Chapter 6.2, Problem 58E

To determine

To prove: To prove ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx=\frac{1}{3}$ using Riemann Sum. $$

Explanation of Solution

Given information:

• Given function is $f\left(x\right)=x^2$ and we have to find ${\displaystyle \underset{0}{\overset{1}{\int }}f\left(x\right)}dx$
• Sum of squares of first n natural numbers is: ${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$is proved using Principal of Mathematical Induction.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ , and:

1. The interval $\left[a,b\right]$ is divided into n sub-intervals of equal width , $\Delta x$ with $\Delta x=\frac{b-a}{n}$
2. The endpoints of these sub-intervals as $a=x_0,x_1,x_2\mathrm{...................}{x}_n=b$
3. $x^{*}{}_1,x^{*}{}_2,x_3*,\mathrm{................}{x}^{*}{}_n$ are any sample points in these sub-intervals, then the definite integral of f from $x=a\text{to }x=b$ is;

${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x^{*}{}_i\right)}\Delta x$ , provided the limit exists.

Proof:

To integrate $f\left(x\right)=x^2$ from $x=0\text{to}x=1$ is same as evaluating the limit of Riemann Sum of $f\left(x\right)=x^2$ from $x=0\text{to}x=1$ using n sub-intervals.

Following are the steps to find the Integral using Riemann Sum:

Step 1:

Finding the Riemann Sum:

Since the integral has to be find from $x=0\text{to}x=1$ , the interval $\left[0,1\right]$ is partitioned into n sub-intervals each of length $\frac{1}{n}$ .That is, $\Delta x=\frac{1}{n}$ and $x_k=\frac{k}{n}$

Now Riemann Sum = ${\left(\frac{1}{n}\right)}^2\cdot \frac{1}{n}+{\left(\frac{2}{n}\right)}^2\cdot \frac{1}{n}+\mathrm{.............}+{\left(\frac{n-1}{n}\right)}^2\cdot \frac{1}{n}+{\left(\frac{n}{n}\right)}^2\cdot \frac{1}{n}$

  $={\displaystyle \sum _{k=1}^n{x}_k}\Delta x$

  $={\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$…….. (1)

Step 2:

Simplifying the equation (1)

  ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}={\displaystyle \sum _{k=1}^n\left(\frac{k^2}{n^2}\cdot \frac{1}{n}\right)}$

  $$

  $={\displaystyle \sum _{k=1}^n\frac{k^2}{n^3}}$

  $=\frac{1}{n^3}{\displaystyle \sum _{k=1}^n{k}^2}$……. (2)

Step 3:

Applying the formula for sum of first n natural numbers to (2):

Sum of squares of first n natural numbers is: ${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

∴ (2) $\Rightarrow$

  $\frac{1}{n^3}{\displaystyle \sum _{k=1}^n{k}^2}=\frac{1}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}$

   = $\frac{1}{n^2}\cdot \frac{\left(n+1\right)\left(2n+1\right)}{6}$

   = $\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$ …….. (3)

Step (4):

Taking the limit as $n\to \infty$on the Riemann Sum in (1):

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

  $=\underset{n\to \infty }{\lim }\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}=\underset{n\to \infty }{\lim }\frac{2n^2+2n+1}{6n^2}$

  $=\underset{n\to \infty }{\lim }\left(\frac{2n^2}{6n^2}+\frac{2n}{6n^2}+\frac{1}{6n^2}\right)$

  $=\underset{n\to \infty }{\lim }\left(\frac{1}{3}+\frac{1}{3n}+\frac{1}{6n^2}\right)$

  $=\underset{n\to \infty }{\lim }\frac{1}{3}+\underset{n\to \infty }{\lim }\frac{1}{3n}+\underset{n\to \infty }{\lim }\frac{1}{6n^2}$

  $=\frac{1}{3}+0$

  $=\frac{1}{3}$ …… (4)

Step (4):

Applying the property that definite integral can be find using limit of Riemann Sum:

From (1) in step1, the Riemann approximation for ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx$ is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$

The exact value of ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx$ = $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$

From (4) $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$

  $=\frac{1}{3}$

∴ ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx=\frac{1}{3}$