Question

The integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{-\pi / 3}^{\pi / 3}(2-\sec x) d x $$

Short answer

The desired area is represented by the region enclosed between the horizontal line y = 2 and the curve of \(\sec{x}\) within the interval [-π/3, π/3].

Step-by-step solution

Understand the integral

The given definite integral is \(\int_{-\pi / 3}^{\pi / 3}(2-\sec{x}) d x\). The integrand consists of two functions: the constant function 2 and the secant function \(\sec{x}\). The integral calculates the area under the curve of the difference of these two functions on the interval [-π/3, π/3].

Sketch graph of constant function

The constant function 2 is a horizontal straight line passing through the y-coordinate 2. To create the graph, just draw a horizontal line through the y-coordinate 2 and extend it to encompass the defined interval [-π/3, π/3].

Sketch graph of secant function

The secant function, \(\sec{x}\), is the reciprocal of the cosine function. To create the graph, note that the function is undefined for x=±π/2 + kπ, where k is any integer. Thus, vertical asymptotes are found at these x-values. Between these vertical asymptotes, \(\sec{x}\) increases from -∞ to 1, then decreases from 1 to -∞. Over the interval [-π/3, π/3] remember to keep track of the secant's behavior to graph it correctly.

Shade the relevant area

Recall that the integrand of the definite integral is (2 - sec(x)). This means the area we're interested in is the area between the graph of the constant function 2 and the graph of the function \(\sec{x}\). In this case, over the interval [-π/3, π/3], this area is given by the difference of the integrals of the two functions: ∫2 dx - ∫\(\sec{x}\) dx. Thus, the area to be shaded is under the graph y=2 and above the graph \(\sec{x}\) from -π/3 to π/3.