How to find the upper Riemann integral of following function?












2












$begingroup$


$f(x)$ is defined on $[a,b]$ as -



$= 0$ if $x ∈ [a, b] ∩ Q$



$= x$ otherwise



Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.






real-analysis integration riemann-integration







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$endgroup$












  • $begingroup$
    Bound it above by a step function.
    $endgroup$
    – Lord Shark the Unknown
    Dec 1 '18 at 17:36










  • $begingroup$
    @LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:14


















2












$begingroup$


$f(x)$ is defined on $[a,b]$ as -



$= 0$ if $x ∈ [a, b] ∩ Q$



$= x$ otherwise



Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.






real-analysis integration riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    Bound it above by a step function.
    $endgroup$
    – Lord Shark the Unknown
    Dec 1 '18 at 17:36










  • $begingroup$
    @LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:14
















2












2








2


2



$begingroup$


$f(x)$ is defined on $[a,b]$ as -



$= 0$ if $x ∈ [a, b] ∩ Q$



$= x$ otherwise



Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.






real-analysis integration riemann-integration







share|cite|improve this question









$endgroup$




$f(x)$ is defined on $[a,b]$ as -



$= 0$ if $x ∈ [a, b] ∩ Q$



$= x$ otherwise



Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.






real-analysis integration riemann-integration




real-analysis integration riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 17:26









BhowmickBhowmick

1438




1438












  • $begingroup$
    Bound it above by a step function.
    $endgroup$
    – Lord Shark the Unknown
    Dec 1 '18 at 17:36










  • $begingroup$
    @LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:14




















  • $begingroup$
    Bound it above by a step function.
    $endgroup$
    – Lord Shark the Unknown
    Dec 1 '18 at 17:36










  • $begingroup$
    @LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:14


















$begingroup$
Bound it above by a step function.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 17:36




$begingroup$
Bound it above by a step function.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 17:36












$begingroup$
@LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:14






$begingroup$
@LordSharktheUnknown after solving that i get $<=$ inequality and not $=$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:14












2 Answers
2






active

oldest

votes


















3












$begingroup$

Since the irrationals are dense, we have $sup_{x in [x_{j-1},x_j]}f(x) = x_j$.



Hence, for any partition, $a = x_0 < x_1 < ldots < x_n = b$, the upper Darboux sum is



$$U(P,f) = sum_{j=1}^n x_j(x_j - x_{j-1}) = frac{1}{2} sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \ = frac{1}{2} sum_{j=1}^n(x_j^2 - x_{j-1}^2) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{b^2-a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2, $$



Since the last term on the RHS is nonnegative we have, for the upper integral,
$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) geqslant frac{b^2 - a^2}{2}$$



To prove we actually have $inf_{P}U(P,f) = frac{b^2 - a^2}{2}$, we show that for any $epsilon > 0$ there exists a partition such that



$$U(P,f) < frac{b^2 - a^2}{2} + epsilon$$



For a uniform partition where $x_j - x_{j-1} = frac{b-a}{n}$ for $j=1,2, ldots, n$ we have



$$frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{1}{2} nfrac{(b-a)^2}{n^2} = frac{(b-a)^2}{2n}$$



Now choose $n > frac{(b-a)^2}{2epsilon}$ and we have



$$begin{align}U(P,f) &=frac{b^2 - a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2\ &= frac{b^2 - a^2}{2}+ frac{(b-a)^2}{2n} \&< frac{b^2 - a^2}{2} + epsilonend{align}$$



Therefore,



$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) = frac{b^2 - a^2}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
    $endgroup$
    – Paramanand Singh
    Dec 2 '18 at 3:31



















1












$begingroup$

For any partition $a=x_0<x_1<ldots <x_n=b$, the Riemann upper sum is
$$
U_{mathbf x}=sum_{i=1}^n(x_i-x_{i-1})sup_{x_{i-1}<x<x_i}f(x)
$$
In each interval $(x_{i-1},x_i)$ there is some real numbers $rin(x_{i-1},x_i)$ and hence the supremum is $sup_{x_{i-1}<x<x_i}f(x)ge f(x_{i-1})=x_{i-1}$.



Therefore
$$
U_{mathbf x}ge sum_{i=1}^n(x_i-x_{i-1})x_{i-1}
$$
Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+frac{(b-a)i}{n}=a+ih$



begin{align}
sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=hsum_{i=1}^n(a+(i-1)h) \
&=frac{1}{2}h(2a+h(n-1))n \
&=frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})
end{align}
With
$$
lim_{ntoinfty}frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})=frac{1}{2}(b^2-a^2)
$$
Hence you have your result,




  • as $ntoinfty$, the upper bound is $U_xgefrac{1}{2}(b-a)^2$

  • as upper bound is greater that lower bound, the function is not Riemann integrable






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But how to prove that upper integral is equal to $(b*b - a*a)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:42










  • $begingroup$
    it is upper bound when $n->infty$, as described in the answer, is it ok or not?
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:55










  • $begingroup$
    but i want upper bound equal to $(b^2-a^2)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    read question once again
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    i have to go right now, read it when back sorry. hold on
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:56











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Since the irrationals are dense, we have $sup_{x in [x_{j-1},x_j]}f(x) = x_j$.



Hence, for any partition, $a = x_0 < x_1 < ldots < x_n = b$, the upper Darboux sum is



$$U(P,f) = sum_{j=1}^n x_j(x_j - x_{j-1}) = frac{1}{2} sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \ = frac{1}{2} sum_{j=1}^n(x_j^2 - x_{j-1}^2) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{b^2-a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2, $$



Since the last term on the RHS is nonnegative we have, for the upper integral,
$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) geqslant frac{b^2 - a^2}{2}$$



To prove we actually have $inf_{P}U(P,f) = frac{b^2 - a^2}{2}$, we show that for any $epsilon > 0$ there exists a partition such that



$$U(P,f) < frac{b^2 - a^2}{2} + epsilon$$



For a uniform partition where $x_j - x_{j-1} = frac{b-a}{n}$ for $j=1,2, ldots, n$ we have



$$frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{1}{2} nfrac{(b-a)^2}{n^2} = frac{(b-a)^2}{2n}$$



Now choose $n > frac{(b-a)^2}{2epsilon}$ and we have



$$begin{align}U(P,f) &=frac{b^2 - a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2\ &= frac{b^2 - a^2}{2}+ frac{(b-a)^2}{2n} \&< frac{b^2 - a^2}{2} + epsilonend{align}$$



Therefore,



$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) = frac{b^2 - a^2}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
    $endgroup$
    – Paramanand Singh
    Dec 2 '18 at 3:31
















3












$begingroup$

Since the irrationals are dense, we have $sup_{x in [x_{j-1},x_j]}f(x) = x_j$.



Hence, for any partition, $a = x_0 < x_1 < ldots < x_n = b$, the upper Darboux sum is



$$U(P,f) = sum_{j=1}^n x_j(x_j - x_{j-1}) = frac{1}{2} sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \ = frac{1}{2} sum_{j=1}^n(x_j^2 - x_{j-1}^2) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{b^2-a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2, $$



Since the last term on the RHS is nonnegative we have, for the upper integral,
$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) geqslant frac{b^2 - a^2}{2}$$



To prove we actually have $inf_{P}U(P,f) = frac{b^2 - a^2}{2}$, we show that for any $epsilon > 0$ there exists a partition such that



$$U(P,f) < frac{b^2 - a^2}{2} + epsilon$$



For a uniform partition where $x_j - x_{j-1} = frac{b-a}{n}$ for $j=1,2, ldots, n$ we have



$$frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{1}{2} nfrac{(b-a)^2}{n^2} = frac{(b-a)^2}{2n}$$



Now choose $n > frac{(b-a)^2}{2epsilon}$ and we have



$$begin{align}U(P,f) &=frac{b^2 - a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2\ &= frac{b^2 - a^2}{2}+ frac{(b-a)^2}{2n} \&< frac{b^2 - a^2}{2} + epsilonend{align}$$



Therefore,



$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) = frac{b^2 - a^2}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
    $endgroup$
    – Paramanand Singh
    Dec 2 '18 at 3:31














3












3








3





$begingroup$

Since the irrationals are dense, we have $sup_{x in [x_{j-1},x_j]}f(x) = x_j$.



Hence, for any partition, $a = x_0 < x_1 < ldots < x_n = b$, the upper Darboux sum is



$$U(P,f) = sum_{j=1}^n x_j(x_j - x_{j-1}) = frac{1}{2} sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \ = frac{1}{2} sum_{j=1}^n(x_j^2 - x_{j-1}^2) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{b^2-a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2, $$



Since the last term on the RHS is nonnegative we have, for the upper integral,
$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) geqslant frac{b^2 - a^2}{2}$$



To prove we actually have $inf_{P}U(P,f) = frac{b^2 - a^2}{2}$, we show that for any $epsilon > 0$ there exists a partition such that



$$U(P,f) < frac{b^2 - a^2}{2} + epsilon$$



For a uniform partition where $x_j - x_{j-1} = frac{b-a}{n}$ for $j=1,2, ldots, n$ we have



$$frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{1}{2} nfrac{(b-a)^2}{n^2} = frac{(b-a)^2}{2n}$$



Now choose $n > frac{(b-a)^2}{2epsilon}$ and we have



$$begin{align}U(P,f) &=frac{b^2 - a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2\ &= frac{b^2 - a^2}{2}+ frac{(b-a)^2}{2n} \&< frac{b^2 - a^2}{2} + epsilonend{align}$$



Therefore,



$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) = frac{b^2 - a^2}{2}$$






share|cite|improve this answer











$endgroup$



Since the irrationals are dense, we have $sup_{x in [x_{j-1},x_j]}f(x) = x_j$.



Hence, for any partition, $a = x_0 < x_1 < ldots < x_n = b$, the upper Darboux sum is



$$U(P,f) = sum_{j=1}^n x_j(x_j - x_{j-1}) = frac{1}{2} sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \ = frac{1}{2} sum_{j=1}^n(x_j^2 - x_{j-1}^2) + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{b^2-a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2, $$



Since the last term on the RHS is nonnegative we have, for the upper integral,
$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) geqslant frac{b^2 - a^2}{2}$$



To prove we actually have $inf_{P}U(P,f) = frac{b^2 - a^2}{2}$, we show that for any $epsilon > 0$ there exists a partition such that



$$U(P,f) < frac{b^2 - a^2}{2} + epsilon$$



For a uniform partition where $x_j - x_{j-1} = frac{b-a}{n}$ for $j=1,2, ldots, n$ we have



$$frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2 = frac{1}{2} nfrac{(b-a)^2}{n^2} = frac{(b-a)^2}{2n}$$



Now choose $n > frac{(b-a)^2}{2epsilon}$ and we have



$$begin{align}U(P,f) &=frac{b^2 - a^2}{2} + frac{1}{2}sum_{j=1}^n(x_j-x_{j-1})^2\ &= frac{b^2 - a^2}{2}+ frac{(b-a)^2}{2n} \&< frac{b^2 - a^2}{2} + epsilonend{align}$$



Therefore,



$$overline{int_a^b} f(x) , dx = inf_{P}U(P,f) = frac{b^2 - a^2}{2}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 3:36

























answered Dec 1 '18 at 20:01









RRLRRL

50k42573




50k42573












  • $begingroup$
    We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
    $endgroup$
    – Paramanand Singh
    Dec 2 '18 at 3:31


















  • $begingroup$
    We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
    $endgroup$
    – Paramanand Singh
    Dec 2 '18 at 3:31
















$begingroup$
We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
$endgroup$
– Paramanand Singh
Dec 2 '18 at 3:31




$begingroup$
We can use the theorem that if $P_n$ is a sequence of partitions such that norm/mesh of $P_n$ tends to $0$ then $U(P_n, f) to overline{int_{a} ^{b} }f(x) , dx$ and then take $P_n$ as uniform partition. +1 as your answer proves this theorem for the given function.
$endgroup$
– Paramanand Singh
Dec 2 '18 at 3:31











1












$begingroup$

For any partition $a=x_0<x_1<ldots <x_n=b$, the Riemann upper sum is
$$
U_{mathbf x}=sum_{i=1}^n(x_i-x_{i-1})sup_{x_{i-1}<x<x_i}f(x)
$$
In each interval $(x_{i-1},x_i)$ there is some real numbers $rin(x_{i-1},x_i)$ and hence the supremum is $sup_{x_{i-1}<x<x_i}f(x)ge f(x_{i-1})=x_{i-1}$.



Therefore
$$
U_{mathbf x}ge sum_{i=1}^n(x_i-x_{i-1})x_{i-1}
$$
Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+frac{(b-a)i}{n}=a+ih$



begin{align}
sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=hsum_{i=1}^n(a+(i-1)h) \
&=frac{1}{2}h(2a+h(n-1))n \
&=frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})
end{align}
With
$$
lim_{ntoinfty}frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})=frac{1}{2}(b^2-a^2)
$$
Hence you have your result,




  • as $ntoinfty$, the upper bound is $U_xgefrac{1}{2}(b-a)^2$

  • as upper bound is greater that lower bound, the function is not Riemann integrable






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But how to prove that upper integral is equal to $(b*b - a*a)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:42










  • $begingroup$
    it is upper bound when $n->infty$, as described in the answer, is it ok or not?
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:55










  • $begingroup$
    but i want upper bound equal to $(b^2-a^2)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    read question once again
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    i have to go right now, read it when back sorry. hold on
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:56
















1












$begingroup$

For any partition $a=x_0<x_1<ldots <x_n=b$, the Riemann upper sum is
$$
U_{mathbf x}=sum_{i=1}^n(x_i-x_{i-1})sup_{x_{i-1}<x<x_i}f(x)
$$
In each interval $(x_{i-1},x_i)$ there is some real numbers $rin(x_{i-1},x_i)$ and hence the supremum is $sup_{x_{i-1}<x<x_i}f(x)ge f(x_{i-1})=x_{i-1}$.



Therefore
$$
U_{mathbf x}ge sum_{i=1}^n(x_i-x_{i-1})x_{i-1}
$$
Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+frac{(b-a)i}{n}=a+ih$



begin{align}
sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=hsum_{i=1}^n(a+(i-1)h) \
&=frac{1}{2}h(2a+h(n-1))n \
&=frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})
end{align}
With
$$
lim_{ntoinfty}frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})=frac{1}{2}(b^2-a^2)
$$
Hence you have your result,




  • as $ntoinfty$, the upper bound is $U_xgefrac{1}{2}(b-a)^2$

  • as upper bound is greater that lower bound, the function is not Riemann integrable






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But how to prove that upper integral is equal to $(b*b - a*a)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:42










  • $begingroup$
    it is upper bound when $n->infty$, as described in the answer, is it ok or not?
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:55










  • $begingroup$
    but i want upper bound equal to $(b^2-a^2)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    read question once again
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    i have to go right now, read it when back sorry. hold on
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:56














1












1








1





$begingroup$

For any partition $a=x_0<x_1<ldots <x_n=b$, the Riemann upper sum is
$$
U_{mathbf x}=sum_{i=1}^n(x_i-x_{i-1})sup_{x_{i-1}<x<x_i}f(x)
$$
In each interval $(x_{i-1},x_i)$ there is some real numbers $rin(x_{i-1},x_i)$ and hence the supremum is $sup_{x_{i-1}<x<x_i}f(x)ge f(x_{i-1})=x_{i-1}$.



Therefore
$$
U_{mathbf x}ge sum_{i=1}^n(x_i-x_{i-1})x_{i-1}
$$
Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+frac{(b-a)i}{n}=a+ih$



begin{align}
sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=hsum_{i=1}^n(a+(i-1)h) \
&=frac{1}{2}h(2a+h(n-1))n \
&=frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})
end{align}
With
$$
lim_{ntoinfty}frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})=frac{1}{2}(b^2-a^2)
$$
Hence you have your result,




  • as $ntoinfty$, the upper bound is $U_xgefrac{1}{2}(b-a)^2$

  • as upper bound is greater that lower bound, the function is not Riemann integrable






share|cite|improve this answer











$endgroup$



For any partition $a=x_0<x_1<ldots <x_n=b$, the Riemann upper sum is
$$
U_{mathbf x}=sum_{i=1}^n(x_i-x_{i-1})sup_{x_{i-1}<x<x_i}f(x)
$$
In each interval $(x_{i-1},x_i)$ there is some real numbers $rin(x_{i-1},x_i)$ and hence the supremum is $sup_{x_{i-1}<x<x_i}f(x)ge f(x_{i-1})=x_{i-1}$.



Therefore
$$
U_{mathbf x}ge sum_{i=1}^n(x_i-x_{i-1})x_{i-1}
$$
Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+frac{(b-a)i}{n}=a+ih$



begin{align}
sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=hsum_{i=1}^n(a+(i-1)h) \
&=frac{1}{2}h(2a+h(n-1))n \
&=frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})
end{align}
With
$$
lim_{ntoinfty}frac{1}{2}(b-a)(2a+frac{(b-a)(n-1)}{n})=frac{1}{2}(b^2-a^2)
$$
Hence you have your result,




  • as $ntoinfty$, the upper bound is $U_xgefrac{1}{2}(b-a)^2$

  • as upper bound is greater that lower bound, the function is not Riemann integrable







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 19:55

























answered Dec 1 '18 at 18:32









Picaud VincentPicaud Vincent

1,49439




1,49439












  • $begingroup$
    But how to prove that upper integral is equal to $(b*b - a*a)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:42










  • $begingroup$
    it is upper bound when $n->infty$, as described in the answer, is it ok or not?
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:55










  • $begingroup$
    but i want upper bound equal to $(b^2-a^2)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    read question once again
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    i have to go right now, read it when back sorry. hold on
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:56


















  • $begingroup$
    But how to prove that upper integral is equal to $(b*b - a*a)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:42










  • $begingroup$
    it is upper bound when $n->infty$, as described in the answer, is it ok or not?
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:55










  • $begingroup$
    but i want upper bound equal to $(b^2-a^2)/2$
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    read question once again
    $endgroup$
    – Bhowmick
    Dec 1 '18 at 18:56










  • $begingroup$
    i have to go right now, read it when back sorry. hold on
    $endgroup$
    – Picaud Vincent
    Dec 1 '18 at 18:56
















$begingroup$
But how to prove that upper integral is equal to $(b*b - a*a)/2$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:42




$begingroup$
But how to prove that upper integral is equal to $(b*b - a*a)/2$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:42












$begingroup$
it is upper bound when $n->infty$, as described in the answer, is it ok or not?
$endgroup$
– Picaud Vincent
Dec 1 '18 at 18:55




$begingroup$
it is upper bound when $n->infty$, as described in the answer, is it ok or not?
$endgroup$
– Picaud Vincent
Dec 1 '18 at 18:55












$begingroup$
but i want upper bound equal to $(b^2-a^2)/2$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:56




$begingroup$
but i want upper bound equal to $(b^2-a^2)/2$
$endgroup$
– Bhowmick
Dec 1 '18 at 18:56












$begingroup$
read question once again
$endgroup$
– Bhowmick
Dec 1 '18 at 18:56




$begingroup$
read question once again
$endgroup$
– Bhowmick
Dec 1 '18 at 18:56












$begingroup$
i have to go right now, read it when back sorry. hold on
$endgroup$
– Picaud Vincent
Dec 1 '18 at 18:56




$begingroup$
i have to go right now, read it when back sorry. hold on
$endgroup$
– Picaud Vincent
Dec 1 '18 at 18:56


















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