Q. 54 - 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]. 

$f\left(x\right)=x^2+1$, $\left[a,b\right]=\left[0,4\right]$

Step-by-Step Solution

Answer

The approximate value is $\begin{array}{rcl}& & 16.819\end{array}$.

Step 1. Given information.

Consider the function is $f\left(x\right)=x^2+1$, $\left[a,b\right]=\left[0,4\right]$.

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.

$\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_0^4\sqrt{1+{\left(\frac{d}{dx}\left(x^2+1\right)\right)}^2}dx\\ & =& {\int }_0^4\sqrt{1+{\left(2x+0\right)}^2}dx\\ & =& {\int }_0^4\sqrt{1+4x^2}dx\end{array}$

Step 4. Use graphing calculator.

Find the approximate value of definite integral $\begin{array}{rcl}& & {\int }_0^4\sqrt{1+4x^2}dx\end{array}$ with the help of graphing calculator.

$\begin{array}{rcl}{\int }_0^4\sqrt{1+4x^2}dx& \approx & 16.819\end{array}$

The area under the definite integral between the interval $\left[0,4\right]$ in the graphing calculator is represented as follows.