Chapter 13.5, Problem 11E

a.

To determine

To find the area of the graph of $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

Answer to Problem 11E

The area of the graph of $f\left(x\right)=1+x^2$ is $8\text{ and }\mathrm{6.875.}$

Explanation of Solution

Given function $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

Calculation:

Let, $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{2-\left(-1\right)}{3}\\ =1\end{array}$

Graph and the rectangles as shown below:

  

The value $A_1$ underestimates the area as all rectangles are under the function’s graph.

Use the right endpoints.

  $\begin{array}{l}{A}_1={\displaystyle {\sum }_{K=1}^{3}f}\left(x_k\right)\Delta x\\ =\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right]\Delta x\\ =\left(1+0^2+1+1^2+1+2^2\right)\cdot 1\\ =8\end{array}$

Having 6 rectangles:

  $\begin{array}{l}\Delta x=\frac{2-\left(-1\right)}{6}\\ =0.5\end{array}$

Use the left endpoints.

  

  $\begin{array}{l}{A^{\prime }}_1={\displaystyle {\sum }_{K=1}^{6}f}\left(x_k\right)\Delta x\\ =\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+f\left(x_5\right)+f\left(x_6\right)\right]\Delta x\\ =\left(1+{\left(-0.5\right)}^2+1+0^2+1+{0.5}^2+1+1^2+1+{1.5}^2+1+2^2\right)\cdot 0.5\\ =6.875\end{array}$

Thus, the area of the graph of $f\left(x\right)=1+x^2$ is $8\text{ and }\mathrm{6.875.}$

b.

To determine

To find the area of the graph of $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

Answer to Problem 11E

The area of the graph of $f\left(x\right)=1+x^2$ is $5,\text{ and }\mathrm{5.375.}$

Explanation of Solution

Given function $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

Calculation:

Let, $f\left(x\right)=1+x^2,x\in \left[-1,2\right].$

  

The value $A_2$ overestimates the area as all rectangles have a region above the function’s graph.

Use the left endpoints.

  $\begin{array}{l}{A}_2={\displaystyle {\sum }_{K=1}^{3}f}\left(x_k\right)\Delta x\\ =\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right]\Delta x\\ =\left(1+{\left(-1\right)}^2+1+0^2+1+1^2\right)\cdot 1\\ =5\end{array}$

Having 6 rectangles:

Having 6 rectangles:

  $\begin{array}{l}\Delta x=\frac{2-\left(-1\right)}{6}\\ =0.5\end{array}$

  

  $\begin{array}{l}{A^{\prime }}_2={\displaystyle {\sum }_{K=1}^{6}f}\left(x_k\right)\Delta x\\ =\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+f\left(x_5\right)+f\left(x_6\right)\right]\Delta x\\ =\left(1+{\left(-1\right)}^2+1+{\left(-0.5\right)}^2+1+{0.5}^2+1+0^2+1+{0.5}^2+1+1^2+1+{1.5}^2\right)\cdot 0.5\\ =5.375\end{array}$

Thus, the area of the graph of $f\left(x\right)=1+x^2$ is $5\text{ and }\mathrm{5.375.}$