Answer
The given integral is
$$
\begin{aligned} \int_{2}^{5}(4-2 x) d x &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=-9 \end{aligned}
$$
Work Step by Step
$$
\int_{2}^{5} (4-2x)d x
$$
The width of the subintervals is
$$
\Delta x=(5-2) / n=\frac{3}{n},
$$
and
$$
x_{i}=2+i \Delta x=2+\frac{3i}{n}.
$$
The given integral is
$$
\begin{aligned} \int_{2}^{5}(4-2 x) d x &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(2+\frac{3 i}{n}\right) \frac{3}{n} \\
&=\lim _{n \rightarrow \infty} \frac{3}{n} \sum_{i=1}^{n}\left[4-2\left(2+\frac{3 i}{n}\right)\right] \\
&=\lim _{n \rightarrow \infty} \frac{3}{n} \sum_{i=1}^{n}\left[-\frac{6 i}{n}\right] \\
&=\lim _{n \rightarrow \infty} \frac{3}{n}\left(-\frac{6}{n}\right) \sum_{i=1}^{n} i \\
&=\lim _{n \rightarrow \infty}\left(-\frac{18}{n^{2}}\right)\left[\frac{n(n+1)}{2}\right] \\
&=\lim _{n \rightarrow \infty}\left(-\frac{18}{2}\right)\left(\frac{n+1}{n}\right) \\
&=-9 \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\\
&=-9(1)\\
&=-9 \end{aligned}
$$
