Chapter 6, Problem 5RE

To determine

The area R by using the fundamental theorem of calculus.

Answer to Problem 5RE

The value is ${\displaystyle \underset{0}{\overset{2}{\int }}\left(4x-x^3\right)}dx=3.75$ .

Explanation of Solution

Given information:

The graph of the function as given below,

  $y=4x-x^3$

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

The required graph is

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of f on [a, b] can be approximated by a straight line segment at each subinterval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases".

That is,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=h\left(\frac{y_0+y_1}{2}\right)+h\left(\frac{y_1+y_2}{2}\right)+\mathrm{....}+h\left(\frac{y_{n-1}+y_n}{2}\right)\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\frac{h}{2}\left(y_0+2y_1+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

  $\begin{array}{l}{y}_0=f\left(a\right)\\ y_1=f\left(x_1\right)\\ .\\ .\\ y_{n-1}=f\left(x_{n-1}\right)\\ y_n=f\left(b\right)\end{array}$

Now, apply the trapezoidal rule with

  $\begin{array}{l}n=4\\ f\left(x\right)=4x-x^3\\ \left[a,b\right]=\left[0,2\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{2-0}{4}\\ h=\frac{1}{2}\end{array}$

The table shows

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\left(4x-x^3\right)}dx=\text{total area}\\ \text{=}\frac{h}{2}\left(y_0+2y_1+2y_2+2y_3+y_4\right)\\ =\frac{\frac{1}{2}}{2}\left(0+2\left(\frac{15}{8}\right)+2\left(3\right)+2\left(\frac{21}{8}\right)+0\right)\\ {\displaystyle \underset{0}{\overset{2}{\int }}\left(4x-x^3\right)}dx=3.75\end{array}$

Conclusion:

The value is

  ${\displaystyle \underset{0}{\overset{2}{\int }}\left(4x-x^3\right)}dx=3.75$