Problem 10 - Calculus Early Transcendentals

Question

Sketch the region and find its area. The region bounded by \(y=\cos x\) and \(y=\sin x\) between \(x=\pi / 4\) and \(x=5 \pi / 4\)

Step-by-Step Solution

Answer

Answer: The area of the enclosed region is 2 square units.

Step 1: Graphing the functions

Plot the two functions, \(y=\cos x\) and \(y=\sin x\), on the same coordinate plane. You will notice that the curves intersect at an angle of 45 degrees since the cosine curve starts at the peak and sine curve starts at the trough.

Step 2: Identifying the points of intersection

Observe that the region of interest is enclosed between \(x=\frac{\pi}{4}\) and \(x=\frac{5\pi}{4}\), and \(y=\cos x\) is above \(y=\sin x\) in this region. We already know that the points of intersection are at \(x=\frac{\pi}{4}\) and \(x=\frac{5\pi}{4}\).

Step 3: Sketching the region

With the above information, you can now sketch the region enclosed by the functions (make sure to shade the area). This region is a shape that resembles a kite.

Step 4: Setting up the integral for the area and calculating its value

To find the area, you will want to subtract the lower function (\(y=\sin x\)) from the upper function (\(y=\cos x\)) and integrate with respect to \(x\), between the given limits: Area = \(\int_{\pi/4}^{5\pi/4} (\cos x - \sin x) dx\) Now, solve the integral: \(\int (\cos x - \sin x) dx = \sin x + \cos x + C\) Apply the Fundamental Theorem of Calculus and substitute the limits: Area = (\(\sin(\frac{5\pi}{4}) + \cos(\frac{5\pi}{4})\)) - (\(\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})\)) Area = [(\((2^{-\frac{1}{2}}) + (-2^{-\frac{1}{2}})\)] - [\((2^{-\frac{1}{2}}) + (2^{-\frac{1}{2}})\)]) = -2 Taking the absolute value, since area can't be negative: Area = \(|-2| = 2\) square units Therefore, the area of the enclosed region is 2 square units.

Key Concepts

Integration

Integration is the process of finding the integral of a function, which can be thought of as the reverse operation of differentiation. It's a core concept in calculus, used to determine quantities like area, volume, and other cumulative totals. When we integrate a function over a particular interval, we essentially "add up" infinitely small quantities to find a total.

• Indefinite Integral: Represents a family of functions and includes a constant of integration, denoted by "C". It is expressed as \( \int f(x) \, dx = F(x) + C \).
• Definite Integral: Provides the actual number that represents the total change or area under a curve between two bounds. It's expressed as \( \int_a^b f(x) \, dx \).

Understanding the rules and techniques of integration is crucial in solving problems involving areas under curves, which is what we applied in the exercise by subtracting the sine function from the cosine function and integrating the result.

Definite Integral

The definite integral is a powerful tool in calculus for calculating the value that represents accumulation of quantities, such as areas between curves. This is achieved by evaluating the integral from a starting point to an endpoint.
In practical terms, consider it like adding up an infinite number of infinitesimally small pieces or slices of the area under a curve. The definite integral is evaluated between two limits, using the notation:\[ \int_a^b f(x) \, dx \]where \( a \) and \( b \) are the lower and upper bounds, respectively. By finding the difference between the integral values at these points, we obtain the number representing the area or total accumulation.

• Fundamental Theorem of Calculus: This theorem connects differentiation with integration, stating that if \( F \) is an antiderivative of \( f \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).
• Application: In the solved exercise, we applied this concept by calculating the integral of \( \cos x - \sin x \) from \( \frac{\pi}{4} \) to \( \frac{5\pi}{4} \), giving us the area between those curves.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics and are extensively used in various fields including calculus. These functions relate angles of a triangle to the lengths of its sides and are represented by periodic functions like sine and cosine.
In calculus, trigonometric functions are often integrated and differentiated, presenting unique behaviors due to their periodic nature.

• Sine Function \( \sin x \): A function that starts at zero, reaches its peak at \( \frac{\pi}{2} \), returns to zero at \( \pi \), and continues its cycle. It is odd, meaning \( \sin(-x) = -\sin x \).
• Cosine Function \( \cos x \): Starts at its maximum value of 1, drops to zero at \( \frac{\pi}{2} \), and reaches its minimum at \( \pi \). It is even, meaning \( \cos(-x) = \cos x \).

In the context of finding areas, knowing when one of these functions is above or below another over a particular interval is key, as seen in our exercise where \( \cos x \) is above \( \sin x \) between \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \). Interpreting these positions helps set up the integrals correctly to find the area between curves.