Chapter 5.2, Problem 6E

a.

To determine

To estimate the right endpoints.

Answer to Problem 6E

The value of right endpoints is $-0.5$ .

Explanation of Solution

Given information:

The equation is

  ${\int }_{-3}^{3}g\left(x\right)dx$ with $n=6$

Calculation:

The right endpoints can be estimate as

  ${\int }_{-3}^{3}g\left(x\right)dx$

The endpoints of the 6 subintervals are $\left[-3,-2\right],\left[-2,-1\right],\left[-1,0\right],\left[0,1\right],\left[1,2\right],\left[2,3\right]$

So, the right endpoints are $-2,-1,0,1,2,3$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{3-\left(-3\right)}{6}\left[\text{since, }b=3,a=-3,n=6\right]\\ \Delta x=\frac{3+3}{6}\\ \Delta x=\frac{6}{6}\\ \Delta x=1\left[\text{divede numerator and denominator by 6}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}g\left(-2\right)=1\\ g\left(-1\right)=-0.5\\ g\left(0\right)=-1.5\\ g\left(1\right)=-1.5\\ g\left(2\right)=-0.5\\ g\left(3\right)=2.5\end{array}$

Therefore,

The right endpoint rule gives

  ${\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-2\right)+g\left(-1\right)+g\left(0\right)+g\left(1\right)+g\left(2\right)+g\left(3\right)\right]$

Put the values of $\begin{array}{l}\Delta x=1,g\left(-2\right)=1,g\left(-1\right)=-0.5,g\left(0\right)=-1.5,\\ g\left(1\right)\text{=}-\text{1}\text{.5,g}\left(2\right)\text{=}-\text{0}\text{.5 and }g\left(3\right)=2.5\end{array}$ on the above equation

  $\begin{array}{l}{\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-2\right)+g\left(-1\right)+g\left(0\right)+g\left(1\right)+g\left(2\right)+g\left(3\right)\right]\\ \text{ =1}\left(1+\left(-0.5\right)+\left(-1.5\right)+\left(-1.5\right)+\left(-0.5\right)+2.5\right)\\ \text{ =1}\left(1-0.5-1.5-1.5-0.5+2.5\right)\\ \text{ =1}\left(3.5-4\right)\\ \text{ =1}\left(-0.5\right)\\ \text{ =}-\text{0}\text{.5}\end{array}$

Hence,

The value of right endpoints is $-0.5$ .

b.

To determine

To estimate the left endpoints.

Answer to Problem 6E

The value of left endpoints is $-1$ .

Explanation of Solution

Given information:

The equation is

  ${\int }_{-3}^{3}g\left(x\right)dx$ with $n=6$

Calculation:

The left endpoints can be estimate as

  ${\int }_{-3}^{3}g\left(x\right)dx$

The endpoints of the 6 subintervals are $\left[-3,-2\right],\left[-2,-1\right],\left[-1,0\right],\left[0,1\right],\left[1,2\right],\left[2,3\right]$

So, the left endpoints are $-3,-2,-1,0,1,2$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{3-\left(-3\right)}{6}\left[\text{since, }b=3,a=-3,n=6\right]\\ \Delta x=\frac{3+3}{6}\\ \Delta x=\frac{6}{6}\\ \Delta x=1\left[\text{divede numerator and denominator by 6}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}g\left(-3\right)=2\\ g\left(-2\right)=1\\ g\left(-1\right)=-0.5\\ g\left(0\right)=-1.5\\ g\left(1\right)=-1.5\\ g\left(2\right)=-0.5\end{array}$

Therefore,

The right endpoint rule gives

  ${\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-3\right)+g\left(-2\right)+g\left(-1\right)+g\left(0\right)+g\left(1\right)+g\left(2\right)\right]$

Put the values of $\begin{array}{l}\Delta x=1,g\left(-3\right)=2,g\left(-2\right)=1,g\left(-1\right)=-0.5,\\ g\left(0\right)\text{=}-\text{1}\text{.5,g}\left(1\right)\text{=}-1.\text{5 and }g\left(2\right)=-0.5\end{array}$ on the above equation

  $\begin{array}{l}{\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-3\right)+g\left(-2\right)+g\left(-1\right)+g\left(0\right)+g\left(1\right)+g\left(2\right)\right]\\ \text{ =1}\left(2+1+\left(-0.5\right)+\left(-1.5\right)+\left(-1.5\right)+\left(-0.5\right)\right)\\ \text{ =1}\left(2+1-0.5-1.5-1.5-0.5\right)\\ \text{ =1}\left(3-4\right)\\ \text{ =1}\left(-1\right)\\ \text{ =}-1\end{array}$

Hence,

The value of left endpoints is $-1$ .

c.

To determine

To estimate the midpoints.

Answer to Problem 6E

The value of midpoints is $-1.75$ .

Explanation of Solution

Given information:

The equation is

  ${\int }_{-3}^{3}g\left(x\right)dx$ with $n=6$

Calculation:

The midpoints can be estimate as

  ${\int }_{-3}^{3}g\left(x\right)dx$

The endpoints of the 6 subintervals are $\left[-3,-2\right],\left[-2,-1\right],\left[-1,0\right],\left[0,1\right],\left[1,2\right],\left[2,3\right]$

So, the midpoints are $-2.5,-1.5,-0.5,0.5,1.5,2.5$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{3-\left(-3\right)}{6}\left[\text{since, }b=3,a=-3,n=6\right]\\ \Delta x=\frac{3+3}{6}\\ \Delta x=\frac{6}{6}\\ \Delta x=1\left[\text{divede numerator and denominator by 6}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}g\left(-2.5\right)=1.5\\ g\left(-1.5\right)=0\\ g\left(-0.5\right)=-1\\ g\left(0.5\right)=-1.75\\ g\left(1.5\right)=-1\\ g\left(2.5\right)=0.5\end{array}$

Therefore,

The midpoint rule gives

  ${\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-2.5\right)+g\left(-1.5\right)+g\left(-0.5\right)+g\left(0.5\right)+g\left(1.5\right)+g\left(2.5\right)\right]$

Put the values of $\begin{array}{l}\Delta x=1,g\left(-2.5\right)=1.5,g\left(-1.5\right)=0.5,g\left(-0.5\right)=-1,\\ g\left(0.5\right)\text{=}-\text{1}\text{.75,g}\left(1.5\right)\text{=}-1\text{ and }g\left(2.5\right)=0.5\end{array}$ on the above equation

  $\begin{array}{l}{\int }_{-3}^{3}g\left(x\right)dx=\Delta x\left[g\left(-2.5\right)+g\left(-1.5\right)+g\left(-0.5\right)+g\left(0.5\right)+g\left(1.5\right)+g\left(2.5\right)\right]\\ \text{ =1}\left(1.5+0+\left(-1\right)+\left(-1.75\right)+\left(-1\right)+0.5\right)\\ \text{ =1}\left(1.5+0-1-1.75-1+0.5\right)\\ \text{ =1}\left(2-3.75\right)\\ \text{ =1}\left(-1.75\right)\\ \text{ =}-1.75\end{array}$

Hence,

The value of midpoints is $-1.75$ .