Answer
(a) $\frac{352}{105}$
(b) $\frac{25}{12}$
(C) $\frac{496}{315}$
Work Step by Step
$\Delta x=\frac{-a+b}{n}=\frac{-1+9}{4}=2$
(a) Left end points: $x_{k}^{*}=x_{k-1}=(-1+k) \Delta x+a=1+(-1+k) \times 2=-1+2 k$
$\therefore f\left(x_{k}^{*}\right)=\frac{1}{-1+2 k}$
$A=\left[\sum_{k=1}^{4} \frac{1}{-1+2 k}\right] \cdot 2=\left[\frac{1}{5}+\frac{1}{7}+\frac{1}{3}+1\right] \cdot 2=\left[\frac{352}{105}\right]$
(b) Midpoints:
$x_{k}^{*}=a+\left(-\frac{1}{2}+k\right) \Delta x=1+\left(-\frac{1}{2}+k\right) \times 2=2 k$
$\therefore f\left(x_{k}^{*}\right)=\frac{1}{2 k}$
$A=\left[\sum_{k=1}^{4} \frac{1}{2 k}\right] \cdot 2=\left[\frac{1}{14}+\frac{1}{6}+\frac{1}{8}+\frac{1}{2}\right] \cdot 2=\left[\frac{25}{12}\right]$
(c) Right end points:
$x_{k}^{*}=x_{k}=a+k \Delta x=k+1 \times 2=2 k+1$
$\therefore f\left(x_{k}^{*}\right)=\frac{1}{1+2 k}$
$A=\left[\sum_{k=1}^{4} \frac{1}{2 k+1}\right] \cdot 2=\left[\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right]\cdot 2={\frac{496}{315}}$
