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I'm given the following Riemann Sum to convert to Integral:

$$\lim_{n\to \infty}\sum_{i=1}^n {1\over n} \sqrt {i-1\over n}$$

I first did the following:

$\Delta x$ = $\dfrac{1}{n},\;$ $f(x)=\sqrt{x},\;$ and $x_i= \dfrac{i-1}{n}$

I tried to solve $a$ by doing:

$a+i \Delta x =\dfrac{i-1}{n}$

$a = \dfrac{-1}{n}$

And now I'm stuck at that. I do not know if I could take the limit of that and assume that $a = 0$ and get $b = 1$, or is that now how you solve it?

Emi Matro
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Phillip
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    When $i=0$ we are taking the square root of $0$ while at $i=n$ we are taking the square root of nearly $1$, so it is a left Riemann sum for $\int_0^1\sqrt{x},dx$. – André Nicolas Feb 08 '14 at 04:55

2 Answers2

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Hint: For all $n>1,$ we have $$\begin{align}\sum_{i=1}^n\frac1n\sqrt{\frac{i-1}n} &= \sum_{i=2}^n\frac1n\sqrt{\frac{i-1}n}\\ &= \sum_{i=1}^{n-1}\frac1n\sqrt{\frac{i}n}\\ &= -\frac1n+ \sum_{i=1}^n\frac1n\sqrt{\frac{i}n}.\end{align}$$

Cameron Buie
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redefine $x-1=t$ to get $$ \lim_{n \to \infty} \sum_{t=0}^{n}\frac{1}{n} \sqrt{\frac{t}{n}} $$ Can you handle from here?

Alex
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  • Let me try this out! – Phillip Feb 08 '14 at 04:54
  • If I follow what you have written, then I believe I will get a = 0 and b = 1 where it should be the integral of sqrt(t) with the parameters of a = 0 and b = 1? However do I not need to convert it back to x? – Phillip Feb 08 '14 at 05:00
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    $t$ is the index for summation; you set $\frac{t}{n}=x$ to get $\int_{0}^{1}\sqrt{x}dx$ – Alex Feb 08 '14 at 05:00