$. Then
$$ exists n in mathbb{N}:npileq x< (n+1)pi.$$
Thus
$$2n=int_0^{npi}|sin t|{rm d}tleq int_0^x|sin t|{rm d}t<int_0^{(n+1)pi}|sin t|{rm d}t=2(n+1).$$
Further
$$frac{2n}{(n+1)pi}<dfrac{int_0^{x}|sin t|{rm d}t}{x}<frac{2(n+1)}{npi}.$$
When $x to +infty$，$n to infty$. We may obatin
$$frac{2n}{(n+1)pi},frac{2(n+1)}{npi} to frac{2}{pi}(n to infty).$$
By the squeeze theorem, it follows that
$$lim_{x to +infty}dfrac{int_0^{x}|sin t|{rm d}t}{x}=frac{2}{pi}.$$
limits proof-verification definite-integrals
limits proof-verification definite-integrals
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
edited Dec 12 '18 at 14:49
mengdie1982
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
asked Dec 12 '18 at 14:44
mengdie1982mengdie1982
4,814618
4,814618
1
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48
add a comment |
1
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48
1
1
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48
add a comment |
2 Answers
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As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence
$$frac1piint_0^pi|sin x|,dx=frac2pi.$$
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
add a comment |
Yes, your solution is correct.
(Answering as community wiki so this question can be closed as answered.)
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
As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence
$$frac1piint_0^pi|sin x|,dx=frac2pi.$$
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
add a comment |
As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence
$$frac1piint_0^pi|sin x|,dx=frac2pi.$$
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
add a comment |
As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence
$$frac1piint_0^pi|sin x|,dx=frac2pi.$$
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
As the function is periodic, the limit is also the average value over a single period (you sum arbitrarily many whole periods plus a single incomplete one, which is bounded), hence
$$frac1piint_0^pi|sin x|,dx=frac2pi.$$
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:13
Yves DaoustYves Daoust
124k671222
124k671222
add a comment |
add a comment |
Yes, your solution is correct.
(Answering as community wiki so this question can be closed as answered.)
add a comment |
Yes, your solution is correct.
(Answering as community wiki so this question can be closed as answered.)
add a comment |
Yes, your solution is correct.
(Answering as community wiki so this question can be closed as answered.)
Yes, your solution is correct.
(Answering as community wiki so this question can be closed as answered.)
answered Dec 12 '18 at 17:07
community wiki
Carl Schildkraut
add a comment |
add a comment |
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1
Yes. $ $ $ $ $ $
– Did
Dec 12 '18 at 14:48