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1st Year Analysis Question

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.
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)=24−14=15/nU(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.
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
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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