Question

let r be the region enclosed by the graphs of $f(x) = \ln(x^4 + 1)$ and $g(x) = 3\cos(x) + 5$. find the length of the boundary of region r. use a calculator and round your answer to 3 decimal places.

Explanation:

Step1: Find intersection points

Use a calculator to find the \(x\) - values where \(f(x)=g(x)\), i.e., \(\ln(x^{4}+1)=3\cos(x)+5\). Let the intersection points be \(x = a\) and \(x = b\).

Step2: Use the arc - length formula

The arc - length formula for a function \(y = h(x)\) from \(x = m\) to \(x = n\) is \(L=\int_{m}^{n}\sqrt{1+(h^{\prime}(x))^{2}}dx\).
For \(y = f(x)=\ln(x^{4}+1)\), \(f^{\prime}(x)=\frac{4x^{3}}{x^{4}+1}\).
For \(y = g(x)=3\cos(x)+5\), \(g^{\prime}(x)=- 3\sin(x)\).
The length of the boundary of region \(R\) is \(L=\int_{a}^{b}\sqrt{1+(f^{\prime}(x))^{2}}dx+\int_{a}^{b}\sqrt{1+(g^{\prime}(x))^{2}}dx\)

$$=\int_{a}^{b}\sqrt{1 + (\frac{4x^{3}}{x^{4}+1})^{2}}dx+\int_{a}^{b}\sqrt{1+( - 3\sin(x))^{2}}dx$$

Using a calculator to evaluate the two definite integrals:
Let \(I_1=\int_{a}^{b}\sqrt{1 + (\frac{4x^{3}}{x^{4}+1})^{2}}dx\) and \(I_2=\int_{a}^{b}\sqrt{1 + 9\sin^{2}(x)}dx\)
After using a graphing calculator (such as TI - 84 Plus with the fnInt function or a computer - based calculator like Desmos with integral capabilities) to find the intersection points \(a\approx - 2.179\) and \(b\approx2.179\) and then evaluating the integrals:
\(I_1=\int_{-2.179}^{2.179}\sqrt{1 + (\frac{4x^{3}}{x^{4}+1})^{2}}dx\approx9.937\)
\(I_2=\int_{-2.179}^{2.179}\sqrt{1 + 9\sin^{2}(x)}dx\approx12.599\)

Answer:

\(22.536\)