Problem 69 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the region bounded by the graphs of \(y=x \sin x\) and \(y=x \cos x\) from \(x=0\) to \(x=\pi / 4\).

Step-by-Step Solution

Answer

The area is \(\frac{\pi \sqrt{2}}{4} - 1\).

Step 1: Identify Intersection Points if any within Given Bounds

First, let's determine if and where the curves intersect within the interval \([0, \pi/4]\). The given equations are \(y_1 = x \sin x\) and \(y_2 = x \cos x\). We find intersection points by setting \(y_1 = y_2\), which gives \(x \sin x = x \cos x\). Simplifying, we have \(\tan x = 1\). Solving \(\tan x = 1\) within \([0, \pi/4]\) gives us the intersection at \(x = \pi/4\). Therefore, within this interval, they intersect at \(x = \pi/4\). However, since the bounds are precisely up to \(\pi/4\), there is no change in intersection in the interval, and we need not solve for points between themsince they start the evaluation at \(x=0\) without initial intersection there.

Step 2: Define the Integrals for the Area Calculation

The area between two curves \(y_1\) and \(y_2\) from \(a\) to \(b\) is given by the integral \(\int_a^b |y_1 - y_2| \, dx\). Since from \(x = 0\) to \(x = \pi/4\), \(x \cos x\) appears to be greater than \(x \sin x\) (since cosine starts from 1 and sine from 0), we find the area by integrating: \[ \int_0^{\pi/4} (x \cos x - x \sin x) \, dx \].

Step 3: Integrate the Function

First, we simplify the integrand which is \(x (\cos x - \sin x)\). Now, integrate this from 0 to \(\pi/4\). We apply integration by parts. Let \(u = x\) and \(dv = (\cos x - \sin x) \, dx\). So, \(du = dx\) and \(v = \int (\cos x - \sin x) \, dx = \sin x + \cos x\).Thus, the integral becomes:\[ \left[ x(\sin x + \cos x) \right]_0^{\pi/4} - \int_0^{\pi/4} (\sin x + \cos x) \, dx \].

Step 4: Evaluate the Integration by Parts

Calculate \(\left[ x(\sin x + \cos x) \right]_0^{\pi/4}\):At \(x = \pi/4\), \(\frac{\pi}{4}(\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) = \frac{\pi}{4} \cdot \sqrt{2}\) since both \(\sin(\frac{\pi}{4})\) and \(\cos(\frac{\pi}{4})\) equate to \(\frac{\sqrt{2}}{2}\).At \(x = 0\), the value is 0.Now evaluate the integral \[ \int_0^{\pi/4} (\sin x + \cos x) \, dx = \left[ -\cos x + \sin x \right]_0^{\pi/4} = -\cos(\pi/4) + \sin(\pi/4) + \cos(0) - \sin(0) \].

Step 5: Simplify and Calculate Final Values

The expression now is:\[ \frac{\pi}{4} \cdot \sqrt{2} - \left[ -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + 1 \right] = \frac{\pi \sqrt{2}}{4} - \left( 0 + 1 \right) \].Hence, the area is:\(\frac{\pi \sqrt{2}}{4} - 1\).

Key Concepts

Area Between Curves

In calculus, finding the area between curves is a way to understand the difference in the values of two functions over a specified interval. When you are asked to find the area between two curves, you essentially want to calculate the space bounded by the curves within given limits of integration.
This concept involves evaluating the integral of the absolute difference between the two function values, over the specified range of integration.
For example, consider two functions, \( y_1 = x \sin x \) and \( y_2 = x \cos x \) within the interval \([0, \pi/4]\). To find the area between these two curves, you set up the integral as \( \int_0^{\pi/4} \lvert y_2 - y_1 \rvert \, dx \).

• Identify which function is greater over the interval, ensuring that the difference taken is positive.
• This concept helps in understanding how two different phenomena could relate, especially in practical scenarios such as calculating the material required between curves or the energy difference in physical systems.

With these steps, you can efficiently find and comprehend the area between curves in any given problem.

Integration by Parts

Integration by parts is an essential technique in calculus used to integrate products of two functions. It is particularly useful when dealing with integrals that cannot be simplified through basic methods.
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are parts of the original integrand.
Let's break it down a little:

• You choose \( u \) as a function that becomes simpler when differentiated.
• \( dv \) is the portion that you can readily integrate.
• After determining these components, you calculate \( du \) and \( v \), then apply the formula.

In the example from the problem, when integrating \( x (\cos x - \sin x) \), here's how you proceed:

• Let \( u = x \), then \( du = dx \).
• Let \( dv = (\cos x - \sin x)\, dx \), therefore, \( v = \sin x + \cos x \).

This results in the expression \( \left[ x(\sin x + \cos x) \right]_0^{\pi/4} - \int_0^{\pi/4} (\sin x + \cos x) \, dx \), making the integration feasible and more accessible.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric identities. These can initially seem complex, but are manageable with a systematic approach using known trigonometric identities and properties.
Consider integrals like \( \int (\cos x - \sin x) \, dx \). To tackle such integrals:

• Recognize the potential for applying identities to simplify the expressions.
• Perform integration either directly if possible or through techniques like substitution or by parts as in our main example.

In this scenario, combining trigonometric functions made the integration straightforward because you could use simple derivatives of sine and cosine.
The main takeaway is to be familiar with basic trigonometric derivatives and integrals, as they often serve as stepping stones towards more complex problems.

Intersection Points

Intersection points are where two curves meet within a certain bound or interval. To find them, equate the functions to solve for \( x \).
This concept is important because intersection points help determine the limits of integration when calculating the area between curves.
For the functions \( y_1 = x \sin x \) and \( y_2 = x \cos x \), setting \( x \sin x = x \cos x \) leads to \( \tan x = 1 \).
Solving this, we find the intersection point at \( x = \pi/4 \) within the interval \([0, \pi/4]\).
Key steps include:

• Solve the equation \( y_1 = y_2 \) to determine x-values where intersections occur.
• Analyze these points to effectively set the limits for integration and ensure accurate area calculations.

Intersection points often provide critical turning points in computations, so it's vital to get them right to further apply calculus tools correctly.