Problem 56 - Essential Calculus

Question

(a) Show that \(\int_{0}^{1} \frac{4}{1+x^{2}} d x=\pi\). (b) Approximate the number \(\pi\) using Simpson's Rule (with \(n=6\) ) and the integral in part (a). (c) Approximate the number \(\pi\) by using the integration capabilities of a graphing utility.

Step-by-Step Solution

Answer

The integral of \(\frac{4}{{1+x^{2}}}\) from 0 to 1 equals \(\pi\). The approximation of \(\pi\) using Simpson's rule and the integration capabilities of a graphing utility will depend on the precision of the methods but should be fairly close to the actual value of \(\pi\), which is approximately 3.14159.

Step 1: Integral Calculation

First, perform the integral calculation of \(\frac{4}{{1+x^{2}}}\) from 0 to 1. Using the well-known integral \(\int \frac{1}{{1+x^{2}}} dx = \arctan(x) + C\), the integral can be computed like so: \(\int_{0}^{1} \frac{4}{{1+x^{2}}} dx = 4[\arctan(x)]_{0}^{1} = 4(\arctan(1) - \arctan(0)) = 4(\frac{\pi}{4} - 0) = \pi\). Therefore, the integral of \(\frac{4}{{1+x^{2}}}\) from 0 to 1 equals \(\pi\).

Step 2: Simpsons Rule

Now, we need to approximate \(\pi\) using Simpson's Rule with \(n=6\) and the integral computed in step 1. Simpson's rule is a method for numerical integration that approximates the integral of a function using quadratic polynomials. The approximation is given by: \(\int_{a}^{b} f(x) dx\approx \frac{(b-a)}{6n} [f(a) + 4f(a+h) + 2f(a+2h) +...+ 4f(b-h) + f(b)]\), where \(n\) is the number of subdivisions and \(h = \frac{(b-a)}{n}\). For the given question, \(f(x)=\frac{4}{{1+x^{2}}}\), \(a=0\), \(b=1\), \(h=\frac{1}{6}\), and \(n=6\). Applying these values into Simpson's Rule formula will give the approximation of \(\pi\).

Step 3: Use of a graphing utility

Next, the value of \(\pi\) is approximated by using the integration capabilities of a graphing utility like Desmos or GeoGebra. Once the function \(f(x)=\frac{4}{{1+x^{2}}}\) is entered into the graphing utility, the integral from 0 to 1 is calculated using the tool's integral function. The accuracy of the result depends on the precision of the graphing utility.