Question

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

Short answer

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

Step-by-step solution

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.

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.

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.

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 \[A = 2/5 - 1/2 = 1/10\].