use left and right endpoints and the given number of rectangles to find two approximations of the area of…

Explanation:

Step1: Calculate the width of each rectangle

The interval is $[a,b]=[0,\pi]$ and $n = 6$. The width of each rectangle $\Delta x=\frac{b - a}{n}=\frac{\pi-0}{6}=\frac{\pi}{6}$.

Step2: Find the left - endpoints approximation

The left - endpoints of the subintervals are $x_i=a + i\Delta x$ for $i = 0,1,\cdots,5$. $x_0=0,x_1=\frac{\pi}{6},x_2=\frac{\pi}{3},x_3=\frac{\pi}{2},x_4=\frac{2\pi}{3},x_5=\frac{5\pi}{6}$ The left - endpoints approximation $L_6=\sum_{i = 0}^{5}g(x_i)\Delta x=\Delta x\left[g(x_0)+g(x_1)+g(x_2)+g(x_3)+g(x_4)+g(x_5)\right]$ $=\frac{\pi}{6}\left[9\sin(0)+9\sin\left(\frac{\pi}{6}\right)+9\sin\left(\frac{\pi}{3}\right)+9\sin\left(\frac{\pi}{2}\right)+9\sin\left(\frac{2\pi}{3}\right)+9\sin\left(\frac{5\pi}{6}\right)\right]$ $=\frac{\pi}{6}\left[0 + \frac{9}{2}+\frac{9\sqrt{3}}{2}+9+\frac{9\sqrt{3}}{2}+\frac{9}{2}\right]=\frac{\pi}{6}\left(18 + 9\sqrt{3}\right)\approx14.1372$

Step3: Find the right - endpoints approximation

The right - endpoints of the subintervals are $x_i=a+(i + 1)\Delta x$ for $i=0,1,\cdots,5$. $x_1=\frac{\pi}{6},x_2=\frac{\pi}{3},x_3=\frac{\pi}{2},x_4=\frac{2\pi}{3},x_5=\frac{5\pi}{6},x_6=\pi$ The right - endpoints approximation $R_6=\sum_{i = 0}^{5}g(x_{i + 1})\Delta x=\Delta x\left[g(x_1)+g(x_2)+g(x_3)+g(x_4)+g(x_5)+g(x_6)\right]$ $=\frac{\pi}{6}\left[9\sin\left(\frac{\pi}{6}\right)+9\sin\left(\frac{\pi}{3}\right)+9\sin\left(\frac{\pi}{2}\right)+9\sin\left(\frac{2\pi}{3}\right)+9\sin\left(\frac{5\pi}{6}\right)+9\sin(\pi)\right]$ $=\frac{\pi}{6}\left[\frac{9}{2}+\frac{9\sqrt{3}}{2}+9+\frac{9\sqrt{3}}{2}+\frac{9}{2}+0\right]=\frac{\pi}{6}\left(18 + 9\sqrt{3}\right)\approx14.1372$

Answer:

left endpoints approximation Area: $14.1372$ right endpoints approximation Area: $14.1372$