## Calculus: Early Transcendentals 8th Edition

$-9$
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 $\sum_{i=1}^{\ n} i= \frac{n(n+1)}{2}$ $\lim_{n\to\infty} (-18/n^2)\frac{n(n+1)}{2}$ $\lim_{n\to\infty} \frac{-18}{2}\frac{n(n+1)}{n^2}$ $\lim_{n\to\infty} -9\frac{n+1}{n}$ $\lim_{n\to\infty} -9(1+\frac{1}{n})$ $\lim_{n\to\infty} -9(1+0)$ $\lim_{n\to\infty} -9=-9$