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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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