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.

Verified step by step guidance

1. Identify the function and the interval: The function given is \(y = x \cos x\), and the interval is \(\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}\).
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.
3. 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 \(\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]\).
4. Set up the integral for the area: \(\text{Area} = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -x \cos x \, dx\).
5. 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: \(\int u \, dv = uv - \int v \, du\).

Key Concepts

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.

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.

Splitting the Integral at Zeros of the Function

To find the total area enclosed by the curve and the x-axis when the function changes sign, split the integral at the zeros of the function. Calculate the integral over each sub-interval and take the absolute value before summing to get the total area.