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Find, with the definition of a definite integral, where $\bar{x}i$ is the right sum of each subinterval. $$ \int_2^4 (3x^2-2)dx $$ So I start here... $$ \Delta xi = 4-2/n = \frac{2}{n} $$ For the right sum: $$ \bar{x}i= 1 +i\Delta xi = 1+\frac{2i}{n} $$ We have: $$ f(\bar{x}i) = (3x^2-2) = f(\bar{x}i) = (3(1+\frac{2i}{n})^2-2) $$

$$ f(\bar{x}i) = (\frac{12i^2}{n^2}+\frac{12i}{n}+1) $$

With the sum of Riemann: $$ SR= \sum f(\bar{x}i)\Delta xi = \sum (\frac{12i^2}{n^2}+\frac{12i}{n}+1)* \frac{2}{n} $$ $$ SR = \sum (\frac{24i^2}{n^3}+\frac{24i}{n^2}+\frac{2}{n}) $$

$$ SR= \frac{12}{n^2}\sum i^2 + \frac{12}{n}\sum i + \frac{2}{n}\sum 1 $$ $$ SR= \frac{12}{n^2}(\frac{n(n+1)(2n+1)}{6}) + \frac{12}{n}(\frac{(n(n+1))}{2}) + \frac{2}{n} (n) $$

$$ SR= \frac{8n^2+8n+4}{n^2} + \frac{12n+12}{n} + 2 $$ $$ SR= \frac{4}{n^2} + \frac{20}{n} + 22 $$ And finally: $$\lim_{n\rightarrow \infty}\sum f(\bar{x})\Delta xi=\frac{4}{n^2} + \frac{20}{n} + 22 = 22$$

This is my answer. But the answer from online calculators are different (52).. Can anyone spot my mistakes? Thank you.

$$$$ Edit: Okay thank you guys! My mistake right at the beginning... where it should have started at 2 and not 1. Thanks again.

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    $\begingroup$ $x_i=2+\frac{2i}{n}$ $\endgroup$ Commented Jun 16, 2016 at 20:58
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    $\begingroup$ Because you start at $2$ then pick $f(2+n \delta (x)$ to sum up until you get to $4$. $\endgroup$ Commented Jun 16, 2016 at 20:59

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\begin{align} & f({{x}_{i}})=3{{\left( 2+\frac{2i}{n} \right)}^{2}}-2=12\left( 1+\frac{2i}{n}+\frac{{{i}^{2}}}{{{n}^{2}}} \right)-2=10+\frac{24i}{n}+\frac{12{{i}^{2}}}{{{n}^{2}}} \\ & \Delta x\sum\limits_{i=1}^{n}{f({{x}_{i}})=\frac{2}{n}}\left( 10n+12(n+1)+\frac{2(n+1)(2n+1)}{n} \right) \\ & \underset{n\to \infty }{\mathop{\lim }}\,\Delta x\sum\limits_{i=1}^{n}{f({{x}_{i}})=20+24+}8=52 \\ \end{align}

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