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    Could someone tell me whether I am answering this correctly please?:
    Pn denotes the partition pn=(1,1+1/n,1+2/n,..,2)
    Write upper and lower Riemann sums or the function f(x)=x^4 with respect to pn.
    Use to show x^4 Riemann integrable on (1,2).

    For the sums I got L(f,pn)=(1/n)(1^4,(1+1/n)4,...,(1+(n-1)/n)^4)
    U(f,pn)=(1/n)((1+1/n)^4, (1+2/n)4,...,2^4)

    So U(f,pn)-L(f,pn)=2^4-1^4=15/n

    Approaches 0 as n approaches infinity so is Riemann Integrable.
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    (Original post by Harriettttttt1)
    Could someone tell me whether I am answering this correctly please?:
    Pn denotes the partition pn=(1,1+1/n,1+2/n,..,2)
    Write upper and lower Riemann sums or the function f(x)=x^4 with respect to pn.
    Use to show x^4 Riemann integrable on (1,2).

    For the sums I got L(f,pn)=(1/n)(1^4,(1+1/n)4,...,(1+(n-1)/n)^4)
    U(f,pn)=(1/n)((1+1/n)^4, (1+2/n)4,...,2^4)

    So U(f,pn)-L(f,pn)=2^4-1^4=15/n

    Approaches 0 as n approaches infinity so is Riemann Integrable.
    1. Latex is your friend - this is close to unreadable.

    2. You haven't written down any sums; you seem to have given a list of values for the upper and lower sums.

    3. Your final calculation (I latexed it) says U(f,pn)-L(f,pn)=2^4-1^4=15/n - this clearly isn't right.

    So, sadly, I'd say you need to fix this up a lot.
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    Sorry I didn't write this out very carefully! The commas are supposed to be pluses and the last part should read
    (2^4)/n-(1^4)/n=15/n is this last part for the differences between the sums right is what I was trying to ask
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    (Original post by Harriettttttt1)
    Sorry I didn't write this out very carefully! The commas are supposed to be pluses and the last part should read
    (2^4)/n-(1^4)/n=15/n is this last part for the differences between the sums right is what I was trying to ask
    Are you for real? Just latex the problem or post a photo. Unfortunately you can't expect people to decipher what you write.
 
 
 
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