Problem 82 - Calculus Early Transcendentals

Question

Find the area of the following regions. The region bounded by the graph of \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) and the \(x\) -axis between \(x=4\) and \(x=5\)

Step-by-Step Solution

Answer

Answer: The area of the region is \(4 - \sqrt{7}\).

Step 1: Identify the function to be integrated

The function to integrate is f(x) = \(\frac{x}{\sqrt{x^{2}-9}}\), with respect to x, from 4 to 5.

Step 2: Setup the integral

The area of the region can be found by integrating the function over the given interval. Set up the integral: \(\int_{4}^{5} \frac{x}{\sqrt{x^{2}-9}} dx\)

Step 3: Evaluate the integral

To evaluate the integral, we can use the substitution method. Let \(u = x^{2} - 9\). Then, \(\frac{du}{dx} = 2x\) and \(dx = \frac{du}{2x}\). Now, our integral becomes: \(\int_{13}^{16} \frac{x}{\sqrt{u}} \cdot \frac{du}{2x}\). Simplify the integral: \(\frac{1}{2} \int_{13}^{16} \frac{1}{\sqrt{u}} du\)

Step 4: Evaluate the simplified integral

Now, evaluate the integral: \(\frac{1}{2} \int_{13}^{16} \frac{1}{\sqrt{u}} du = \frac{1}{2} [2\sqrt{u}]_{13}^{16}\). Simplify the expression: \([\sqrt{u}]_{13}^{16}\)

Step 5: Substitute back the initial variable and evaluate

Replace u with the initial variable: \([\sqrt{x^2 - 9}]_{4}^{5}\). Evaluate the expression: \(\sqrt{5^2 - 9} - \sqrt{4^2 - 9} = \sqrt{16} - \sqrt{7} = 4 - \sqrt{7}\)

Step 6: State the final answer

The area of the region bounded by the graph of \(f(x) = \frac{x}{\sqrt{x^2 - 9}}\) and the x-axis between x=4 and x=5 is \(4 - \sqrt{7}\).