Chapter 5.2, Problem 5E

a.

To determine

To estimate the right endpoints.

Answer to Problem 5E

The value of right endpoints is 4.

Explanation of Solution

Given information:

The equation is

  ${\int }_0^{8}f\left(x\right)dx$ with $n=4$

Calculation:

The right endpoints can be estimate as

  ${\int }_0^{8}f\left(x\right)dx$

The endpoints of the 4 subintervals are $\left[0,2\right],\left[2,4\right],\left[4,6\right],\left[6,8\right]$

So, the right endpoints are $2,4,6,8$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{8-0}{4}\left[\text{since, }b=8,a=0,n=4\right]\\ \Delta x=\frac{8}{4}\\ \Delta x=2\left[\text{divede numerator and denominator by 4}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}f\left(2\right)=1\\ f\left(4\right)=2\\ f\left(6\right)=-2\\ f\left(8\right)=1\end{array}$

Therefore,

The right endpoint rule gives

  ${\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(2\right)+f\left(4\right)+f\left(6\right)+f\left(8\right)\right]$

Put the values of $\Delta x=2,f\left(2\right)=1,f\left(4\right)=2,f\left(6\right)=-2\text{ and }f\left(8\right)=1$ on the above equation

  $\begin{array}{l}{\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(2\right)+f\left(4\right)+f\left(6\right)+f\left(8\right)\right]\\ \text{ =2}\left(1+2+\left(-2\right)+1\right)\\ \text{ =2}\left(1+2-2+1\right)\\ \text{ =2}\left(2\right)\\ \text{ =4}\end{array}$

Hence,

The value of right endpoints is 4.

b.

To determine

To estimate the left endpoints.

Answer to Problem 5E

The value of left endpoints is 6.

Explanation of Solution

Given information:

The equation is

  ${\int }_0^{8}f\left(x\right)dx$ with $n=4$

Calculation:

The left endpoints can be estimate as

  ${\int }_0^{8}f\left(x\right)dx$

The endpoints of the 4 subintervals are $\left[0,2\right],\left[2,4\right],\left[4,6\right],\left[6,8\right]$

So, the left endpoints are $0,2,4,6$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{8-0}{4}\left[\text{since, }b=8,a=0,n=4\right]\\ \Delta x=\frac{8}{4}\\ \Delta x=2\left[\text{divede numerator and denominator by 4}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}f\left(0\right)=2\\ f\left(2\right)=1\\ f\left(4\right)=2\\ f\left(6\right)=-2\end{array}$

Therefore,

The right endpoint rule gives

  ${\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(0\right)+f\left(2\right)+f\left(4\right)+f\left(6\right)\right]$

Put the values of $\Delta x=2,f\left(0\right)=2,f\left(2\right)=1,f\left(4\right)=2\text{ and }f\left(6\right)=-2$ on the above equation

  $\begin{array}{l}{\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(0\right)+f\left(2\right)+f\left(4\right)+f\left(6\right)\right]\\ \text{ =2}\left(2+1+2+\left(-2\right)\right)\\ \text{ =2}\left(2+1+2-2\right)\\ \text{ =2}\left(3\right)\\ \text{ =6}\end{array}$

Hence,

The value of left endpoints is 6.

c.

To determine

To estimate the midpoints.

Answer to Problem 5E

The value of midpoints is 10.

Explanation of Solution

Given information:

The equation is

  ${\int }_0^{8}f\left(x\right)dx$ with $n=4$

Calculation:

The midpoints can be estimate as

  ${\int }_0^{8}f\left(x\right)dx$

The endpoints of the 4 subintervals are $\left[0,2\right],\left[2,4\right],\left[4,6\right],\left[6,8\right]$

So, the midpoints are $1,3,,5,7$

The width of the subintervals is

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{8-0}{4}\left[\text{since, }b=8,a=0,n=4\right]\\ \Delta x=\frac{8}{4}\\ \Delta x=2\left[\text{divede numerator and denominator by 4}\right]\end{array}$

From the given graph in the question we can observed that

  $\begin{array}{l}f\left(1\right)=3\\ f\left(3\right)=2\\ f\left(5\right)=1\\ f\left(7\right)=-1\end{array}$

Therefore,

The midpoint rule gives

  ${\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(1\right)+f\left(3\right)+f\left(5\right)+f\left(7\right)\right]$

Put the values of $\Delta x=2,f\left(1\right)=3,f\left(3\right)=2,f\left(5\right)=1\text{ and }f\left(7\right)=-1$ on the above equation

  $\begin{array}{l}{\int }_0^{8}f\left(x\right)dx=\Delta x\left[f\left(1\right)+f\left(3\right)+f\left(5\right)+f\left(7\right)\right]\\ \text{ =2}\left(3+2+1+\left(-1\right)\right)\\ \text{ =2}\left(3+2+1-1\right)\\ \text{ =2}\left(5\right)\\ \text{ =10}\end{array}$

Hence,

The value of midpoints is 10.