Question

In Exercises \(31-36,\) use an iterated integral to find the area of the region bounded by the graphs of the equations. $$ y=x, \quad y=2 x, \quad x=2 $$

Short answer

The area of the region bounded by the graphs of the given equations is 2.

Step-by-step solution

Sketch the region

Firstly, needing to sketch the region by three equations \( y=x \), \( y=2x \), and \( x=2 \). The region is a triangular area between these three lines.

Set up the iterated integral

The function to be integrated for finding the area is \( f(x,y)=1 \). The limits of the integral are determined by the region. The variable of the outer integral corresponds to the horizontal line \( x=2 \), going from its intersection point with \( y=x \) to the intersection with \( y=2x \), so \( x \) goes from \(0\) to \(2\). The variable of the inner integral corresponds to the bounds \( y=x \) and \( y=2x \), which determine \( y \)'s range for each \( x \). The iterated integral is \(\int_{0}^{2}\int_{x}^{2x} dy \, dx \).

Evaluate the inner integral

First, need to evaluate the inner integral \(\int_{x}^{2x} dy \). The antiderivative of the function \(1\) with respect to \(y\) is \( y \), resulting in the value \( [y]_{x}^{2x} = 2x - x = x \).

Evaluate the outer integral

Next, need to evaluate the outer integral \( \int_{0}^{2} x \, dx \). The antiderivative of \( x \) with respect to \( x \) is \(\frac{x^2}{2}\), resulting in the expression \( \left[\frac{x^2}{2}\right]_{0}^{2} = 2 - 0 = 2 \).