Problem 23 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ y=\frac{8}{x}, y=x^{2}, y=0, x=1, x=4 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs is \(8ln4 - \frac{64}{3} - 8ln1 + \frac{1}{3}\) square units.

Step 1: Sketch the Graphs

First, sketch the graphs of the functions. The function \(y = \frac{8}{x}\) is an inverse variation function that passes through points \((1, 8)\) and \((4, 2)\). The function \(y = x^{2}\) is a basic parabolic function which intersects with \(y = \frac{8}{x}\) at two points, \((2, 4)\) and \((4, 16)\). The lines \(y = 0\), \(x = 1\), and \(x = 4\) are the boundaries of the region.

Step 2: Define Bounded Region

After sketching the graphs, identify the bounded region. That is the area enclosed between \(y=\frac{8}{x}\), \(y=x^{2}\) respect to the x-axis formed on the interval from \(x = 1\) to \(x = 4\). These functions represent the top and bottom curves that bound the region we are interested in.

Step 3: Set up the Definite Integral

Set up the definite integral that will calculate the area. We subtract the y-values of the bottom function (\(y=x^2\)) from the top function (\(y=\frac{8}{x}\)) and integrate over the interval from 1 to 4. Accordingly, the integral becomes \(\int_1^4 (\frac{8}{x} - x^2) dx\).

Step 4: Evaluate the Integral

To find the area, compute the integral. First, calculate the antiderivative of the functions: \(8ln|x| - \frac{x^3}{3}\). Then, using the Fundamental Theorem of Calculus, plug in the bounds of 1 and 4: \([8ln|4| - \frac{4^3}{3}]-[8ln|1| - \frac{1^3}{3}]\). Simplify this to find the area.