Approximate the area $A$ under the graph of $f(x) = x^{2}$ from $0$ to $10$ by using lower sums $s_{n}$ (rectangles that lie under the graph) for:

Solution (a) For $n=2$, we partition the closed interval $[0,10]$ into two subintervals $[0,5]$ and $[5,10]$, each of length $\Delta x=\dfrac{10-0}{2}=5$. See Figure 9(a). To compute $s_{2}$, we need to know where $f$ attains its minimum value in each subinterval. Since $f$ is an increasing function, the absolute minimum is attained at the left endpoint of each subinterval. So, for $n=2,$ the minimum of $f$ on $[0,5]$ occurs at $0$ and the minimum of $f$ on $[5,10]$ occurs at 5. The lower sum $s_{2}$ is

Figure 9

347

$$s_{2}=\sum\limits_{i=1}^{2}~f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{2}f(c_{i}) \underset{\underset{\underset{\color{#0066A7}{f\left(c_{1}\right) =f\left( 0\right);f\left( c_{2}\right) =f\left(5\right)}}{\color{#0066A7}{\Delta x=5}}}{\color{#0066A7}{\uparrow}}}{=}5\,\left[ ~f(0)+f(5)\right] \underset{\underset{\underset{\color{#0066A7}{f(5)=25}}{\color{#0066A7}{f(0)=0}}}{\color{#0066A7}{\uparrow}}}{=}5(0+25)=125$$

(b) For $n=5$, partition the interval $[0,10]$ into five subintervals $[0,2] ,$ $[2,4] ,$ $[4,6] ,$ $[6,8] ,$ $[8,10]$, each of length $\Delta x = \dfrac{10-0}{5}=2$. See Figure 9(b). The lower sum $s_{5}$ is $$\begin{eqnarray*} s_{5} &=&\sum\limits_{i=1}^{5} f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{5}f(c_{i})=2 [ f(0)+f(2)+f(4) +f(6)+f(8)] \\ &=&2(0+4+16+36+64)=240 \end{eqnarray*}$$

(c) For $n=10$, partition $[0,10]$ into $10$ subintervals, each of length $\Delta x=\dfrac{10-0}{10}=1$. See Figure 9(c). The lower sum $s_{10}$ is $$\begin{eqnarray*} s_{10} &=&\sum\limits_{i=1}^{10} f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{10}f(c_{i})=1[ f(0)+f(1)+f(2)+\cdots +f(9) ] \\ &=&0+1+4+9+16+25+36+49+64+81=285 \end{eqnarray*}$$