Answer
(a) $46=A$
(b) $52=A$
$(\mathrm{c}) \mathrm{A}=58$
Work Step by Step
$\Delta x=\frac{-a+b}{n}=\frac{-2+6}{4}=1$
(a) Left end points: $x_{k}^{*}=x_{k-1}=a+(-1+k) \Delta x=2+(-1+k) \times 1=k+1$
$\therefore f\left(x_{k}^{*}\right)=1+3(1+k)=3 k+4$
$A=\left[\sum_{k=1}^{4}(4+3 k)\right] \cdot 1=10+13+16+7=46$
(b) Midpoints: $x_{k}^{*}=x_{k-1}=a+\left(-\frac{1}{2}+k\right) \Delta x=2+\left(-\frac{1}{2}+k\right) \times 1=\frac{3}{2}+k$
$\therefore f\left(x_{k}^{*}\right)=3\left(\frac{3}{2}+k\right)+1=\frac{11}{2}+3 k$
$A=\left[\sum_{k=1}^{4}\left(\frac{11}{2}+3 k\right)\right] \cdot 1=\frac{17}{2}+\frac{23}{2}+\frac{29}{2}+\frac{35}{2}=52$
(c) Right end points:
$x_{k}^{*}=x_{k-1}=a+k \Delta x=2+k \times 1=2+k$
$\therefore f\left(x_{k}^{*}\right)=1+(2+k)3=7+3 k$
$A=\left[\sum_{k=1}^{4}(7+3 k)\right] \cdot 1=13+16+19+10= 58$
