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Chapter 5: Problem 7

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

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

Step by step solution

01

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 [-蟺/3, 蟺/3].
02

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 [-蟺/3, 蟺/3].
03

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=卤蟺/2 + k蟺, where k is any integer. Thus, vertical asymptotes are found at these x-values. Between these vertical asymptotes, \(\sec{x}\) increases from -鈭� to 1, then decreases from 1 to -鈭�. Over the interval [-蟺/3, 蟺/3] remember to keep track of the secant's behavior to graph it correctly.
04

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 [-蟺/3, 蟺/3], this area is given by the difference of the integrals of the two functions: 鈭�2 dx - 鈭玕(\sec{x}\) dx. Thus, the area to be shaded is under the graph y=2 and above the graph \(\sec{x}\) from -蟺/3 to 蟺/3.

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