Question

Finding Area Find the area of the region enclosed by the \(x\) -axis and the curve \(y=x \sin x\) for (a) \(0 \leq x \leq \pi\) (b) \(\pi \leq x \leq 2 \pi\) (c) \(0 \leq x \leq 2 \pi\)

Short answer

The areas enclosed by the x-axis and the curve \(y=x \sin x\) are \(π\) square units for 0 ≤ x ≤ π and π ≤ x ≤ 2π, and \(2π\) square units for 0 ≤ x ≤ 2π.

Step-by-step solution

Find the Indefinite Integral

To find the area, first we need to find the indefinite integral of \(x \sin x\). Using integration by parts, let \(u=x\) and \(v=\sin x\), then \(du=dx\) and \(dv=\cos x dx\). Thus, the integral of \(x \sin x\) becomes \(-x \cos x + \int \cos x \, dx\), which simplifies to \(-x \cos x + \sin x + C\).

Applying Limits for 0 ≤ x ≤ π

The definite integral from 0 to π is now evaluated by taking the antiderivative at π and subtracting the antiderivative at 0. This gives \( [-π \cos(π) + \sin(π)] - [-0 \cos(0) + \sin(0)]\). Because \(\cos(π) = -1\), \(\sin(π)=0\), \(\cos(0) = 1\), and \(\sin(0) =0\), the evaluation yields \( π + 0 - 0 = π\). Since the function is all above the x-axis for the defined limit, there is no need for absolute values.

Applying Limits for π ≤ x ≤ 2π

Now, we apply the limits from π to 2π to the antiderivative. This gives \([-2π \cos(2π) + \sin(2π)] - [-π \cos(π) + \sin(π)]\). Because \(\cos(2π) = 1\), \(\sin(2π)=0\), \(\cos(π) = -1\), and \(\sin(π) =0\), the evaluation yields \( 2π - (π) = π\). However, the function is below the x-axis in this interval, so we need to take the absolute value and our answer is \(|π| = π\).

Adding the Areas for 0 ≤ x ≤ 2π

The total area enclosed by the x-axis and the curve \(y=x \sin x\) from 0 to 2π is the sum of the absolute values of the areas calculated in steps 2 and 3. This gives \(π + π = 2π\).