Proving Bounds for a Riemann Sum: Part II

ronho1234
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this is a riemann sum question and i need help with part 2

let Sn denote the finite sum 1+2^ 3/2 +...+n^ 3/2

i) use suitable upper and lower riemann sums for the function f(x)=x^3/2 on the interval [0,100] to prove that S99<J<100

ummm i did this and found 40000<J<41000

II) hence, or otherwise, find integer lower and upper bounds, no more than 1000 units apart, for S100

ummm i don't understand what the question is asking me...
 
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Do you mean S99< J< S100 rather than S99< J< 100? And what is 'J'? You don't seem to have defined it anywhere.
 
yes i meant S100 and J is from the first part
Calculate J= the integral from 0-100 x^3/2dx which i found to be 40000 i think which leads onto the next two questions
 
Well, from your description, it seems you already did #2 , by finding a lower

bound of 40000 and an upper bound of 41000, since their difference satisfies

the condition.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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