Problem 33 - Calculus: Graphical, Numerical, Algebraic

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) \(0 \leq x \leq 2 \pi\)

Step-by-Step Solution

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 1: 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 \(-x \cos x + \sin x + C\).

Step 2: 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.

Step 3: 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 \(|π| = π\).

Step 4: 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 \(π + π = 2π\).

Key Concepts

Definite Integral

A definite integral is a way of calculating the net area between a function and the x-axis within a specific interval. Unlike the indefinite integral, which produces a general formula,we solve the definite integral by evaluating it with upper and lower limits to find a numerical value.Key steps to evaluate a definite integral:

• Find the indefinite integral or the antiderivative of the function. This is the reverse process of differentiation.
• Substitute the upper and lower bounds of the interval into the antiderivative.
• Subtract the value at the lower bound from the upper bound to find the net area.

In the exercise where the area is enclosed by the curve and the x-axis using the function \(y = x \sin x\),we applied these steps first for the interval \(0 \leq x \leq \pi\) and found the area to be \(\pi\). Then, for \(\pi \leq x \leq 2\pi\),the calculation showed an area of \(\pi\), highlighting the importance of definite integrals in finding precise area measures.

Indefinite Integral

An indefinite integral, also known as an antiderivative, represents a family of functions and includes a constant \(C\).It shows the primitive form of a function, without specified limits, thus not giving a precise area but a general expression.Integration by parts is a common technique to find indefinite integrals, especially when dealing with products of functions.For the function \(x \sin x\), we use the formula:\[ \int u \, dv = uv - \int v \, du \]This technique entails:

• Selecting parts of the function to differentiate and integrate: Here, \(u = x\) leads to \(du = dx\), and \(dv = \sin x \, dx\) gives \(v = -\cos x\).
• Substituting back to find the antiderivative: The integration becomes \(-x \cos x + \sin x + C\).

This expression is essential for solving the definite integral later,illustrating why understanding the indefinite integral helps in such problems.

Area Under a Curve

Finding the area under a curve is a fundamental concept of integration.It involves calculating how much space is "trapped" between the curve and the x-axis across an interval.For calculating this area, we use definite integrals.In our exercise, the curve \(y = x \sin x\) forms sections that either ascend above or descend below the x-axis.For the interval \(0 \leq x \leq \pi\), the curve lies entirely above the x-axis,giving a straightforward calculation of the area as \(\pi\).For \(\pi \leq x \leq 2\pi\), the curve is below the x-axis,which means the calculated area is negative (\(-\pi\)), but as area cannot be negative, we take the absolute value.Ultimately, calculating the area under a curve gives insight into the total "size" of a function's graph over an interval.This is why it's so vital in various real-world applications like physics and economics, where understanding the space a function occupies can reveal important information.