Chapter 6, Problem 12RE

(a)

To determine

To find: $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(c_k\right)}^3}\Delta x$ as a definite interval.

Answer to Problem 12RE

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(c_k\right)}^3}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}{x}^{3}dx}$

Explanation of Solution

Given information:

The interval $\left[0,10\right]$ is partitioned into n subintervals of length $\Delta x=\frac{10}{n}$ .

The Riemann Sum ${\displaystyle \sum _{k=1}^n{\left(c_k\right)}^3}\Delta x$ is formed by choosing each $c_k$ in the $k^{th}$ subinterval.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ then a limit of the form $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x$ on the interval $\left[a,b\right]$ can be written as the integral ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

That is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$……… (1)

Limit of the given Riemann Sum is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(c_k\right)}^3}\Delta x$

Comparing it with standard equation in (1)

  $f\left(c_k\right)=c_k{}^3$

That is, $f\left(x\right)=x^3$

Since the given interval is $\left[0,10\right]$ then $a=0$ and $b=10$

∴ $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(c_k\right)}^3}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}{x}^{3}dx}$

(b)

To determine

To find: $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k\left(\sin c_k\right)}\Delta x$ as a definite interval.

Answer to Problem 12RE

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k\left(\sin c_k\right)}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}x\sin xdx}$

Explanation of Solution

Given information:

The interval $\left[0,10\right]$ is partitioned into n sub-intervals of length $\Delta x=\frac{10}{n}$ .

The Riemann Sum ${\displaystyle \sum _{k=1}^n{c}_k\left(\sin c_k\right)}\Delta x$ is formed by choosing each $c_k$ in the $k^{th}$ subinterval.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ then a limit of the form $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x$ on the interval $\left[a,b\right]$ can be written as the integral ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

That is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$……… (1)

Limit of the given Riemann Sum is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k\left(\sin c_k\right)}\Delta x$

Comparing it with standard equation in (1)

  $f\left(c_k\right)=c_k\left(\sin c_k\right)$

That is, $f\left(x\right)=x\sin x$

Since the given interval is $\left[0,10\right]$ then $a=0$ and $b=10$

∴ $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k\left(\sin c_k\right)}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}x\sin xdx}$

(c)

To determine

To find: $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k{\left(3c_k-2\right)}^2}\Delta x$ as a definite interval.

Answer to Problem 12RE

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k{\left(3c_k-2\right)}^2}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}x{\left(3x-2\right)}^{2}dx}$

Explanation of Solution

Given information:

The interval $\left[0,10\right]$ is partitioned into n sub-intervals of length $\Delta x=\frac{10}{n}$ .

The Riemann Sum ${\displaystyle \sum _{k=1}^n{c}_k{\left(3c_k-2\right)}^2}\Delta x$ is formed by choosing each $c_k$ in the $k^{th}$ subinterval.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ then a limit of the form $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x$ on the interval $\left[a,b\right]$ can be written as the integral ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

That is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$……… (1)

Limit of the given Riemann Sum is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k{\left(3c_k-2\right)}^2}\Delta x$

Comparing it with standard equation in (1)

  $f\left(c_k\right)=c_k{\left(3c_k-2\right)}^2$

That is, $f\left(x\right)=x{\left(3x-2\right)}^2$

Since the given interval is $\left[0,10\right]$ , $a=0$ and $b=10$

∴ $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{c}_k{\left(3c_k-2\right)}^2}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}x{\left(3x-2\right)}^{2}dx}$

(d)

To determine

To find: $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(1+c_k{}^2\right)}^{-1}}\Delta x$ as a definite interval.

Answer to Problem 12RE

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(1+c_k{}^2\right)}^{-1}}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}{\left(1+x^2\right)}^{-1}dx}$

Explanation of Solution

Given information:

The interval $\left[0,10\right]$ is partitioned into n sub-intervals of length $\Delta x=\frac{10}{n}$ .

The Riemann Sum ${\displaystyle \sum _{k=1}^n{\left(1+c_k{}^2\right)}^{-1}}\Delta x$ is formed by choosing each $c_k$ in the $k^{th}$ subinterval.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ then a limit of the form $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x$ on the interval $\left[a,b\right]$ can be written as the integral ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

That is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$……… (1)

Limit of the given Riemann Sum is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(1+c_k{}^2\right)}^{-1}}\Delta x$

Comparing it with standard equation in (1)

  $f\left(c_k\right)={\left(1+c_k{}^2\right)}^{-1}$

That is, $f\left(x\right)={\left(1+x^2\right)}^{-1}$

Since the given interval is $\left[0,10\right]$ , $a=0$ and $b=10$

∴ $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(1+c_k{}^2\right)}^{-1}}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}{\left(1+x^2\right)}^{-1}dx}$

(e)

To determine

To find: $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)}\Delta x$ as a definite interval.

Answer to Problem 12RE

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}\pi \left(9-{\sin }^2\left(\frac{\pi x}{10}\right)\right)dx}$

Explanation of Solution

Given information:

The interval $\left[0,10\right]$ is partitioned into n sub-intervals of length $\Delta x=\frac{10}{n}$ .

The Riemann Sum ${\displaystyle \sum _{k=1}^n\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)}\Delta x$ is formed by choosing each $c_k$ in the

  $k^{th}$ subinterval.

Theorem Used:

Definite Integral as limit of Riemann Sum:

If $f\left(x\right)$ is continuous on $\left[a,b\right]$ then a limit of the form $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x$ on the interval $\left[a,b\right]$ can be written as the integral ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

That is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(c_k\right)}\Delta x={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$……… (1)

Limit of the given Riemann Sum is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)}\Delta x$

Comparing it with standard equation in (1)

  $f\left(c_k\right)=\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)$

That is, $f\left(x\right)=\pi \left(9-{\sin }^2\left(\frac{\pi x}{10}\right)\right)$

Since the given interval is $\left[0,10\right]$ , $a=0$ and $b=10$

∴ $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\pi \left(9-{\sin }^2\left(\frac{\pi c_k}{10}\right)\right)}\Delta x={\displaystyle \underset{0}{\overset{10}{\int }}\pi \left(9-{\sin }^2\left(\frac{\pi x}{10}\right)\right)dx}$