We need to calculate the arc length over the given integral.

Question

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Answer 1

Question

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

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Answer 2

Question

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

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Answer 3

Question

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

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Answer 4

Question

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

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Answer 5

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Answer 6

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$

Thus, the arc length is $s = \frac{779}{240}$

Conclusion:

Hence, the arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Answer 7

Chapter 8, Problem 1CRE

To determine

We need to calculate the arc length over the given integral.

Answer 8

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The arc length over the interval $[1, 2]$ is $s = \frac{779}{240}$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The following function is given

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6} \dots (1)$

over the interval $[1, 2]$

Formula Used:

$s = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} d x \dots (2)$

Where s is the arc length over $[a, b]$ for graph $f (x)$ .

Calculation:

Consider the given equation

$y = \frac{x^{5}}{10} + \frac{x^{- 3}}{6}$

Now, differentiating it with respect to x, we get

$\begin{matrix} y' = \frac{5 x^{4}}{10} + (\frac{- 3 x^{- 4}}{6}) \\ = \frac{x^{4}}{2} - \frac{x^{- 4}}{2} \\ = \frac{1}{2} (x^{4} - x^{- 4}) \end{matrix}$

Now,

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 + {[y']}^{2} \\ = 1 + {[\frac{1}{2} (x^{4} - x^{- 4})]}^{2} \\ = 1 + \frac{1}{4} {(x^{4} - x^{- 4})}^{2} \\ = 1 + \frac{1}{4} (x^{8} + x^{- 8} - 2) \end{matrix}$

$\begin{matrix} 1 + {[f' (x)]}^{2} = 1 - \frac{1}{2} + \frac{1}{4} (x^{8} + x^{- 8}) \\ = \frac{1}{2} + \frac{1}{4} x^{8} + \frac{1}{4} x^{- 8} \\ = {(\frac{x^{8}}{2} + \frac{x^{- 8}}{2})}^{2} \end{matrix}$

Now, from the equation (2), the arc length over the interval $[1, 2]$ is:

$\begin{matrix} s = \int_{1}^{2} \sqrt{{(\frac{x^{4}}{2} + \frac{x^{- 4}}{2})}^{2}} d x \\ = \int_{1}^{2} (\frac{x^{4}}{2} + \frac{x^{- 4}}{2}) d x \\ = {[\frac{x^{5}}{10} - \frac{x^{- 3}}{6}]}_{1}^{2} \\ = (\frac{2^{5}}{10} - \frac{2^{- 3}}{6}) - (\frac{1^{5}}{10} - \frac{1^{- 3}}{6}) \end{matrix}$

$\begin{matrix} s = \frac{32}{10} - \frac{1}{48} - \frac{1}{10} + \frac{1}{6} \\ = \frac{31}{10} + \frac{7}{48} \\ = \frac{744 + 35}{240} \\ s = \frac{779}{240} \end{matrix}$