Chapter 5.1, Problem 22E

(a)

To determine

To find: The expression for area under the curve $y=x^3$.

Answer to Problem 22E

The area under the curve $y=x^3$ is expressed as $\underset{\_}{\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}}$.

Explanation of Solution

Given:

The function is $y=f\left(x\right)=x^3$.

The upper limit is $b=1$ and lower limit is $a=0$.

Formula used:

The area A of the region (S) under the graph f of a continuous function is the sum of area of the approximating rectangles is given by,

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x$ (1)

Calculation:

Find the width of the interval $\left(\Delta x\right)$ using the relation:

$\Delta x=\frac{b-a}{n}$ (2)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

Substitute 1 for b and 0 for a in Equation (2).

$\begin{array}{c}\Delta x=\frac{1-0}{n}\\ =\frac{1}{n}\end{array}$

Find the value of $x_i$ using the relation:

$x_i=a+i\Delta x$ (3)

Substitute 0 for a and $\frac{1}{n}$ for $\Delta x$ in Equation (3).

$\begin{array}{c}{x}_i=0+i\frac{1}{n}\\ =\frac{i}{n}\end{array}$

Use definition 2 to obtain the expression for area under the curve.

Substitute $x_i^3$ for $f\left(x_i\right)$ and $\frac{1}{n}$ for $\Delta x$ in equation (1).

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{x}_i^3}\cdot \frac{1}{n}$ (4)

Substitute $\frac{i}{n}$ for $x_i$ in equation (4).

$A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}$

Therefore, the expression for the area under the curve is $\underset{\_}{\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}}$.

(b)

To determine

To evaluate: The limit $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}$.

Answer to Problem 22E

The value of the limit is $\underset{\_}{\frac{1}{4}}$.

Explanation of Solution

Calculation:

The general expression of sum of cubes for the first n integers is shown below:

$1^3+2^3+3^3+\mathrm{...}+n^3={\left[\frac{n\left(n+1\right)}{2}\right]}^2$

Rearrange the limit function as shown below.

$\begin{array}{c}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{i^3}{n^3}}\cdot \frac{1}{n}\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\end{array}$ (1)

Apply the general expression of sum of cubes in equation (1).

$\begin{array}{c}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}=\underset{n\to \infty }{\lim }\frac{1}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}\left[\frac{n^2{\left(n+1\right)}^2}{4}\right]\\ =\frac{1}{4}\underset{n\to \infty }{\lim }\left[\frac{{\left(n+1\right)}^2}{n^2}\right]\\ =\frac{1}{4}\underset{n\to \infty }{\lim }\left[\frac{n^2+2n+1}{n^2}\right]\end{array}$

On further simplification,

$\begin{array}{c}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\cdot \frac{1}{n}=\frac{1}{4}\underset{n\to \infty }{\lim }\left[1+\frac{2}{n}+\frac{1}{n^2}\right]\\ =\frac{1}{4}\underset{n\to \infty }{\lim }\left[{\left(1+\frac{1}{n}\right)}^2\right]\\ =\frac{1}{4}{\left(1+\frac{1}{\infty }\right)}^2\\ =\frac{1}{4}{\left(1+0\right)}^2\end{array}$

$=\frac{1}{4}$

Therefore, the value of the limit is $\underset{\_}{\frac{1}{4}}$.