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In Exercises \(69-72\) , find the area of the region bounded by the graphs of the equations. Use a graphing utility to verify your result. $$y=2 \sec \frac{\pi x}{6}, \quad x=0, \quad x=2, \quad y=0$$

Short Answer

Expert verified
The area of the region bounded by the graphs of the given equations is given by the absolute value of \(2[\ln|\sec(\frac{\pi * 2}{6}) + \tan(\frac{\pi * 2}{6})| - \ln|\sec(\frac{\pi * 0}{6}) + \tan(\frac{\pi * 0}{6})|]\)

Step by step solution

01

Set up the Integral

The given function is \(y = 2 \sec(\frac{\pi x}{6})\). The area of the region bounded by this curve and the x-axis between \(x = 0\) and \(x = 2\) can be found by setting up the integral \(\int_0^2 |2 \sec(\frac{\pi x}{6})| dx\). The absolute value is important since this will ensure that the area is non-negative.
02

Evaluate the Integral

Using the secant integral rule, \(\int \sec(u) du = \ln | \sec(u) + \tan(u) | + C\), we can now evaluate it: \(\int_0^2 2 \sec(\frac{\pi x}{6}) dx = 2[\ln | \sec(\frac{\pi x}{6}) + \tan(\frac{\pi x}{6}) |]_0^2\).
03

Calculate the Area

Substitute the upper and lower limits of the integral: \(2[\ln|\sec(\frac{\pi * 2}{6}) + \tan(\frac{\pi * 2}{6})| - \ln|\sec(\frac{\pi * 0}{6}) + \tan(\frac{\pi * 0}{6})|]\). Simplifying this will give the area of the region bounded by the function and the x-axis between \(x = 0\) and \(x = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
In calculus, the definite integral is a fundamental concept that refers to the accumulation of quantities, which can represent areas under curves, total displacement, or the amount of a substance dissolved in a solution. To understand this, imagine you're trying to find out how much paint is needed to cover a specific area on a canvas that has varying width along its height. Calculating the definite integral is like summing up all the little bits of paint for each tiny strip of the canvas' width from the bottom to the top.

For the given exercise, the definite integral is used to find the area under the curve of the function between two points on the x-axis, which in this case are 0 and 2. The integral is constructed as \(\int_0^2 |2 \sec(\frac{\pi x}{6})| dx\). This notation represents the total area under the curve of \(y=2 \sec(\frac{\pi x}{6})\) from \(x=0\) to \(x=2\), by summing infinitesimally small rectangles under the curve.
Trigonometric Functions
Trigonometric functions are essential tools in calculus for dealing with problems that involve angles and lengths, such as the oscillations of waves, circles, and periodic phenomena. The function \(y = 2 \sec(\frac{\pi x}{6})\), as seen in our exercise, involves the secant function, which is one over cosine. It stretches and compresses as the angle \(\frac{\pi x}{6}\) changes, creating a wave-like graph.

Understanding trigonometric functions like sine, cosine, secant, and their reciprocal functions is critical in calculus, especially when they are combined with other operations like differentiation and integration. The secant function, in particular, has a unique property of having asymptotes and being undefined where the cosine function is zero, leading to a characteristic shape in its graph that impacts the way we interpret the integral and calculate areas under the curve.
Graphing Utility Verification
A graphing utility is a powerful tool for visualizing functions and their behaviors, which can be extremely helpful when learning calculus. By graphing the function \(y = 2 \sec(\frac{\pi x}{6})\), students can verify the area they calculated with the integral by seeing it visually as the region under the curve.

Verifying Solutions

Using a graphing utility, students can confirm their calculations by ensuring that the shaded area matches their computed result. For example, after calculating the definite integral, a student could graph the function and use features within the graphing utility to shade the area under the curve between \(x=0\) and \(x=2\). If the visual representation aligns with the calculated area, it increases the student's confidence in their understanding and computation.

Secant Integral Rule
The secant integral rule is an essential tool for solving integrals involving the secant function. It states that \(\int \sec(u) du = \ln | \sec(u) + \tan(u) | + C\), where \(C\) is the constant of integration. This rule is applied to solve definite and indefinite integrals featuring the secant function.

Applying the Rule

In the context of the given exercise, the secant integral rule helps in calculating the area under the curve of \(2 \sec(\frac{\pi x}{6})\). After setting up the integral, the rule tells us how to integrate secant, which leads to finding the natural logarithm of the absolute value of the sum of secant and tangent evaluated at the upper and lower limits of integration. The final result from applying this rule provides the exact area sought in the problem.

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Most popular questions from this chapter

In Exercises \(5-28\) , find the indefinite integral. $$\int \frac{5}{x} d x$$

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In Exercises 55 and 56, a differential equation, a point, and a slope field are given. (a) Sketch two approximate solutions of the differential equation on the slope field, one of which passes through the given point. (To print an enlarged copy of the graph, go to MathGraphs.com.) (b) Use integration and the given point to find the particular solution of the differential equation and use a graphing utility to graph the solution. Compare the result with the sketch in part (a) that passes through the given point. $$\frac{d y}{d x}=\frac{2}{9+x^{2}}, \quad(0,2)$$

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