Problem 87 - Thomas Calculus

Question

Area Find the area of the region bounded above by \(y=2 \cos x\) and below by \(y=\sec x,-\pi / 4 \leq x \leq \pi / 4\)

Step-by-Step Solution

Answer

The area is \( 2\sqrt{2} - \ln \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \).

Step 1: Understand the Problem

The problem asks us to find the area between two curves: the upper curve is defined by \( y = 2\cos x \), and the lower curve is defined by \( y = \sec x \). The area is bounded horizontally between \( x = -\pi/4 \) and \( x = \pi/4 \).

Step 2: Set Up the Integral Expression

First, express the area between the curves as an integral. The area \( A \) is the integral from \( -\pi/4 \) to \( \pi/4 \) of the difference between the two functions. Thus, the integral is:\[A = \int_{-\pi/4}^{\pi/4} (2\cos x - \sec x)\, dx\]

Step 3: Evaluate \( \int 2\cos x\, dx \)

Find the integral of \( 2\cos x \). The integral of \( \cos x \) is \( \sin x \), so:\[ \int 2\cos x\, dx = 2\sin x + C\]

Step 4: Evaluate \( \int \sec x\, dx \)

To find the integral of \( \sec x \), use the formula \( \int \sec x\, dx = \ln |\sec x + \tan x| + C \).

Step 5: Solve the Definite Integral

Evaluate the definite integral:\[\int_{-\pi/4}^{\pi/4} (2\cos x - \sec x)\, dx = \left[2\sin x - \ln |\sec x + \tan x| \right]_{-\pi/4}^{\pi/4}\]

Step 6: Plug in the Upper and Lower Limits

Substitute \( x = \pi/4 \) and \( x = -\pi/4 \) into the antiderivative and subtract:Upper limit: \[ 2\sin(\pi/4) - \ln |\sec(\pi/4) + \tan(\pi/4)| \]\[ = 2\left(\frac{\sqrt{2}}{2}\right) - \ln |\sqrt{2} + 1| \]\[ = \sqrt{2} - \ln(\sqrt{2} + 1) \]Lower limit:\[ 2\sin(-\pi/4) - \ln |\sec(-\pi/4) + \tan(-\pi/4)| \]\[ = 2\left(-\frac{\sqrt{2}}{2}\right) - \ln |\sqrt{2} - 1| \]\[ = -\sqrt{2} - \ln(\sqrt{2} - 1) \]

Step 7: Compute the Difference

Subtract the lower limit result from the upper limit result:\[ (\sqrt{2} - \ln(\sqrt{2} + 1)) - (-\sqrt{2} - \ln(\sqrt{2} - 1)) \]Simplify:\[ 2\sqrt{2} - \ln \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \]

Step 8: Finalize the Result

Finally, simplify the expression, if possible. Ensure all logarithms are computed properly and constants are calculated correctly.The area of the region is:\[ 2\sqrt{2} - \ln \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \]

Key Concepts

Area Between Curves

When working with definite integration, one common application is finding the area between two curves. In our problem, we have two functions: the upper curve defined by \( y = 2\cos x \) and the lower curve given by \( y = \sec x \). The goal is to calculate the region area between these curves from \( x = -\pi/4 \) to \( x = \pi/4 \).

Generally, the area between two curves can be found by taking the integral of the difference between the two functions, over a specified interval. This can be expressed with the integral:

• \( A = \int_{a}^{b} [f(x) - g(x)]\, dx \)

For our scenario:

• Upper curve: \( f(x) = 2\cos x \)
• Lower curve: \( g(x) = \sec x \)
• Bounded from \( x = -\pi/4 \) to \( x = \pi/4 \)

The definite integral setup looks like this: \( A = \int_{-\pi/4}^{\pi/4} (2\cos x - \sec x)\, dx \). This integral will then give us the "net" or actual area between the two curves over this interval, by accounting for the height difference (\( f(x) - g(x) \)) at each point \( x \).

Trigonometric Integrals

Dealing with trigonometric integrals is essential in calculus, particularly when working with functions like sine, cosine, tangent, and secant. In this exercise, we're integrating \( 2\cos x \) and \( \sec x \).

For \( 2\cos x \):

• The antiderivative of \( \cos x \) is \( \sin x \).
• Thus, the integral of \( 2\cos x \) becomes \( 2\sin x + C \), where \( C \) represents the constant of integration.

For \( \sec x \), it's a bit more complex. The integral formula you need is:

• \( \int \sec x\, dx = \ln |\sec x + \tan x| + C \).

It helps to memorize these basic integration results and recognize when to apply them. Knowing the antiderivatives is crucial, and for functions involving trigonometric identities, familiarity and practice with these formulas will make solving such integrals more manageable.

Definite Integral Evaluation

After setting up and evaluating indefinite integrals, the next step is to find the definite integral, which involves actual computation over specified bounds. For this exercise, we have:

• Upper limit: \( x = \pi/4 \)
• Lower limit: \( x = -\pi/4 \)

Using the antiderivatives obtained:

• \( 2\sin x - \ln |\sec x + \tan x| \)

You then substitute these limits into the result of the indefinite integral. This requires calculating the value of the antiderivative at the upper limit, subtracting the value of the antiderivative at the lower limit:

Upper Limit Calculation:

• \( 2\sin(\pi/4) - \ln |\sec(\pi/4) + \tan(\pi/4)| \)

Lower Limit Calculation:

• \( 2\sin(-\pi/4) - \ln |\sec(-\pi/4) + \tan(-\pi/4)| \)

Subtract the lower calculation from the upper to find the total area:

• \( (\sqrt{2} - \ln(\sqrt{2} + 1)) - (-\sqrt{2} - \ln(\sqrt{2} - 1)) \)

This yields: \( 2\sqrt{2} - \ln \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \). Understanding the steps involved in evaluating definite integrals is key for solving many calculus problems.