这是一份umass麻省大学 MATH 623作业代写的成功案例
Proof. We can suppose $x<y$. By the ‘additivity’ proved in
$$
\int_{a}^{y} f=\int_{a}^{x} f+\int_{x}^{y} f,
$$
thus
$$
\int_{x}^{y} f=F(y)-F(x) .
$$
If $m^{\prime}$ and $M^{\prime}$ are the infimum and supremum of $f$ on the interval $[x, y]$, we have $m \leq m^{\prime} \leq M^{\prime} \leq M$; citing
$$
m(y-x) \leq m^{\prime}(y-x) \leq \int_{x}^{y} f \leq M^{\prime}(y-x) \leq M(y-x)
$$
MATH623 COURSE NOTES ：
$$
s(\sigma)=\sum_{\nu=1}^{n} f\left(a_{\nu-1}\right) \mathrm{e}{\nu}, \quad S(\sigma)=\sum{\nu=1}^{n} f\left(a_{\nu}\right) e_{\nu}
$$
$$
S(\sigma)-s(\sigma)=\sum_{\nu=1}^{n}\left[f\left(a_{\nu}\right)-f\left(a_{\nu-1}\right)\right] e_{\nu}
$$
Now assume that the points of $\sigma$ are equally spaced, so that
$$
e_{\nu}=\frac{1}{n}(b-a) \quad(\nu=1, \ldots, n) ;
$$