Q. 42 - Calculus

Question

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

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

Step-by-Step Solution

Answer

The arc length is $2.957$.  

Step 1. Given information.

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

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 the arc length.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_{-1}^1\sqrt{1+{\left(\frac{d}{dx}\left(x^2\right)\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+{\left(2x\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+4x^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+4x^2}dx\\ & =& {\left[\frac{arc\sin h\left(2x\right)}{4}+\frac{x\sqrt{1+4x^2}}{2}\right]}_{-1}^1\\ & =& \left[\frac{arc\sin h\left(2\right(1\left)\right)}{4}+\frac{\left(1\right)\sqrt{1+4{\left(1\right)}^2}}{2}\right]-\left[\frac{arc\sin h\left(2\right(-1\left)\right)}{4}+\frac{(-1)\sqrt{1+4{(-1)}^2}}{2}\right]\\ & =& \left[\frac{arc\sin h\left(2\right)}{4}+\frac{\left(1\sqrt{5}\right)}{2}\right]-\left[-\frac{arc\sin h\left(2\right)}{4}-\frac{(-\sqrt{5}}{2}\right]\\ & =& \frac{arc\sin h\left(2\right)}{2}+\sqrt{5}\\ & \approx & 2.957885\end{array}$