Chapter 11, Problem 15CT

To determine

To Approximate: the area of the integral bounded by the given graph using the rectangles of equal width.

Answer to Problem 15CT

The approximated area of the region bounded by the graph $f\left(x\right)=8-2x^2$ is $\boxed{12.5}$ unit square.

Explanation of Solution

Given information:

The graph of $f\left(x\right)=8-2x^2$ in the interval $\left[0,2\right]$ is

  

Formula used:

Let n be the number of partition. Here $n=4$ .

Also $\left[a,b\right]=\left[0,2\right]$

So, the formula for step length will be $h=\frac{b-a}{n}$

The area will be calculated by ${\displaystyle \sum _{i=0}^{n-1}hf\left(hi\right)}$

Calculation:

Put $x=hi$ in $f\left(x\right)=8-2x^2$ for $f\left(hi\right)$

So, $f\left(hi\right)=8-2{\left(hi\right)}^2$

Since $h=\frac{1}{2}$ , so $f\left(\frac{i}{2}\right)=8-2{\left(\frac{i}{2}\right)}^2$

Now

  $\begin{array}{c}{\displaystyle \sum _{i=0}^{n-1}hf\left(hi\right)}={\displaystyle \sum _{i=0}^{4-1}\frac{1}{2}\left(8-2{\left(\frac{i}{2}\right)}^2\right)}\\ ={\displaystyle \sum _{i=0}^{3}4}-{\displaystyle \sum _{i=0}^3{\left(\frac{i}{2}\right)}^2}\\ ={\displaystyle \sum _{i=0}^{3}4}-{\displaystyle \sum _{i=0}^3{\left(\frac{i}{2}\right)}^2}\\ =4\left(3+1\right)-\frac{1}{4}{\displaystyle \sum _{i=0}^3{\left(i\right)}^2}\left(\because {\displaystyle \sum _{i=0}^{n}k}=k\left(n+1\right)\right)\\ =4\left(3\right)-\frac{1}{4}\left(\frac{3\left(3+1\right)\left(2\left(3\right)+1\right)}{6}\right)\left(\because {\displaystyle \sum _{i=0}^n{\left(i\right)}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\\ =16-\frac{1}{4}\left(\frac{3\left(4\right)\left(7\right)}{6}\right)\\ =16-\frac{7}{2}\\ =\frac{32-7}{2}\\ =\frac{25}{2}\\ =12.5\end{array}$

Conclusion:

The approximated area of the region bounded by the graph $f\left(x\right)=8-2x^2$ is $\boxed{12.5}$ unit square.