Chapter 7.4, Problem 27E

To determine

To calculate: In this exercise, you must determine the length of the provided curve $y=x^3+5\left|x\right|$ where is $-2\le x\le 1$ .

Answer to Problem 27E

There are no straightforward techniques to calculate the previously integrals; instead, utilise Wolfram Alpha to evaluate.

  $L\approx 13.1318$

Explanation of Solution

Given Information: The widely used equation for determining the length of a smooth curve (page 421). However, because the form of the supplied function includes absolute value, you should note that our curve is not smooth. As a result, measure the length of the curve first over the range $\left[-2,0\right]$ and then over the $\left[0,1\right]$ this indicates that there is a corner at $x=0.$

Calculation:

Verify our assertion by drawing the curve in Geiger.

  

Additionally, let's rewrite the curve as

  $x^3+5\left|x\right|=\left\{{}_{x^3-5x,x<0}^{x^3+5x,x\ge 0}\right\}\mathrm{.......}(1)$

Consequently, the curve's length is

  $L\approx 13.1318$

Previously, integrals could distinguish the $\left(1\right).$ So,

  $\begin{array}{c}\frac{d}{dx}\left(x^3+5x\right)=3x^2+5\\ \frac{d}{dx}\left(x^3-5x\right)=3x^2-5\end{array}$

Returning to the formula, obtain

  $\begin{array}{c}L={\displaystyle \underset{-2}{\overset{0}{\int }}\sqrt{1+{\left(3x^2-5\right)}^2}dx+{\displaystyle \underset{0}{\overset{1}{\int }}\begin{array}{l}{\sqrt{1+\left(3x^2+5\right)}}^{2}dx\\ \end{array}}}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}\sqrt{9x^4-30x^2+26}}dx+{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{9x^4+30x^2+26}dx}\end{array}$

Because there are no easy ways to calculate the previously integrals, evaluate using Wolfram Alpha. Therefore

  $L\approx 13.1318$