Chapter 12, Problem 15CT

To determine

Find the area of the region bounded by the graph of the function.

Answer to Problem 15CT

The area of region bounded by graph is $\boxed{\frac{25}{2}=12.5}$

Explanation of Solution

Given information:

Approximate the area of the region bounded by the graph of $f\left(x\right)=8-2x^2$ shown at the left using the indicated number of rectangles of equal width.

  

Calculation:

Consider the given function and its graph,

  $f\left(x\right)=5-\frac{2}{5}x$

  

To calculate the area of shaded region by using limits,

First of all find the interval on $x-axis$ of shaded region which are $\left(0,2\right)$ .

Now estimate the number of rectangles for iteration $\left(i\right)$ which are $4$

Now divide the number of intervals by number of rectangles to obtain the width.

  $h=\frac{2-0}{4}=0.5$

Now find $f\left(hi\right)$ for the value of $h$

  $\begin{array}{l}f\left(hi\right)=f\left(\frac{i}{2}\right)\\ =8-2{\left(\frac{i}{2}\right)}^2\\ =8-\frac{i^2}{2}\end{array}$

Now find submission,

  $\begin{array}{l}{\displaystyle \sum _{i=0}^{h-1}hf}\left(hi\right)={\displaystyle \sum _{i=0}^3\frac{1}{2}}\left[8-\frac{i^2}{2}\right]\\ =\frac{1}{2}{\displaystyle \sum _{i=0}^3\left[8-\frac{i^2}{2}\right]}\\ =\frac{1}{2}{\displaystyle \sum _{i=0}^{3}8-\frac{1}{2}{\displaystyle \sum _{i=0}^3\left[\frac{i^2}{2}\right]}}\\ =\frac{1}{2}{\displaystyle \sum _{i=0}^{3}8-\frac{1}{4}{\displaystyle \sum _{i=0}^3{i}^2}}\end{array}$

With the help of standard submission formula,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}k=kn}\\ {\displaystyle \sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}\\ so,\\ {\displaystyle \sum _{i=1}^{h}hf}\left(hi\right)=\frac{1}{2}\left[8\times 4\right]-\frac{1}{4}\left[\frac{3\left(3+1\right)\left(6+1\right)}{6}\right]\\ =16-\frac{1}{4}\left[\frac{3\left(4\right)\left(7\right)}{6}\right]\\ =16-\frac{7}{2}\\ =\frac{25}{2}\end{array}$

Hence the area of region bounded by graph is $\boxed{\frac{25}{2}=12.5}$