# Using the definition of a definite integral find $\int_2^4 (3x^2-2)dx$

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.

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