Calculate definite integral using the definition (Riemann Sum)

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I would like to know how to use the Riemann Sum (With Unequal width) to calculate $\displaystyle f(x) = \sqrt{x}$ from $[1,2]$

I know how to calculate it when the interval starts with 0, for example [0,1], I will do the following way:

• $\displaystyle c_i = \frac{i^2}{n^2}$ and $\displaystyle \Delta x_i = \frac{i^2}{n^2}-\frac{(i-1)^2}{n^2}=\frac{2i-1}{n^2}$
• $\displaystyle \lim_{n\to\infty}\sum_{i=1}^nf(c_i)\Delta x_i = \lim_{n\to\infty}\sum_{i=1}^n = \sqrt{\frac{i^2}{n^2}}\left(\frac{2i-1}{n^2} \right) = \frac{2}{3}$

I understand that by choosing a partition of $\displaystyle\frac{i^2}{n^2}$, it will facilitate our calculation of summation because we have the following formula

• $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$

However, if the partition starts with 1 or other numbers, we would have something like $\displaystyle \sum_{i=1}^n \sqrt{1+\frac{i^2}{n^2}}$ which we don't have a directly formula to use.

Is it possible for me to do such Riemann Sum ($\displaystyle\sqrt{x}$ when interval doesn't start with 0) with what I have learned from First Year Calculus in university?

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2 Answers

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I see two potential approaches:

1. Try computing both integrals between $[0, 2]$ and $[0, 1)$, and subtract afterwards.

2. If you want to stick to a partition of $[1, 2]$ you could still make the partition follow the same pattern. Try $\frac{i^2}{n^2}$ for $i$ going from $n$ to $\sqrt{2}n$, for example.

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Let $$c_i=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2$$

Hence,\begin{align}\Delta x_i&=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2-\left(1+ \frac{(\sqrt2-1)(i-1)}{n}\right)^2\\ &=\left( 2+\frac{(\sqrt2-1)(2i-1)}{n}\right)\left(\frac{\sqrt{2}-1}{n} \right)\end{align}

and the integral is

\begin{align} \lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x_i &= \lim_{n \to \infty} \sum_{i=1}^n \left(1+ \frac{(\sqrt2-1)i}{n}\right)\left( 2+\frac{(\sqrt2-1)(2i-1)}{n}\right)\left(\frac{\sqrt{2}-1}{n} \right) \\ &= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3} \sum_{i=1}^n [n+(\sqrt2-1)i][2n+(\sqrt2-1)(2i-1)] \\ &= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3} \sum_{i=1}^n(2n^2+n(\sqrt2-1)(2i-1)+2(\sqrt2-1)in + (\sqrt2-1)^2i(2i-1))\\ &= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3}\left(2n^3 +n \sum_{i=1}^n 4(\sqrt2-1)i + 2\sum_{i=1}^n (\sqrt2-1)^2i^2 \right) \\ &=\lim_{n \to \infty } \frac{\sqrt2-1}{n^3} \left(2n^3 + 2n^3(\sqrt2-1)+\frac23(\sqrt2-1)^2n^3 \right) \\ &= (\sqrt2-1)\left(2\sqrt2+ \frac23(3-2\sqrt2) \right) \\ &= \frac13(\sqrt2-1)(2\sqrt2+6) \\ &= \frac23(\sqrt2-1)(\sqrt2+3) \\ &= \frac23 (2\sqrt2-1) \end{align}

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