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Chapter 6: Problem 9

Sketch the region enclosed by the curves and find its area. $$ y=\cos 2 x, y=0, x=\pi / 4, x=\pi / 2 $$

Short Answer

Expert verified
The area of the region is \( \frac{1}{2} \).

Step by step solution

01

Identify the Region

First, identify the region enclosed by the given curves. You have the curve \( y = \cos(2x) \), the line \( y = 0 \) which is the x-axis, and the vertical lines \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{2} \). The region of interest is the area under \( y = \cos(2x) \) from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \).
02

Set Up the Integral

To find the area of the region, set up the definite integral of \( y = \cos(2x) \) with respect to \( x \) from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \). This gives the integral:\[ \int_{\pi/4}^{\pi/2} \cos(2x) \, dx \]
03

Integrate the Function

To perform the integration, use the substitution \( u = 2x \), then \( du = 2dx \) or \( dx = \frac{du}{2} \). Change the bounds: when \( x = \frac{\pi}{4} \), \( u = \frac{\pi}{2} \), and when \( x = \frac{\pi}{2} \), \( u = \pi \). The integral then becomes:\[ \frac{1}{2} \int_{\pi/2}^{\pi} \cos(u) \, du \]
04

Evaluate the Integral

Integrate \( \cos(u) \) to get \( \sin(u) \). Then, evaluate the definite integral:\[ \frac{1}{2} \left[ \sin(u) \right]_{\pi/2}^{\pi} \] This gives:\[ \frac{1}{2} (\sin(\pi) - \sin(\pi/2)) \] Since \( \sin(\pi) = 0 \) and \( \sin(\pi/2) = 1 \), this becomes:\[ \frac{1}{2} (0 - 1) = -\frac{1}{2} \]
05

Interpret the Result

The result \( -\frac{1}{2} \) indicates the area is below the x-axis. Since area cannot be negative, take the absolute value to find the area of the enclosed region. Thus, the area is \( \frac{1}{2} \).

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Key Concepts

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

Definite Integral
A definite integral is a way to calculate the net area under a curve over a specified interval. It is essential in determining the total accumulation of quantities such as area, volume, and other physical concepts. To find a definite integral, you evaluate the function at two endpoints and find the difference between these values.
In the exercise provided, the definite integral takes the trigonometric function \(y = \cos(2x)\) and integrates it over the bounds \(x = \frac{\pi}{4}\) to \(x = \frac{\pi}{2}\).
  • Start by setting up the integral \( \int_{\pi/4}^{\pi/2} \cos(2x) \, dx \).
  • Perform integration, possibly using substitution to simplify calculations.
  • Calculate the result by evaluating the antiderivative at the specified limits.
Definite integrals are widely used in calculus to find areas between curves and axes, among other applications.
Area Under a Curve
The concept of finding the area under a curve is pivotal in integral calculus to measure the total space enclosed by the function graph and the x-axis within given limits. For the integral \( \int_{\pi/4}^{\pi/2} \cos(2x) \, dx \), we are interested in the area trapped between \(y = \cos(2x)\), the x-axis, and the vertical lines \(x = \frac{\pi}{4}\) and \(x = \frac{\pi}{2}\).
  • The shape of the curve determines whether the area is above or below the x-axis.
  • If the curve is below the x-axis, the integral naturally computes a negative value, so the absolute value represents the actual area.
  • The resulting absolute area from the example was \(\frac{1}{2}\), due to the negative outcome reflecting a part of the curve below the x-axis.
Remember, the integral provides the signed area, and when calculating physical space, the focus is on the non-negative area measure.
Trigonometric Functions
Trigonometric functions like sine and cosine are frequently explored in calculus, particularly when examining periodic and oscillating behaviors. In the problem, the function \(y = \cos(2x)\) showcases these features by transforming a standard cosine wave.
  • The function \(\cos(2x)\) adjusts the frequency of the typical cosine curve, effectively making it oscillate faster within the same interval.
  • Trigonometric integrals may require substitutions to simplify computations, as seen with \( u = 2x \) in our problem.
  • Understanding properties such as \(\cos(\pi) = 0\) and \(\cos(\frac{\pi}{2}) = 1\) facilitates easier evaluation of definite integrals.
In essence, trigonometric functions play a vital role in modeling and solving calculus problems through their wave-like characteristics.

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