Chapter 5.10, Problem 3E

To determine

To find the area under the curve $y=\frac{1}{x^3}$ from $x=1\text{to}x=t$

Answer to Problem 3E

The Area under the curve for different values of $t$ is:

1. When $t\text{=10}$ , $\text{area = 0}\text{.495}$ square units
2. When $t\text{=100}$ , $\text{area = 0}\text{.4999}$ square units
3. When $t\text{=1000}$ , $\text{area = 0}\text{.49995}$ square units
4. When $t>1$ , $\text{area = 0}\text{.5}$ square units

Explanation of Solution

Given information:

Given curve is $y\text{(}x\text{)}\text{=}\frac{1}{x^3}$

Calculation:

  $\begin{array}{l}\text{The area under the curve = }{\displaystyle \underset{1}{\overset{t}{\int }}y(x)}dx\\ ={\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^3}}dx={\left[-\frac{1}{2x^2}\right]}_1^t\\ =-\frac{1}{2}\left[\frac{1}{t^2}-1\right]=\frac{1}{2}\left[1-\frac{1}{t^2}\right]\end{array}$

1. When $t\text{ =10, area under the curve = }\frac{99}{200}$ square units
2. When $t\text{=100, area under the curve = }\frac{9999}{20000}$ square units
3. When $t\text{=1000, area under the curve = }\frac{999999}{2000000}$ square units

Also, find the area of the curve, when $t>1\text{or}t\in [1,\infty ]$

Thus, $\text{area under the cuve}\text{=}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^3}}dx={\left[-\frac{1}{2x^2}\right]}_1^{\infty }=\frac{1}{2}$

Hence, the area under the curve for the different values of $t$ is:

1. When $t\text{=10}$ , $\text{area = 0}\text{.495}$ square units
2. When $t\text{=100}$ , $\text{area = 0}\text{.4999}$ square units
3. When $t\text{=1000}$ , $\text{area = 0}\text{.49995}$ square units
4. When $t>1$ , $\text{area = 0}\text{.5}$ square units