Chapter 11.5, Problem 20E

To determine

To calculate:

The approximate area value of the following region:

  $f\left(x\right)=\frac{1}{2}{\left(x-1\right)}^3$

Answer to Problem 20E

The approximate area of the region is $4.125$ Sq units.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{2}{\left(x-1\right)}^3$

  

Length of the interval along x-axis: $L=2$ units.

No of rectangles: $4$ rectangles

Width of the rectangle: $W=\frac{2}{4}=\frac{1}{2}$ .

Formula Used:

By using this formula:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}}\\ {\displaystyle \sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}\\ {\displaystyle \sum _{i=1}^n{i}^3=\frac{n^2{\left(n+1\right)}^2}{4}}\\ {\displaystyle \sum _{i=1}^{n}c=nc}\end{array}$

Calculation:

Consider this equation:

  $f\left(x\right)=\frac{1}{2}{\left(x-1\right)}^3$

The four intervals are $\left(\frac{1}{2},1\right),\left(1,\frac{3}{2}\right),\left(\frac{3}{2},2\right),\left(2,\frac{5}{2}\right),\left(\frac{5}{2},3\right)$ .

The sum of areas of the four rectangles:

  $\begin{array}{l}{\displaystyle \sum _{i=2}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)=}{\displaystyle \sum _{i=2}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}\\ {\displaystyle \sum _{i=2}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)=}{\displaystyle \sum _{i=2}^{6}f\left(\frac{{\left(\frac{1}{2}i-1\right)}^3}{2}\right)\left(\frac{1}{2}\right)}\\ {\displaystyle \sum _{i=2}^{6}f\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)=}\frac{1}{32}\left({\displaystyle \sum _{i=2}^6{i}^3}-6i^2+12i+8\right)\\ {\displaystyle \sum _{i=2}^{6}f\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)=}\frac{1}{32}\left({\displaystyle \sum _{i=2}^6{i}^3}-6i^2+12i+8-{\displaystyle \sum _{i=0}^1{i}^3}-6i^2+12i+8\right)\end{array}$

By using this formula:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}}\\ {\displaystyle \sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}\\ {\displaystyle \sum _{i=1}^n{i}^3=\frac{n^2{\left(n+1\right)}^2}{4}}\\ {\displaystyle \sum _{i=1}^{n}c=nc}\end{array}$

Comparing with this formula:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)=}\frac{1}{32}\left({\left(\frac{6\times 7}{2}\right)}^2-\left(\frac{6\times 7\times 13}{1}\right)+12\left(\frac{6\times 7}{2}-15\right)\right)\\ {\displaystyle \sum _{i=1}^{n}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)=}4.125\end{array}$

The approximate area of the region is $4.125$ Sq units.