Finding area
Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:
b. π ≤ x ≤ 2π.

Verified step by step guidance

1. Identify the region whose area is to be found: the curve is given by \(y = x \sin x\) and the interval is \(\pi \leq x \leq 2\pi\). From the graph, note that the curve lies below the x-axis in this interval, so the function values are negative.
2. Set up the integral for the area between the curve and the x-axis. Since the curve is below the x-axis, the area is given by the integral of the negative of the function: \(\text{Area} = \int_{\pi}^{2\pi} -x \sin x \, dx\).
3. Use integration by parts to evaluate the integral \(\int x \sin x \, dx\). Let \(u = x\) and \(dv = \sin x \, dx\), then $du = dx$ and \(v = -\cos x\). Applying integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
4. Substitute back into the definite integral and evaluate the resulting expression at the limits \(x = \pi\) and \(x = 2\pi\).
5. Take the absolute value of the result if necessary to ensure the area is positive, as area cannot be negative.

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 dips below the x-axis, the integral yields a negative value, so the absolute value or splitting the integral at zeros is necessary to find the total enclosed area.

Behavior of the Function y = x sin(x)

The function y = x sin(x) oscillates with increasing amplitude as x increases. It crosses the x-axis at multiples of π, creating regions above and below the axis. Understanding where the function is positive or negative helps determine how to set up the integral for area.

Splitting the Integral at Zeros of the Function

To find the area between the curve and the x-axis over an interval where the function changes sign, split the integral at points where y = 0. Calculate the integral over each subinterval and take the absolute value of each to sum the total area.