Answer
$-9$
Work Step by Step
We know the following.
$\int_{a}^{b} f(x) dx = \lim_{n\to\infty} \sum_{i=1}^{\ n} f(x_{i}) \Delta x$,
where $x_{i} = a+i\Delta x$ and $ \Delta x= (b- a)/n$.
From our problem we can gather the following.
$a=2$
$b=5$
$f(x)=4- 2x$.
Therefore,
$\Delta x= (5-2)/n= 3/n$
$x_{i}= 2 +(3i/n)$
$f(2 +(3i/n))= 4-2(2 +(3i/n))= 4-4-(6i/n)= -6i/n$.
From here we plug it all in to get the following:
$\lim_{n\to\infty} \sum_{i=1}^{\ n} (-6i/n )(3/n)$
Simplify
$\lim_{n\to\infty} \sum_{i=1}^{\ n} (-18i/n^2)$
Factor
$\lim_{n\to\infty} (-18/n^2)\sum_{i=1}^{\ n} i$
Since $\sum_{i=1}^{\ n} i= \frac{n(n+1)}{2}$
$\lim_{n\to\infty} (-18/n^2)\frac{n(n+1)}{2}$
$\lim_{n\to\infty} \frac{-18}{2}\frac{n(n+1)}{n^2}$
$\lim_{n\to\infty} -9\frac{n+1}{n}$
$\lim_{n\to\infty} -9(1+\frac{1}{n})$
$\lim_{n\to\infty} -9(1+0)$
$\lim_{n\to\infty} -9=-9$
