Chapter 15.3, Problem 15E

To determine

Find the area using the limit definition and sigma notation.

Answer to Problem 15E

The area of the region is $312$ square units.

Explanation of Solution

Given:

The given expressions are

  $y=2x^3$ from $x=1$ to $x=5$ .

Calculation:

The exact area of the region under the graph of the function is

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$ , where

  $\Delta x=\frac{b-a}{n}$

  $x_i=a+i\Delta x$

Graph of the given region is given below.

  

For the given region

Area $={\displaystyle {\int }_1^{5}2x^{3}dx}={\displaystyle {\int }_0^{5}2x^{3}dx}-{\displaystyle {\int }_0^{1}2x^{3}dx}$

For ${\displaystyle {\int }_0^{5}2x^{3}dx}$

  $\Delta x=\frac{5-0}{n}=\frac{5}{n}$

  $x_i=0+i\cdot \frac{5}{n}=\frac{5i}{n}$

  $=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}2{\left(\frac{5i}{n}\right)}^3\frac{5}{n}}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{1250i^3}{n^4}\right)}\\ =1250\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^3}{n^4}\right)}\\ =1250\underset{n\to \infty }{\lim }\left(\frac{1^3+2^3+3^3+---+n^3}{n^4}\right)\\ =1250\underset{n\to \infty }{\lim }\left(\frac{n^2{\left(n+1\right)}^2}{4n^4}\right)\\ =\frac{1250}{4}\underset{n\to \infty }{\lim }\left(\frac{n^4+2n^3+n^2}{n^4}\right)\\ =\frac{1250}{4}\underset{n\to \infty }{\lim }\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\\ =\frac{1250}{4}\left[\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{2}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}\right]\\ =\frac{1250}{4}\left[1+0+0\right]\\ =\frac{1250}{4}\end{array}$

For ${\displaystyle {\int }_0^{1}2x^{3}dx}$

  $\Delta x=\frac{1-0}{n}=\frac{1}{n}$

  $x_i=0+i\cdot \frac{1}{n}=\frac{i}{n}$

  $=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}2{\left(\frac{i}{n}\right)}^3\frac{1}{n}}$

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

Area $=\frac{1250}{4}-\frac{2}{4}=\frac{1250-2}{4}=\frac{1248}{4}=312$

Hence the area of the region is $312$ square units.