Answer
(a) $\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=-2$
(b) $\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=-1$
$(\mathrm{c}) \sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=-2$
Work Step by Step
$\Delta x=\frac{-a+b}{n}=\frac{1+3}{4}=1$
(a) Left end points: $x_{k}^{*}=x_{k-1}=(-1+k) \Delta x+a=-1+(-1+k) \times 1=-2+k$
$\therefore f\left(x_{k}^{*}\right)=2(k-2)-(k-2)^{2}$
$\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\left[\sum_{k=1}^{4}\left[(-2+k)2-(-2+k)^{2}\right] \cdot 1=0+1+0-3=-2\right.$
(b) Midpoints: $x_{k}^{*}=a+\left(k-\frac{1}{2}\right) \Delta x=-1+\left(k-\frac{1}{2}\right) \times 1=-\frac{3}{2}+k$
$f\left(x_{k}^{*}\right)=2\left(-\frac{3}{2}+k\right)-\left(k-\frac{3}{2}\right)^{2}$
$\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\left[\sum_{k=1}^{4}\left[2\left(k-\frac{3}{2}\right)-\left(k-\frac{3}{2}\right)^{2}\right] \cdot 1=\frac{3}{4}+\frac{3}{4}-\frac{5}{4}-\frac{5}{4}+=-1\right.$
(c) Right end points: $x_{k}^{*}=x_{k}=a+k \Delta x=k-1 \times 1=-1+k $
$\therefore f\left(x_{k}^{*}\right)=2(k-1)-(k-1)^{2}$
$\sum_{k=1}^{4} f\left(x_{k} *\right) \Delta x=\left[\sum_{k=1}^{4}\left[2(-1+k)-(-1+k)^{2}\right] \cdot 1=1+0-3+0=-2\right.$
