Q. 53 - Calculus

Question

You may have noticed that even very simple functions give rise to arc length integrals that we have no idea how to compute. Use a graphing calculator to approximate a definite integral that represents the arc length of the given function f(x) on the interval [a, b]. 

$f\left(x\right)=x^3$, $\left[a,b\right]=\left[-1,1\right]$

Step-by-Step Solution

Answer

The approximate value is $3.0957$. 

Step 1. Given information.

Consider the function is $f\left(x\right)=x^3$, $\left[a,b\right]=\left[-1,1\right]$. 

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x=a$ to $x=b$ is given by ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find definite integral for the given function.

Substitute corresponding values into the arc length formula.

$\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_{-1}^1\sqrt{1+{\left(\frac{d}{dx}\left(x^3\right)\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+{\left(3x^2\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+9x^4}dx\end{array}$

Step 4. Use graphing calculator.

Find the approximate value of definite integral $\begin{array}{rcl}& & {\int }_{-1}^1\sqrt{1+9x^4}dx\end{array}$ with the help of graphing calculator.

$\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & {}_{-1}^1\end{array}\begin{array}{rcl}& & \sqrt{1+9x^4}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& \approx & \end{array}3.0957$

The area under the definite integral between the interval $\left[-1,1\right]$ in the graphing calculator is represented as follows.