Chapter 5.2, Problem 63E

To determine

To evaluate: An integral using graph.

Answer to Problem 63E

The value of required integral is $0.72<{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)}dx<1.00$ that is correctly 0.75 units.

Explanation of Solution

Given data: The given integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)dx}$ , and ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^3\right)dx}=\frac{1}{4}$ .

Method/Formula used:

The area between a curve $y=f\left(x\right)$ and x -axis in the interval $a\le x\le b$ is given by

  $A={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ … (1)

Calculations:

The given curve is $y=1-x^3$ and the interval $0\le x\le 1$ is shown in Fig. (1).

  

The value of integral ${\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)}dx$ is calculated as:

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)}dx={\left[x-\frac{x^4}{4}\right]}_0^1\\ =\left[\left(1-\frac{1^4}{4}\right)-\left(0-\frac{0^4}{4}\right)\right]\\ =1-\frac{1}{4}\\ =0.75\end{array}$

Here, ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^3\right)dx}=\frac{1}{4}$ is the unshaded region shown in Fig.1.

From the graph it is observed that area of each small square is $\frac{1\times 1}{25}=0.04\text{units}$ . Total number of full small squares in the shaded region that represents the area between the curve and the x -axis is around 18.

Therefore, total area under the curve that equals the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)}dx$ is $\approx \text{18×0}\text{.04=0}\text{.72}$ , therefore

  $0.72<{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^3\right)}dx<1.00$ .