Chapter 15.3, Problem 7CFU

To determine

Find the value of the given integral using the limit.

Answer to Problem 7CFU

The value of the given integral is $72$ .

Explanation of Solution

Given:

The given integralis ${\displaystyle {\int }_0^6{x}^{2}dx}$ .

Calculation:

For the given integral $y=x^2$ from $x=0$ to $x=6$ .

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

  $\Delta x=\frac{6-0}{n}=\frac{6}{n}$

  $x_i=0+i\cdot \frac{6}{n}=\frac{6i}{n}$

  $\begin{array}{l}{\displaystyle {\int }_0^6{x}^{2}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{6i}{n}\right)}^2\frac{6}{n}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{216i^2}{n^3}}\end{array}$

  $\begin{array}{l}=216\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}\right)}\\ =216\underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+3^2+---+n^2}{n^3}\right)\\ =216\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}\right)\\ =216\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{6n^2}\right)\\ =\frac{216}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2}{n^2}+\frac{3n}{n^2}+\frac{1}{n^2}\right)\\ =\frac{216}{6}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)\\ =36\left(\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }\frac{3}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}\right)\\ =36\left[2+0+0\right]\\ =72\end{array}$

Hence the value of the given integral is $72$ .