Problem 32 - Thomas Calculus

Question

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.

Step-by-Step Solution

Answer

The area for each interval is approximately \(\pi^2/2\), repeating every \(\pi\) radians due to the periodicity of \(\cos x\).

Step 1: Understand the Problem

We need to find the area enclosed by the curve \(y = x \cos x\) and the \(x\)-axis for various intervals. The area is given by the definite integral of \(|y|dx\) over the specified interval, due to the potential for negative values of \( \cos x \), necessitating absolute value consideration.

Step 2: Part A: Set up the Integral

For \(\pi/2 \leq x \leq 3\pi/2\), set up the integral: \[ \int_{\pi/2}^{3\pi/2} |x \cos x| \, dx \]. In this interval, \(\cos x\) is positive for \(\pi/2 \leq x < \pi\) and negative for \(\pi < x \leq 3\pi/2\), hence, \(|x \cos x| = x \cos x\) for \(\pi/2 \leq x \leq \pi\) and \(-x \cos x\) for \(\pi < x \leq 3\pi/2\).

Step 3: Part A: Evaluate the Integral

Split the integral into two parts: \[ \int_{\pi/2}^{\pi} x \cos x \, dx + \int_{\pi}^{3\pi/2} -x \cos x \, dx \]. Solve these using integration by parts or other appropriate methods to find the total area.

Step 4: Part B: Set up the Integral

For \(3\pi/2 \leq x \leq 5\pi/2\), set up the integral: \[ \int_{3\pi/2}^{5\pi/2} |x \cos x| \, dx \]. The behavior of \(\cos x\) flips at \(x = 2\pi\). Use \(|x \cos x| = -x \cos x\) for \(3\pi/2 \leq x < 2\pi\) and \(x \cos x\) for \(2\pi \leq x \leq 5\pi/2\).

Step 5: Part B: Evaluate the Integral

Split and evaluate the integral: \[ \int_{3\pi/2}^{2\pi} -x \cos x \, dx + \int_{2\pi}^{5\pi/2} x \cos x \, dx \]. Solve these integrals similarly as in Part A to find the area.

Step 6: Part C: Set up the Integral

For \(5\pi/2 \leq x \leq 7\pi/2\), the integral is \[ \int_{5\pi/2}^{7\pi/2} |x \cos x| \, dx \]. Analyze \(|x \cos x|\) based on intervals \(5\pi/2 \leq x < 3\pi\) and \(3\pi \leq x \leq 7\pi/2\) and determine where the function is positive or negative to split the integral appropriately.

Step 7: Part C: Evaluate the Integral

Evaluate each relevant portion of the integral, using expression modifications per \(|\cos x|\) as done in Parts A and B. Integrate to find the required area.

Step 8: Part D: Find the Pattern

Analyze the results from parts A, B, and C. Notice each interval \((\pi, 3\pi/2), (3\pi/2, 5\pi/2)\) leads to the same absolute area value reflecting periodicity due to the behavior of the cosine function over \(\pi\) intervals.

Step 9: Part D: Generalize for Arbitrary Interval

Conclude from the pattern that the area between the curve and the \(x\)-axis for any interval \(\left(\frac{2n - 1}{2}\pi, \frac{2n + 1}{2}\pi \right)\) is the same, due to the repetitive zero-crossing and sign-flipping behavior of \(\cos x\) over such ranges. The integrals across any such interval result in an area of approximately \(\pi^2/2\) after calculation.

Key Concepts

Integration by Parts

When tackling problems involving definite integration, especially with products of functions, integration by parts can be a powerful tool. It is derived from the product rule of differentiation. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \], where you choose one function to differentiate (\(u\)) and the other to integrate (\(dv\)).
For example, in the integral \( \int x \cos x \, dx \), you might choose \(u = x\) (making \(du = dx\)) and \(dv = \cos x \, dx\) (making \(v = \sin x\)).
This choice is due to the fact that differentiating \(x\) simplifies the integral, and the integral of \(\cos x\) is straightforward.

• Differentiate \(u\): \(du = dx\).
• Integrate \(dv\): \(v = \sin x\).
• Apply the integration by parts formula.
• Calculate \(uv - \int v \, du\).

This method breaks down the original problem into more manageable parts and helps solve integrals where the functions are too complex to integrate directly.

Trigonomtric Functions

Trigonomtric functions like \(\cos x\) are periodic, meaning they repeat values in regular intervals. Specifically, the cosine function has a period of \(2\pi\), meaning \(\cos(x) = \cos(x + 2n\pi)\) for any integer \(n\). This periodic nature is crucial when calculating areas enclosed by trigonometric curves.
In our exercise, we consider the behavior of \(x\cos x\), affected by the periodicity and sign of \(\cos x\). Notably, \(\cos x\) changes sign between intervals of \(\pi\), which affects whether the product \(x \cos x\) contributes a positive or negative area under the curve.

• \(\cos x\) is positive in \((0, \pi)\) and negative in \((\pi, 2\pi)\).
• The pattern repeats in subsequent intervals.
• This sign change plays a key role in determining the absolute value areas between the function and the x-axis.

Understanding these trends helps predict the behaviour of integrals over different intervals, simplifying complex trigonometric integration tasks.

Area Between Curves

Finding the area between curves involves determining where the curve lies relative to the x-axis and calculating the definite integral over specified bounds. When the function crosses the x-axis, the area is split into regions where the function is above and below the axis.
For the curve \(y = x \cos x\), we first evaluate its behavior over specific intervals to ascertain where it adds positively or negatively to the total area. This requires examining the sign of \(\cos x\).

• Identify where \(x \cos x\) is positive or negative.
• Consider absolute values to ensure all areas contribute positively.
• Split the integrals over sections using different signs for \(x \cos x\) based on \(\cos x\)'s sign.
• Sum these absolute integral values to find the total area.

The concept demonstrates that integrating absolute values can help manage regions where a curve flips across the x-axis, ensuring that calculated areas accurately reflect the true space enclosed.