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Chapter 8: Problem 54

Find the area of the region enclosed by the curve \(y=x \cos x\) and the \(x\) -axis (see the accompanying figure) for a. \(\pi / 2 \leq x \leq 3 \pi / 2\) b. \(3 \pi / 2 \leq x \leq 5 \pi / 2\) c. \(5 \pi / 2 \leq x \leq 7 \pi / 2\) d. What pattern do you see? What is the area between the curve and the \(x\) -axis for $$\left(\frac{2 n-1}{2}\right) \pi \leq x \leq\left(\frac{2 n+1}{2}\right) \pi$$ \(n\) an arbitrary positive integer? Give reasons for your answer.

Short Answer

Expert verified
Each specified region has an area of \(3\pi/4\). The pattern is that the area is the same for each interval described.

Step by step solution

01

Understanding the Problem

We need to find the area enclosed by the curve given by the function \( y = x \cos x \) and the x-axis over the specific intervals provided.
02

Integral for part a

Calculate the integral of \( |y| \) from \( \pi/2 \) to \( 3\pi/2 \). Because the area is above the x-axis from \( \pi/2 \) to \( \pi \) and below from \( \pi \) to \( 3\pi/2 \), we have:\[ A = \int_{\pi/2}^{\pi} x \cos x \, dx - \int_{\pi}^{3\pi/2} x \cos x \, dx. \]
03

Evaluate Integral for xcosx

To solve \( \int x \cos x \, dx \), we use integration by parts. Let \( u = x \) and \( dv = \cos x \, dx \), giving \( du = dx \) and \( v = \sin x \). Then:\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C. \]
04

Calculate part a

Apply the evaluated integral to the bounds \( [\pi/2, \pi] \) and subtract from the evaluation over \( [\pi, 3\pi/2] \). The computed areas give us the total area \[ A = \left[ x \sin x + \cos x \right]_{\pi/2}^{\pi} - \left[ x \sin x + \cos x \right]_{\pi}^{3\pi/2}. \] Calculate and add the results to find the total area for part a, which is \( 3\pi/4 \).
05

Integral for part b

Calculate the integral of \( |y| \) from \( 3\pi/2 \) to \( 5\pi/2 \). Similar to part a, the function changes from below to above the x-axis at \( 2\pi \), so:\[ A = -\int_{3\pi/2}^{2\pi} x \cos x \, dx + \int_{2\pi}^{5\pi/2} x \cos x \, dx. \]
06

Calculate part b

Use the evaluated integral from Step 2 to solve for each part:Apply the evaluated integral over the intervals and compute the areas accordingly. The total area for part b is also found to be \( 3\pi/4 \).
07

Integral for part c

Calculate the integral from \( 5\pi/2 \) to \( 7\pi/2 \), which behaves as:\[ A = \int_{5\pi/2}^{3\pi} x \cos x \, dx - \int_{3\pi}^{7\pi/2} x \cos x \, dx. \]
08

Calculate part c

Use the evaluated integral over the intervals from Step 2 to compute the area:Similar to parts a and b, compute the total area for part c, which results in \( 3\pi/4 \).
09

Detecting the Pattern

The area for each interval is \( 3\pi/4 \), suggesting a pattern where the area stays constant between specific segments defined as \( \left(\frac{2n-1}{2}\right)\pi \leq x \leq \left(\frac{2n+1}{2}\right)\pi \).
10

Generalize for arbitrary n

Conclude that for any \( n \), the area between the curve and the x-axis in the interval \( \left(\frac{2n-1}{2}\right)\pi \leq x \leq \left(\frac{2n+1}{2}\right)\pi \) is \( 3\pi/4 \). This is because the symmetry and periodic behavior of the cosine function combined with integration properties lead to consistent area results for similar integral bounds for each n.

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Key Concepts

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

Area under a Curve
The area under a curve is an important concept in calculus and helps us determine the size of a region under a graph within specified limits. To find this area, we typically consider the function that describes the curve 鈥� in this case, it's given by the function \( y = x \cos x \). Our task is to calculate how much space this curve occupies between specific intervals and the x-axis.
  • We use integration to find the enclosed area.
  • Since the curve can be above or below the x-axis, we consider the absolute value of the function.
This ensures our calculations for the area are always positive, as area cannot be negative. For example, over the interval \( \pi/2 \leq x \leq 3\pi/2 \), the computation involves breaking it into where it is above and below the x-axis, integrating separately, and then summing absolute values.
Definite Integral
A definite integral helps us calculate the exact area under a function between two points on the x-axis. It involves evaluating the integral of the function within given limits to find the accumulated sum over that range. For our exercise, we perform the definite integral of \( y = x \cos x \) over multiple intervals.
  • The chosen intervals are important: they define the specific section of the curve we analyze.
  • In our example, for each part of the exercise, the area between specific bounds like \( \pi/2 \leq x \leq 3\pi/2 \) is evaluated.
To solve the integral of \( x \cos x \), integration by parts is utilized. This technique involves breaking down the integral into simpler parts, making it easier to calculate the definite values between chosen limits.
Trigonometric Integrals
When dealing with functions that include trigonometric terms, such as \( x \cos x \), integrals can become more complex. In such cases, certain techniques and formulas come in handy, especially integration by parts. This process is used to simplify and solve integrals that involve products of functions.
  • Let \( u = x \) and \( dv = \cos x \, dx \); then \( du = dx \) and \( v = \sin x \).
  • Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
By substituting these in, we transform \( \int x \cos x \, dx \) to \( x \sin x - \int \sin x \, dx \), leading to \( x \sin x + \cos x + C \). This simplifies the problem considerably, allowing us to handle the trigonometric integrals effectively and derive the precise area under our specified curve.

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