Question: evaluate riemann sum using partition which does not have equal intervals?

the question is as follow:

The partition does not always have to be equal intervals. Consider evaluating f(x)=x³ between 3 and 5, but splitting up the interval into a partition in which the end points of the subintervals are in a geometric progression. The common ratio r has to be chosen so that 3 is the first term and 5 is the last. Also the subintervals must be capable of getting smaller as n the number of subintervals increases. Check that the geometric series

a, ar, ar², ar³,.....arⁱ, .....arⁿ =b

with r=  and suitable choices for a and b satisfies these criteria. Treating the difference between arⁱ and ar^(i-1) as the width of the subinterval and using the right hand endpoint of the subinterval, evaluate the Riemann sum to n terms for f(x)=x³. Find the limit as n tends to infinity to show that the partition does not affect the result.

here is what i have got so far, can anyone check if im doing it right? thanks

>a:=3:

>b:=a*r^10:

>r:=(5/3)^1/10:

>for i from 0 to 5 do a*r^i end do; -> a list of number appear in sequence ie:3, 3.157...,3.323...3.497...etc

>restart;

>a:=3:

>b:=a*r^100:

>r:=(5/3)^1/100:

>dxj:=a*r^i-a*r^i-1

>xj:=i*dxj+a

>f:=x->x^3

>evalf(sum(f(xj^*)dxj,i=1..100)) -> my value is sth like 162.4788870...

I tried to find the limit, but maple 16 freezed so i think i must have done sth seriously wrong?

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