Question

Use a graphing utility to graph the region bounded by the graphs of the functions. Write the definite integrals that represent the area of the region. (Hint: Multiple integrals may be necessary.) $$ f(x)=2 x, g(x)=4-2 x, h(x)=0 $$

Short answer

The area of the region bounded by the graphs is 4 square units.

Step-by-step solution

Graph the Functions

Use a graphing utility to plot \(f(x)=2x\), \(g(x)=4-2x\), and \(h(x)=0\). Observe that the region of interest is split into two distinct regions by the line \(h(x)=0\). The first region is bounded by \(f(x)\) and \(g(x)\) and lies to the left of \(h(x)\), and the second is also bounded by \(f(x)\) and \(g(x)\), but lies to the right of \(h(x)\).

Identify Intersection Points

Identify the intersection points of \(f(x)\) and \(g(x)\) as they define the limits of integration for the two regions. For intersection points, we set \(f(x)=g(x)\) which yields \(2x = 4-2x\). Solving for \(x\) yields \(x = 1\). Consequently, the limits of integration are \(x=0\) and \(x=1\) for the two regions.

Write Definite Integrals

Write the definite integrals representing the area of the bounded region. For the first region to the left of \(h(x)\), \(g(x)\) is the upper function and \(f(x)\) is the lower function so the integral becomes \(\int_{-1}^{0} [g(x) - f(x)] dx = \int_{-1}^{0} [4 - 2x - 2x] dx\). For the second region to the right of \(h(x)\), \(f(x)\) is the upper function and \(g(x)\) is the lower function, so the integral becomes \(\int_{0}^{1} [f(x) - g(x)] dx = \int_{0}^{1} [2x - (4-2x)] dx\).

Evaluate Definite Integrals

Evaluate the integrals to find the area. For the first integral, the antiderivative of \([4 - 4x]\) is \([4x - 2x^2]\) which evaluated from \(-1\) to \(0\) yields \(-2\) square units. Applying the same steps for the second integral yields \(2\) square units. When calculating areas we discard the sign and consider only the absolute values.