Problem 42 - College Algebra and Calculus: An Applied Approach

Question

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ y=x^{3 / 2}, y=x $$

Step-by-Step Solution

Answer

The area of the region bounded by the given curves is 1/10 square units.

Step 1: Determine the bounding functions

When x increases, \(x^{3/2}\) increases faster than x, so for positive x, \(x^{3/2} > x\). Therefore, \(y = x^{3 / 2}\) is the upper function and \(y = x\) is the lower function.

Step 2: Find the intersection points

The graphs of \(y = x^{3 / 2}\) and \(y = x\) intersect where \(x^{3 / 2} = x\). This results in possible values for x being 0 and 1.

Step 3: Setup the double integral

The area A of the region bounded by the graphs is given by the integral \[ A = \int \left( \int (x^{3 / 2} - x ) dy \right) dx \]. The limits of the outer integral are from x=0 to x=1, as determined by the intersection points. For the inner integral, y varies from the lower function \(y=x\) up to the upper function \(y= x^{3 / 2}\). So, the final integral is: \[ A = \int_{0}^{1} \int_{x}^{x^{3 / 2}} dy dx \]. Note that there's no function to integrate over because we merely want the area, dy serves as the placeholder to represent unit area.

Step 4: Evaluate the double integral

When you integrate dy from \(x\) to \(x^{3/2}\), it simply evaluates to the upper limit minus the lower limit, that is, \[ x^{3 / 2} - x \]. To compute the area, we now need to evaluate the outer integral: \[ A = \int_{0}^{1} (x^{3 / 2} - x) dx \]. This evaluates to \[A = [2/5 x^{5 / 2} - 1/2 x^2]_{0}^{1}\], which reduces to \[A = 2/5 - 1/2 = 1/10\].