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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.2.58a

Finding area
Find the area of the region enclosed by the curve y = x cos(x) and the x-axis (see the accompanying figure) for:
a. π/2 ≤ x ≤ 3π/2.
Graph of y = x cos x from 0 to 7π/2 showing oscillations crossing the x-axis between -10 and 10.

Verified step by step guidance
1
Identify the function and the interval: The function given is \(y = x \cos x\), and the interval is \(\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}\).
Analyze the graph to determine where the function is above or below the x-axis in the given interval. From the graph, \(y = x \cos x\) is negative between \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) because \(\cos x\) is negative in this interval and \(x\) is positive.
Since the function is below the x-axis in this interval, the area enclosed by the curve and the x-axis is the integral of the absolute value of the function. This means we need to integrate \(-x \cos x\) over \(\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]\).
Set up the integral for the area: \(\text{Area} = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -x \cos x \, dx\).
To solve the integral, use integration by parts where you let \(u = x\) and \(dv = \cos x \, dx\). Then compute \(du = dx\) and \(v = \sin x\). Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral for Area Calculation

The definite integral of a function over an interval gives the net area between the curve and the x-axis. When the function crosses the x-axis, the integral sums positive and negative areas, so the absolute value or splitting the integral is needed to find the total enclosed area.
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Behavior of the Function y = x cos(x)

The function y = x cos(x) oscillates and crosses the x-axis at points where cos(x) = 0, such as x = π/2 and 3π/2. Understanding these zeros helps determine the limits for integration and whether the function is above or below the x-axis in the given interval.
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Splitting the Integral at Zeros of the Function

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