Question

In Exercises \(15-22,\) (a) graph the function, highlighting the part indicated by the given interval, (b) find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ y=4-x^{2}, \quad 0 \leq x \leq 2 $$

Short answer

The arc length of the curve on the given interval can't be determined exactly using familiar integration techniques, however it can be approximated numerically using a graphing utility. The numerical value of the integral, as computed by the utility, will give the required approximation for the arc length.

Step-by-step solution

Graph the Function and Highlight the Part Indicated by the Interval

The equation \(y = 4 - x^{2}\) is a parabola that opens downwards. Using a graphing utility, plot the function with \(x\) values ranging from 0 to 2. Highlight the portion of the plot for this range.

Find a Definite Integral Representing the Arc Length of the Curve

The formula for arc length represented by definite integral is \[L = \int_{a}^{b} \sqrt{1 + [f'(x)]^{2}} dx\] where \(f'(x)\) is the derivative of the function. First, find the derivative of the function \(y = 4 - x^{2}\), which is \(f'(x) = -2x\). Then plug \(f'(x)\) into the formula and keep in mind that we are looking at the interval \(0 \leq x \leq 2\), so our integral becomes: \[L = \int_{0}^{2} \sqrt{1 + [-2x]^{2}} dx\]. Observing this integral, it can be seen that it cannot be calculated using elementary functions. Thus, numerical method should be used for the evaluation, which leads us to the next step.

Use a Graphing Utility to Approximate the Arc Length

Using the integration capabilities of a graphing utility, input the integral \[L = \int_{0}^{2} \sqrt{1 + [-2x]^{2}} dx\] and approximate the value for the arc length of the curve over the interval from 0 to 2. The result is a numerical approximation, since the exact integral could not be computed with elementary functions.