Question

Sketch the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=\cos x, \mathrm{~g}(x)=2-\cos x, 0 \leq x \leq 2 \pi $$

Short answer

The area of the region bounded by the two functions is \(4\pi.\)

Step-by-step solution

Plot the Functions

Plot the two functions \(f(x) = cos(x)\) and \(g(x) = 2 - cos(x)\) on the graph for the interval \(0 \leq x \leq 2\pi\). The first function has graph that follows the pattern of a cosine curve whilst the second function graph is just an upside-down cosine curve, shifted upwards by 2 units.

Identify the Region

The region of interest is bounded by these two curves. Note that on the interval \([0, 2\pi]\), the graph of \(g(x) = 2 - cos(x)\) is always above the graph of \(f(x) = cos(x)\).

Compute the Area of the Region

The area of the region can be computed as the integral of the absolute value of the difference between the two functions over the interval \([0, 2\pi]\). Here, the integral turns out to be \(\int_{0}^{2\pi} |g(x) - f(x)| dx = \int_{0}^{2\pi} (2 - cos(x) - cos(x)) dx = \int_{0}^{2\pi} (2 - 2cos(x)) dx.\)

Evaluate the Integral

The integral equals \([2x - 2sin(x)]_{0}^{2\pi},\) which simplifies to \(4\pi.\)