Let $f : [1,2] → mathbb R$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and compute $int_1^2f(x)dx$
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Let $f : [1,2] → mathbb R$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and
compute $$int_1^2f(x)dx$$ as the limit of upper (and lower) sums.
I tried solving this but I don't know if my answer is right.
Let $P_n$ be the uniform partition of $[1,2]$ given by
$$1 lt 1+frac1n lt 1+frac2n lt cdots lt 1+fracn-1nlt2$$
The function $f(x)=x$ is increasing, hence
$$m_i=inf_x in [x_i-1,x_i]f(x)=f(x_i-1)=1+fraci-1n$$
and
$$M_i=sup_x in [x_i-1,x_i]f(x)=f(x_i)=1+fracin$$
$$L(f,P)=sum_1^n(1+fraci-1n)fracin=frac3n-12n$$ and $$U(f,P)=sum_1^n(1+fracin)fracin=frac3n+12n$$
Therefore $$lim_ntoinftyL(f,P_n)=frac32 quad lim_ntoinftyU(f,P_n)=frac32$$
By the Criterion of Integrability $$int_1^2f(x)dx=lim_ntoinftyL(f,P_n)=lim_ntoinftyU(f,P_n)=frac32$$
calculus riemann-integration riemann-sum
edited Aug 16 at 13:48
Jneven
575319
asked Aug 16 at 9:44
J.Dane
187112
add a comment |Â
up vote
2
down vote
favorite
Let $f : [1,2] → mathbb R$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and
compute $$int_1^2f(x)dx$$ as the limit of upper (and lower) sums.
I tried solving this but I don't know if my answer is right.
Let $P_n$ be the uniform partition of $[1,2]$ given by
$$1 lt 1+frac1n lt 1+frac2n lt cdots lt 1+fracn-1nlt2$$
The function $f(x)=x$ is increasing, hence
$$m_i=inf_x in [x_i-1,x_i]f(x)=f(x_i-1)=1+fraci-1n$$
and
$$M_i=sup_x in [x_i-1,x_i]f(x)=f(x_i)=1+fracin$$
$$L(f,P)=sum_1^n(1+fraci-1n)fracin=frac3n-12n$$ and $$U(f,P)=sum_1^n(1+fracin)fracin=frac3n+12n$$
Therefore $$lim_ntoinftyL(f,P_n)=frac32 quad lim_ntoinftyU(f,P_n)=frac32$$
By the Criterion of Integrability $$int_1^2f(x)dx=lim_ntoinftyL(f,P_n)=lim_ntoinftyU(f,P_n)=frac32$$
calculus riemann-integration riemann-sum
edited Aug 16 at 13:48
Jneven
575319
asked Aug 16 at 9:44
J.Dane
187112
The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f : [1,2] → mathbb R$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and
compute $$int_1^2f(x)dx$$ as the limit of upper (and lower) sums.
I tried solving this but I don't know if my answer is right.
Let $P_n$ be the uniform partition of $[1,2]$ given by
$$1 lt 1+frac1n lt 1+frac2n lt cdots lt 1+fracn-1nlt2$$
The function $f(x)=x$ is increasing, hence
$$m_i=inf_x in [x_i-1,x_i]f(x)=f(x_i-1)=1+fraci-1n$$
and
$$M_i=sup_x in [x_i-1,x_i]f(x)=f(x_i)=1+fracin$$
$$L(f,P)=sum_1^n(1+fraci-1n)fracin=frac3n-12n$$ and $$U(f,P)=sum_1^n(1+fracin)fracin=frac3n+12n$$
Therefore $$lim_ntoinftyL(f,P_n)=frac32 quad lim_ntoinftyU(f,P_n)=frac32$$
By the Criterion of Integrability $$int_1^2f(x)dx=lim_ntoinftyL(f,P_n)=lim_ntoinftyU(f,P_n)=frac32$$
calculus riemann-integration riemann-sum
edited Aug 16 at 13:48
Jneven
575319
asked Aug 16 at 9:44
J.Dane
187112
Let $f : [1,2] → mathbb R$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and
compute $$int_1^2f(x)dx$$ as the limit of upper (and lower) sums.
I tried solving this but I don't know if my answer is right.
Let $P_n$ be the uniform partition of $[1,2]$ given by
$$1 lt 1+frac1n lt 1+frac2n lt cdots lt 1+fracn-1nlt2$$
The function $f(x)=x$ is increasing, hence
$$m_i=inf_x in [x_i-1,x_i]f(x)=f(x_i-1)=1+fraci-1n$$
and
$$M_i=sup_x in [x_i-1,x_i]f(x)=f(x_i)=1+fracin$$
$$L(f,P)=sum_1^n(1+fraci-1n)fracin=frac3n-12n$$ and $$U(f,P)=sum_1^n(1+fracin)fracin=frac3n+12n$$
Therefore $$lim_ntoinftyL(f,P_n)=frac32 quad lim_ntoinftyU(f,P_n)=frac32$$
By the Criterion of Integrability $$int_1^2f(x)dx=lim_ntoinftyL(f,P_n)=lim_ntoinftyU(f,P_n)=frac32$$
calculus riemann-integration riemann-sum
edited Aug 16 at 13:48
Jneven
575319
asked Aug 16 at 9:44
J.Dane
187112
edited Aug 16 at 13:48
Jneven
575319
edited Aug 16 at 13:48
Jneven
575319
edited Aug 16 at 13:48
Jneven
575319
575319
asked Aug 16 at 9:44
J.Dane
187112
asked Aug 16 at 9:44
J.Dane
187112
asked Aug 16 at 9:44
J.Dane
187112
187112
The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58
add a comment |Â
The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58
The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Yes, your proof is fine, everything is o.k.
answered Aug 16 at 9:48
Fred
38k1238
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, your proof is fine, everything is o.k.
answered Aug 16 at 9:48
Fred
38k1238
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
 |Â
show 2 more comments
up vote
0
down vote
Yes, your proof is fine, everything is o.k.
answered Aug 16 at 9:48
Fred
38k1238
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
 |Â
show 2 more comments
up vote
0
down vote
up vote
0
down vote
Yes, your proof is fine, everything is o.k.
answered Aug 16 at 9:48
Fred
38k1238
Yes, your proof is fine, everything is o.k.
answered Aug 16 at 9:48
Fred
38k1238
answered Aug 16 at 9:48
Fred
38k1238
answered Aug 16 at 9:48
Fred
38k1238
answered Aug 16 at 9:48
Fred
38k1238
38k1238
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
 |Â
show 2 more comments
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$
– J.Dane
Aug 16 at 9:49
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
Why the downvotes ?????????????
– Fred
Aug 16 at 9:53
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ?
– Fred
Aug 16 at 9:54
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
Is it really sufficient to consider uniform partitions only?
– Martin R
Aug 16 at 10:09
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= frac1n$. If $ epsilon >0$ the choose $n$ such that $ frac1n< epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) to int_1^2f(x)dx$.
– Fred
Aug 16 at 10:14
 |Â
show 2 more comments
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The function is continuous then is Riemann integrable.
– Nosrati
Aug 16 at 9:55
yes but I had to prove as the limit of upper and lower sums
– J.Dane
Aug 16 at 9:58