upper and lower sum answer not matching











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I have to show that f is integrable using the Riemann criterion where
f(x) = x on [0, 1]. I am a bit confused over how my solution differs from the hint to solution of that question I have.
My approach is below:



Let partition P = {0 , 1/n , 2/n , ..... , (n-1)/n , n/n}.



So, U(P,f) = ∑(1/n) . { f(1/n) + f(2/n) + ....... + f(n/n) }

And, L(p,f) = ∑(1/n) . { f(0) + f(2/n) + ....... + f(n-1/n) }

Then, U(P,f) - L(P,f) = ( f(n/n) - f(0) )/n
= 1/n -> 0



Hint to solution given :
link is here



I am very confused over how did they get the result 1/n² , instead of my 1/n . Please help me out.






calculus







share|cite|improve this question




















  • 1




    I believe the hint is wrong.
    – B. Goddard
    Nov 17 at 13:42















up vote
0
down vote

favorite












I have to show that f is integrable using the Riemann criterion where
f(x) = x on [0, 1]. I am a bit confused over how my solution differs from the hint to solution of that question I have.
My approach is below:



Let partition P = {0 , 1/n , 2/n , ..... , (n-1)/n , n/n}.



So, U(P,f) = ∑(1/n) . { f(1/n) + f(2/n) + ....... + f(n/n) }

And, L(p,f) = ∑(1/n) . { f(0) + f(2/n) + ....... + f(n-1/n) }

Then, U(P,f) - L(P,f) = ( f(n/n) - f(0) )/n
= 1/n -> 0



Hint to solution given :
link is here



I am very confused over how did they get the result 1/n² , instead of my 1/n . Please help me out.






calculus







share|cite|improve this question




















  • 1




    I believe the hint is wrong.
    – B. Goddard
    Nov 17 at 13:42













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to show that f is integrable using the Riemann criterion where
f(x) = x on [0, 1]. I am a bit confused over how my solution differs from the hint to solution of that question I have.
My approach is below:



Let partition P = {0 , 1/n , 2/n , ..... , (n-1)/n , n/n}.



So, U(P,f) = ∑(1/n) . { f(1/n) + f(2/n) + ....... + f(n/n) }

And, L(p,f) = ∑(1/n) . { f(0) + f(2/n) + ....... + f(n-1/n) }

Then, U(P,f) - L(P,f) = ( f(n/n) - f(0) )/n
= 1/n -> 0



Hint to solution given :
link is here



I am very confused over how did they get the result 1/n² , instead of my 1/n . Please help me out.






calculus







share|cite|improve this question















I have to show that f is integrable using the Riemann criterion where
f(x) = x on [0, 1]. I am a bit confused over how my solution differs from the hint to solution of that question I have.
My approach is below:



Let partition P = {0 , 1/n , 2/n , ..... , (n-1)/n , n/n}.



So, U(P,f) = ∑(1/n) . { f(1/n) + f(2/n) + ....... + f(n/n) }

And, L(p,f) = ∑(1/n) . { f(0) + f(2/n) + ....... + f(n-1/n) }

Then, U(P,f) - L(P,f) = ( f(n/n) - f(0) )/n
= 1/n -> 0



Hint to solution given :
link is here



I am very confused over how did they get the result 1/n² , instead of my 1/n . Please help me out.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 at 10:42









Parcly Taxel

41k137199




41k137199










asked Nov 17 at 10:42









Kaustav Bhattacharjee

61




61








  • 1




    I believe the hint is wrong.
    – B. Goddard
    Nov 17 at 13:42














  • 1




    I believe the hint is wrong.
    – B. Goddard
    Nov 17 at 13:42








1




1




I believe the hint is wrong.
– B. Goddard
Nov 17 at 13:42




I believe the hint is wrong.
– B. Goddard
Nov 17 at 13:42















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