Chapter 5.2, Problem 70E

(a.)

To determine

To Show: The RRAM Riemann sum for the integral is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$ .

Answer to Problem 70E

It has been shown that the RRAM Riemann sum for the given integral is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$ .

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Concept used:

The RRAM Riemann sum for the integral ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ after partitioning the interval $\left[a,b\right]$ into $n$ equal subintervals is given as ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}$ .

Calculation:

The given integral is ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Comparing ${\displaystyle {\int }_0^1{x}^{2}dx}$ with ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ to get,

  $a=0$ , $b=1$ and $f\left(x\right)=x^2$ .

Partition $\left[0,1\right]$ into $n$ equal subintervals.

Put $a=0$ and $b=1$ in ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}$ to get,

  ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}={\displaystyle \sum _{k=1}^n\left(f\left(0+k\left(\frac{1-0}{n}\right)\right)\left(\frac{1-0}{n}\right)\right)}$

Simplifying,

  ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}={\displaystyle \sum _{k=1}^n\left(f\left(k\left(\frac{1}{n}\right)\right)\left(\frac{1}{n}\right)\right)}$

On further simplification,

  ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}={\displaystyle \sum _{k=1}^n\left(f\left(\frac{k}{n}\right)\cdot \frac{1}{n}\right)}$

Now, $f\left(x\right)=x^2$ . Then, $f\left(\frac{k}{n}\right)={\left(\frac{k}{n}\right)}^2$ .

Put $f\left(\frac{k}{n}\right)={\left(\frac{k}{n}\right)}^2$ in ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}={\displaystyle \sum _{k=1}^n\left(f\left(\frac{k}{n}\right)\cdot \frac{1}{n}\right)}$ to get,

  ${\displaystyle \sum _{k=1}^n\left(f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)\right)}={\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$

This shows that the RRAM Riemann sum for the given integral is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$ .

Conclusion:

It has been shown that the RRAM Riemann sum for the given integral is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$ .

(b.)

To determine

To Show: The sum obtained in part (a) can be written as $\frac{1}{n^3}\cdot {\displaystyle \sum _{k=1}^n{k}^2}$ .

Answer to Problem 70E

It has been shown that the sum obtained in part (a) can be written as $\frac{1}{n^3}\cdot {\displaystyle \sum _{k=1}^n{k}^2}$ .

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Concept used:

Any term not containing the index variable can be taken out of the summation.

Calculation:

As determined previously, the sum obtained in part (a) is ${\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}$ .

Simplifying,

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

On further simplification,

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

Note that the index variable in the above summation is $k$ . So, any term not containing $k$ can be taken out of the summation as follows:

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

This shows that the sum obtained in part (a) can be written as $\frac{1}{n^3}\cdot {\displaystyle \sum _{k=1}^n{k}^2}$ .

Conclusion:

It has been shown that the sum obtained in part (a) can be written as $\frac{1}{n^3}\cdot {\displaystyle \sum _{k=1}^n{k}^2}$ .

(c.)

To determine

To Show: The sum obtained in part (b) can be written as $\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$ .

Answer to Problem 70E

It has been shown that the sum obtained in part (b) can be written as $\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$ .

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Concept used:

It can be shown by mathematical induction that ${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ .

Calculation:

As determined previously, the sum obtained in part (b) is $\frac{1}{n^3}\cdot {\displaystyle \sum _{k=1}^n{k}^2}$ .

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

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

Simplifying,

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

On further simplification,

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

This shows that the sum obtained in part (b) can be written as $\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$ .

Conclusion:

It has been shown that the sum obtained in part (b) can be written as $\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}$ .

(d.)

To determine

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

Answer to Problem 70E

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

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Concept used:

  $\underset{n\to \infty }{\lim }\frac{1}{n}=0$

Calculation:

As determined previously,

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

As determined previously,

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

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

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

Simplifying,

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

On further simplification,

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

Continuing simplification,

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

Taking limit on both sides,

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

Simplifying,

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

On further simplification,

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

Put $\underset{n\to \infty }{\lim }\frac{1}{n}=0$ in the above expression to get,

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

Solving,

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

This is the required proof.

Conclusion:

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

(e.)

To determine

To Explain: Why the equation in part (d) proves that ${\displaystyle {\int }_0^1{x}^{2}dx}=\frac{1}{3}$ .

Answer to Problem 70E

It has been explained why the equation in part (d) implies that ${\displaystyle {\int }_0^1{x}^{2}dx}=\frac{1}{3}$ .

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ .

Concept used:

If the interval $\left[a,b\right]$ is partitioned into $n$ equal subintervals of length $\Delta x$ and $c_k$ is a point chosen from the $k^{\text{th}}$ subinterval, then it follows that $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}={\displaystyle {\int }_a^{b}f\left(x\right)dx}$ .

Calculation:

Partitioning $\left[0,1\right]$ into $n$ equal subintervals, the length of each subinterval is $\Delta x=\frac{1}{n}$ .

The $k^{\text{th}}$ subinterval is given as $\left[\frac{k-1}{n},\frac{k}{n}\right]$ .

Choosing $c_k=\frac{k}{n}$ , it follows that,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(f\left(\frac{k}{n}\right)\left(\frac{1}{n}\right)\right)}$

It can be seen from the given integral ${\displaystyle {\int }_0^1{x}^{2}dx}$ that $f\left(x\right)=x^2$ .

Then, $f\left(\frac{k}{n}\right)={\left(\frac{k}{n}\right)}^2$ .

Put $f\left(\frac{k}{n}\right)={\left(\frac{k}{n}\right)}^2$ in $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(f\left(\frac{k}{n}\right)\left(\frac{1}{n}\right)\right)}$ to get,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\left(\frac{1}{n}\right)\right)}$

Put $f\left(x\right)=x^2$ and $\left[a,b\right]=\left[0,1\right]$ in ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ to get,

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle {\int }_0^1{x}^{2}dx}$

Put $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\left(\frac{1}{n}\right)\right)}$ and ${\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle {\int }_0^1{x}^{2}dx}$ in $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)\Delta x}={\displaystyle {\int }_a^{b}f\left(x\right)dx}$ to get,

  ${\displaystyle {\int }_0^1{x}^{2}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\left(\frac{1}{n}\right)\right)}$

As determined in part (d),

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

Put $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\cdot \frac{1}{n}\right)}=\frac{1}{3}$ in ${\displaystyle {\int }_0^1{x}^{2}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left({\left(\frac{k}{n}\right)}^2\left(\frac{1}{n}\right)\right)}$ to get,

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

This is the required proof.

Conclusion:

It has been explained why the equation in part (d) implies that ${\displaystyle {\int }_0^1{x}^{2}dx}=\frac{1}{3}$ .