For the function $f(x) =x^{2}-3,$ $0\leq x\leq 6,$ partition the interval $[0,6]$ into 4 subintervals $[0,1]$, $[1,2]$, $[2,4]$, $[4,6]$ and form the Riemann sum for which

Figure 16 $f(x) = x^2-3, 0 \leq x \leq 6$

Solution In forming Riemann sums $\sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i}$, $n$ is the number of subintervals in the partition, $f(u_{i})$ is the value of $f$ at the number $u_{i}$ chosen in the $i$th subinterval, and $\Delta x_{i}$ is the length of the $i$th subinterval.

(a) Figure 16 shows the graph of $f,$ the partition of the interval $[0,6]$ into the 4 subintervals, and values of $f(u_{i})$ at the left endpoint of each subinterval, namely, $$\begin{array}{rcl@{\qquad}crcl} f(u_{1}) &=& f(0) =-3 & f(u_{2}) &=&f(1) =-2\\ f(u_{3}) &=& f(2) =1 & f(u_{4}) &=& f(4) = 13 \end{array}$$

The 4 subintervals have length $$\Delta x_{1}=1-0=1 \qquad \Delta x_{2}=2-1=1 \qquad \Delta x_{3}=4-2=2 \qquad \Delta x_{4}=6-4=2$$

The Riemann sum is formed by adding the products $f(u_{i}) \Delta x_{i}$ for $i=1,2,3,4$. $$\sum\limits_{i=1}^{4} f(u_{i}) \Delta x_{i}=-3\cdot 1 + (-2) \cdot 1+1\cdot 2 + 13 \cdot 2=23$$

Figure 17 $f(x) = x^2-3, 0 \leq x \leq 6$

(b) See Figure 17. If $u_{i}$ is chosen as the midpoint of each subinterval, then the value of $f(u_{i})$ at the midpoint of each subinterval is $$\begin{array}{rcl@{\qquad}crcl} f(u_{1}) &=& f\left(\dfrac{1}{2}\right) =-\dfrac{11}{4} & f(u_{2}) &=&f\left( \dfrac{3}{2}\right) =-\dfrac{3}{4}\\ f(u_{3}) &=&f (3) =6 & f(u_{4}) &=& f(5) = 22 \end{array}$$

The Riemann sum formed by adding the products $f(u_{i}) \Delta x_{i}$ for $i=1,2,3,4$ is $$\sum\limits_{i=1}^{4}f(u_{i}) \Delta x_{i}=-\dfrac{11}{4} \cdot 1 + \left(-\dfrac{3}{4}\right) \cdot 1 + 6 \cdot 2 + 22 \cdot 2 =\dfrac{105}{2}=52.5$$