5.4 Integral as a limit of areas

We split up the interval $[a,b]$ into subintervals $I_{j}=[a+(j-1)h,a+jh]$ of length $h$, where $j=1,2,\dots,n$ and $b-a=nh$. Supposing that $m_{j}\leq f(x)\leq M_{j}$ for $x$ in $I_{j}$, we find it evident that

$$\Bigl({\hbox{area of lower rectangle based upon}}\quad I_{j}\Bigr)$$

$$\leq\Bigl({\hbox{area under graph above}}\quad I_{j}\Bigr)$$

$$\leq\Bigl({\hbox{area of upper rectangle based upon}}\quad I_{j}\Bigr).$$

Since the area of a rectangle is simply $(base)\cdot(height)$, we have

$$hm_{j}\leq\bigl({\hbox{area under the graph above}}\quad I_{j}\bigr)\leq hM_{j}.$$

On summing over $j$, we obtain

$$\sum_{j=1}^{n}hm_{j}\leq\bigl({\hbox{area under graph above }}\,[a,b]\bigr)\leq\sum_{j=1}^{n}hM_{j}.$$