Problem 1 - Calculus Early Transcendentals

Question

Draw the graphs of two functions \(f\) and \(g\) that are continuous and intersect exactly twice on \((-\infty, \infty) .\) Explain how to use integration to find the area of the region bounded by the two curves.

Step-by-Step Solution

Answer

Question: Determine the area of the region bounded by the two continuous functions \(f(x) = x^2 - 4\) and \(g(x) = -x^2 + 4\) that intersect exactly twice on \((-\infty, \infty)\). Answer: The area of the region bounded by the two curves is \(\frac{64}{3}\) square units.

Step 1: Draw the functions f and g

First, we need to choose two continuous functions, \(f\) and \(g\), that intersect exactly twice. For example, let's take these functions: \[ f(x) = x^2 - 4 \] and \[ g(x) = -x^2 + 4. \] These functions are both continuous, and their graphs are parabolas with opposite orientations. We can see that they intersect twice: once on the left and once on the right.

Step 2: Find the intersection points of f and g

To find the area between the two curves, we need to determine the points of intersection. To do this, we need to solve the equation \(f(x) = g(x)\). So, we get: \[ x^2 - 4 = -x^2 + 4. \] Solving for \(x\), we find the intersection points at \(x = -2\) and \(x = 2\).

Step 3: Set up the integral for the area between the curves

Now that we have the intersection points, we can set up the integral to calculate the area between the curves. To do this, we need to find the difference between the functions, \(g(x) - f(x)\), and integrate that difference with respect to \(x\) from the left intersection point to the right intersection point. Mathematically, this can be expressed like this: \[ A = \int_{-2}^{2} (g(x) - f(x)) dx. \]

Step 4: Evaluate the integral to find the area

Now that we have the integral, we can evaluate it to find the area between the two curves: \[ A = \int_{-2}^{2} (-x^2 + 4 - (x^2 - 4)) dx. \] Simplify the integrand: \[ A = \int_{-2}^{2} (-2x^2 + 8) dx. \] Now, integrate with respect to \(x\) and evaluate using the endpoints: \[ A = \left[ -\frac{2}{3}x^3 + 8x \right]_{-2}^{2}. \] Finally, substitute the limits and find the difference: \[ A = \left[ -\frac{16}{3} + 16 \right] - \left[ \frac{16}{3} - 16 \right] = \frac{32}{3} + \frac{32}{3} = \frac{64}{3}. \] So, the area of the region bounded by the two curves \(f(x) = x^2 - 4\) and \(g(x) = -x^2 + 4\) is \(\frac{64}{3}\) square units.