Integrating using Riemann sum method

Question

i tried integrating $$\int_4^13x-2\,$$, using Riemann sum by definition $$\int_a^bf(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(xi)\,\Delta x$$ where $$\Delta x =\frac{b-a}n$$ and $xi$ is any point within each interval i take $\Delta x= \frac {3}{n}$ and $xi=\frac {3i}{n}$ which most right point in each interval , this turn general formula into $$\sum_{i=1}^{n} f(\frac {3i}{n)\frac {3}{n}$$

$\sum_{i=1}^{n} \frac {27i-6n}{n^2}$ if we take the sum it become $\frac {27n^2+27n-12n^2}{2n^2}$ defining $i$ as $\frac {n(n+1)}{2}$ and summation of constant as $cn$. and by taking limit to infinity final answer is $7.5$ that obviously wrong , please point out my mistake

It's better to start from$$\int_4^13x-2,=-\space \int_1^43x-2,$$ — Khosrotash, Oct 31 '19 at 07:59
You are missing something important in your integral! As now it makes no sense. — mathreadler, Oct 31 '19 at 08:12
Thanks for all for kind response, i actually made some major mistake above and having hard time expressing myself due my ignorance of latex , but now i understand there is big flaw in my understanding in very basic of the formula — pikarin-g, Oct 31 '19 at 09:57

score 0 · Accepted Answer · edited Oct 31 '19 at 09:12

We have that

$$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} \frac{b-a} {n} \sum_{k=1}^{n}f\left(a+k\cdot\frac {b-a} {n} \right)$$

then

$$\int_4^13x-2 \ dx=-\int_1^43x-2 \ dx=-\lim_{n\to \infty}\frac3n\sum_{k=1}^n \left(3\left(1+\frac {3k} n\right)-2\right)$$

and

$$-\frac3n\sum_{k=1}^n \left(3\left(1+\frac {3k} n\right)-2\right)=-\frac 3 n\left(n+\frac 9n \frac{n(n+1)}2\right)=-3-\frac {27}2\frac{n(n+1)}{n^2}\to -\frac{33}2$$

Refer also to the related

Perfect understanding of Riemann Sums

Integrating using Riemann sum method

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