Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ equal subintervals and taking limit.
Answer:
The length of the interval $ \ [2,4] \ $ is 
$ \ 4-2=2 \ $ , 
Thus subinterval size $ =\Delta x=\frac{2}{n} , \ $
Then the required area is given by 
$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x \\  =  \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\ = -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n} \\ \\ = -\large  \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +..........+\frac{n}{n^2}\} \right) \\ =\large 0-  \lim_{n \to \infty} \frac{12}{n^2} (1+2+3+.....+n) \\ = -\large  \lim_{n \to \infty} \frac{12}{n^2} \cdot \frac{n(n+1)}{2} \\ =-6 $ 
Thus the required area  $ \ =6 \ $
But $ \ \int_{2}^{4} (4-3x) dx=-10 \ $  i.e., the area is $ \ 10 \ $
Why such difference in approximation above.
I thought I have done something wrong in calculation .
Kindly examine my work and help me out
 A: $$\begin{align} \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x& =  \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\&= -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n}\\ &= -\large  \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\} \right) \end{align}$$
The first row is OK, the second one is as well, but the second one is wrong.
What you should get is
$$\sum_{i=1}^n \frac{2}{n}\left(2+i\frac6n\right) = \sum_{i=1}^n\left(\frac4n + i\frac{12}{n^2}\right) \\=\left(\frac4n+1\cdot\frac{12}{n^2}\right) + \left(\frac4n+2\cdot\frac{12}{n^2}\right)+\cdots +\left(\frac{4}n+n\cdot\frac{12}{n^2}\right)$$
which we can reorder into
$$\frac4n + \frac4n + \cdots + \frac4n + 12\left(\frac1{n^2}+\frac12{n^2}+\cdots+\frac{n}{n^2}\right)$$ 
and therefore equals
$$\color{red}{n\cdot} \frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\}$$
whic is close to what you got, i.e.
$$\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +\cdots+\frac{n}{n^2}\}$$
(note the $n$ in front of $n\cdot \frac4n$)
A: There is a mistake in one of your steps:
$$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x  =  \lim_{n \to \infty} \sum_{i=1}^{n} \left\{\left[4-3\left(2+i \frac{2}{n}\right)\right] \frac{2}{n} \right\}= -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n} \\ \\ = -\large  \lim_{n \to \infty} \left(\color{blue}{n\cdot\frac{4}{n}}+12 \left[\frac{1}{n^2} + \frac{2}{n^2} +..........+\frac{n}{n^2}\right] \right) \\ =-  \lim_{n \to \infty} \left(4+\frac{12}{n^2} \sum_{k=1}^{n}k\right) =- 4 -\large  \lim_{n \to \infty} \frac{12}{n^2} \cdot \frac{n(n+1)}{2}  =\color{red}{-10} $$
