Chapter 5.2, Problem 11E

To determine

To find: The limit as a definite integral.

Answer to Problem 11E

The limit as a definite integral is

  ${{\displaystyle \int }}_0^1{x}^{3}dx$

Explanation of Solution

Given information:

The limit is: $\underset{n\to \infty }{\lim }\frac{1+2^3+3^3+4^3+\cdots +n^3}{n^4}$

Calculation:

The given limit is: $\underset{n\to \infty }{\lim }\frac{1+2^3+3^3+4^3+\cdots +n^3}{n^4}$

Rewrite as:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{1+2^3+3^3+4^3+\cdots +n^3}{n^3}\cdot \frac{1}{n}\\ =\underset{n\to \infty }{\lim }\left[\frac{1^3}{n^3}+\frac{2^3}{n^3}+\frac{3^3}{n^3}+\frac{4^3}{n^3}+\cdots +\frac{n^3}{n^3}\right]\cdot \frac{1}{n}\end{array}$

So, rewrite the following summation into an integral:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum }_{k=1}^n{(k/n)}^3\cdot \frac{1}{n}$ $\to$ (Expression 1)

The interval $[a,b]$ into $n$ equal parts and $c_k$ is an element from the $k$ th such interval, then,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum }_{k=1}^{n}f\left(c_k\right)\text{Δ}x={{\displaystyle \int }}_a^{b}f(x)dx$

Assume that, $a=0$

and $b=1$ ,so that

  $\begin{array}{c}\text{Δ}x=\frac{b-a}{n}\\ =\frac{1-0}{n}\\ =1/n\end{array}$

Now replace $1/n$ in (Expression 1) with $\text{Δ}x$ :

  $\underset{n\to \infty }{\lim }{\displaystyle \sum }_{k=1}^n{(k\text{Δ}x)}^3\cdot \text{Δ}x$

Assume that $f(x)=x^3$ ,to write as:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum }_{k=1}^{n}f(k\text{Δ}x)\cdot \text{Δ}x$

Know that, $f(x)=x^3$ and

  $a=0$

  $b=1$

So that the definite integral is:

  ${{\displaystyle \int }}_0^1{x}^{3}dx$

Therefore, the required limit as a definite integral is. ${{\displaystyle \int }}_0^1{x}^{3}dx$