To find the area bounded by the curve \( y = x|\sin x| \) and the X-axis between \( x = 0 \) and \( x = 2\pi \), we must integrate the function with respect to \( x \) over the specified interval. Here's the solution broken down step-by-step:
- Recognize that \( |\sin x| \) is an absolute value function, meaning it changes the sign of portions of the sine function to ensure it is always non-negative.
- The period of \( \sin x \) is \( 2\pi \). Within \( 0 \leq x \leq 2\pi \), the absolute value function \( |\sin x| \) can be split for integration into segments:
- From \( 0 \) to \( \pi \), \( |\sin x| = \sin x \) because sine is positive.
- From \( \pi \) to \( 2\pi \), \( |\sin x| = -\sin x \) because sine is negative.
- Calculate the area by integrating separately over these intervals.
\(A = \int_{0}^{\pi} x\sin x \, dx + \int_{\pi}^{2\pi} x(-\sin x) \, dx\) - Use integration by parts to evaluate these integrals, which is given by the formula:
\(\int u \, dv = uv - \int v \, du\)
- For \( \int_{0}^{\pi} x\sin x \, dx \):
- Let \( u = x \) and \( dv = \sin x \, dx \).
- Then, \( du = dx \) and \( v = -\cos x \).
- The integral becomes \( \left[-x\cos x\right]_{0}^{\pi} + \int_{0}^{\pi} \cos x \, dx \).
- This results in: \( [-\pi \cdot (-1) - 0] + [0 - 1] = \pi - 1 \).
- For \( \int_{\pi}^{2\pi} -x\sin x \, dx \):
- Let \( u = x \) and \( dv = -\sin x \, dx \).
- Then, \( du = dx \) and \( v = \cos x \).
- The integral becomes \( \left[x\cos x\right]_{\pi}^{2\pi} - \int_{\pi}^{2\pi} \cos x \, dx \).
- This results in: \([2\pi \cdot 1 - \pi \cdot(-1)] - [0 + 1] = 3\pi - 1 \).
- Add the two areas together:
\(\text{Total Area} = (\pi - 1) + (3\pi - 1) = 4\pi - 2\)\)
Thus, the area bounded by the curve and X-axis is \(4\pi\) sq units.