Chapter 5, Problem 6RE

(a)

To determine

The sum value of the function using computer algebra system.

Answer to Problem 6RE

The sum value of the function is 5,220.

Explanation of Solution

Given information:

The function as ${\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}$.

The region lies between $x=1$ and $x=5$.

The expression to find the sum value when the sample points at right endpoints are shown below:

${\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$ (1)

Here, the height of sample at the right endpoint is $f\left(x_i\right)$ and the width is $\Delta x$.

Find the width $\left(\Delta x\right)$ using the relation:

$\Delta x=\frac{b-a}{n}$ (2)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

Substitute 5 for b and 1 for a in Equation (2).

$\begin{array}{c}\Delta x=\frac{5-1}{n}\\ =\frac{4}{n}\end{array}$

To find the value of $x_i$ using the relation:

$x_i=a+i\Delta x$ (3)

Substitute 1 for a and $\frac{4}{n}$ for $\Delta x$ in Equation (3).

$\begin{array}{c}{x}_i=1+i\frac{4}{n}\\ =1+\frac{4i}{n}\end{array}$

Substitute $\left(x_i+2x_i^5\right)$ for $f\left(x_i\right)$ in Equation (1).

${\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(x_i+2x_i^5\right)\Delta x}$ (4)

Substitute $\frac{4}{n}$ for $\Delta x$ and $1+\frac{4i}{n}$ for $x_i$ in Equation (4).

$\begin{array}{c}{\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[\left(1+\frac{4i}{n}\right)+2{\left(1+\frac{4i}{n}\right)}^5\right]\left(\frac{4}{n}\right)}\\ =\left(\frac{4}{n}\right)\left[\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(1+\frac{4i}{n}\right)}+\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}2{\left(1+\frac{4i}{n}\right)}^5}\right]\\ \stackrel{\text{CAS}}{=}\underset{n\to \infty }{\lim }\left(\frac{1,305n^4+3,126n^3+2,080n^2-256}{n^3}\right)\left(\frac{4}{n}\right)\\ =4\times \underset{n\to \infty }{\lim }\left(1,305+\frac{3,126}{n}+\frac{2,080}{n^2}-\frac{256}{n^4}\right)\end{array}$

$\begin{array}{c}{\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}=4\times 1,305\\ =5,220\end{array}$

Therefore, the sum value of the function is 5,220.

(b)

To determine

The value of the integral using the fundamental theorem.

Answer to Problem 6RE

The value of the integral is 5,220.

Explanation of Solution

Given information:

The function as ${\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}$.

The region lies between $x=1$ and $x=5$.

The expression to find the value of the integral using the fundamental theorem is shown below:

$\begin{array}{c}{\displaystyle \underset{1}{\overset{5}{\int }}\left(x+2x^5\right)dx}={\left(\frac{1}{2}{x}^2+\frac{2}{6}{x}^6\right)}_1^5\\ =\left(\frac{1}{2}\times 5^2+\frac{2}{6}\times 5^6\right)-\left(\frac{1}{2}\times 1^2+\frac{2}{6}\times 1^6\right)\\ =12.5+5,208.33-0.5-0.33\\ =5,220\end{array}$

Therefore, the value of the integral using the fundamental theorem is 5,220.