91Ó°ÊÓ

In integral calculus, the concept of finding the area under a curve is fundamental. It involves calculating the definite integral of a function over a specific interval. This gives us the accumulated area between the curve and the x-axis.
In the given exercise, we are tasked with finding the area under the curve defined by the function \( y = x \sin x \) over specified intervals. More specifically:

• For the interval \( 0 \leq x \leq \pi \), the area is determined by the integral \( \int_0^\pi x \sin x \, dx \).
• In the interval \( \pi \leq x \leq 2\pi \), since the function appears below the x-axis, the area below is expressed similarly, and we consider the absolute value for uniformity.
• Finally, the interval \( 2\pi \leq x \leq 3\pi \) follows the same principle.

These intervals combined reveal a pattern of area accumulation which the exercise aims to identify. Understanding such derivations through definite integrals is critical in calculus for evaluating areas accurately.

Integration by parts is a technique used in calculus to integrate products of functions. It's derived from the product rule for differentiation and involves breaking down complex integrands into simpler parts.
The formula for integration by parts is \( \int u \cdot dv = uv - \int v \cdot du \), where you choose parts of your integrand as \( u \) and \( dv \).
For the function \( y = x \sin x \):

• We let \( u = x \), which implies \( du = dx \).
• We select \( dv = \sin x \, dx \), leading to \( v = -\cos x \).

Substituting these into the formula gives us \( \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \). Solving this further allows us to complete the integration for the specific intervals mentioned. This technique simplifies complex integrations significantly, especially when the function involves products of a polynomial and a trigonometric function.

Definite integrals represent the net area between a curve and the x-axis over a specific interval. The notation \( \int_a^b f(x) \, dx \) is used, where \( a \) and \( b \) are the bounds of integration.
In this exercise, the definite integral calculations allow us to determine the specific area for each interval:

• From \( x = 0 \) to \( x = \pi \), the integral \( \int_0^\pi x \sin x \, dx \) calculates to \( \pi \).
• From \( x = \pi \) to \( x = 2\pi \), the integral \( \int_\pi^{2\pi} x \sin x \, dx \) also gives an area of \( \pi \) when absolute values are considered.
• For \( x = 2\pi \) to \( x = 3\pi \), we observe the same pattern.

This approach shows a clear pattern of area accumulation per period defined as \( [n\pi, (n+1)\pi] \), indicating a repeating area of \( \pi \). Recognizing such patterns facilitates easier predictions and calculations of integrals in many mathematical and practical applications.