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Magazine Chapter 7: Problem 69
Find the area of the region bounded by the graphs of \(y=x \sin x\) and \(y=x \cos x\) from \(x=0\) to \(x=\pi / 4\).
Short Answer
Expert verified
The area is \(\frac{\pi}{8}\).
Step by step solution
01
Identify the Functions
The exercise involves two functions: the first function is \(y = x \sin x\) and the second function is \(y = x \cos x\). We need to consider both as functions of \(x\).
02
Find Points of Intersection
To find the area between the curves, first determine where the graphs intersect in the interval \([0, \pi/4]\). Set \(x \sin x = x \cos x\) and solve: \(\tan x = 1\), which gives \(x = \pi/4\). Thus, they intersect at \(x = 0\) and \(x = \pi/4\).
03
Set Up the Integral for Area
The area \(A\) between the curves from \(x = 0\) to \(x = \pi/4\) is given by the integral of the difference of the functions: \[ A = \int_{0}^{\pi/4} \left( x \cos x - x \sin x \right) \, dx. \]
04
Simplify the Integrand and Integrate
Simplify the integrand, notice it's of the form \(x(f'(x) - g'(x))\). The derivatives are such that \(f(x) = \sin x\) and \(g(x) = \cos x\). Thus, use integration by parts: \[ \int x \cos x \, dx - \int x \sin x \, dx. \]Each term involves integration by parts: let \(u = x\), \(dv = \cos x \, dx\) for the first and \(u = x\), \(dv = \sin x \, dx\) for the second.
05
Solve the Integrals Using Integration by Parts
Using integration by parts for each:1. For \( \int x \cos x \, dx \): \( du = dx\) and \(v = \sin x\), the integral becomes: \[ x \sin x \bigg|_0^{\pi/4} - \int \sin x \, dx = x \sin x + \cos x. \]2. For \( \int x \sin x \, dx \): \( du = dx\) and \(v = -\cos x\), the integral becomes: \[ -x \cos x \bigg|_0^{\pi/4} + \int \cos x \, dx = -x \cos x + \sin x. \]
06
Evaluate the Definite Integrals
Combine the results from Step 5, evaluate each from \(0\) to \(\pi/4\):1. \(\left[ x \sin x + \cos x \right]_0^{\pi/4}\) yields \(\frac{\sqrt{2}}{2}\pi/4 + \frac{\sqrt{2}}{2} - 1.\)2. \(\left[ -x \cos x + \sin x \right]_0^{\pi/4}\) yields \(-\frac{\sqrt{2}}{2}\pi/4 - \frac{\sqrt{2}}{2} + 0.\)Therefore, total area \(A = 1 - 0 = \frac{\pi}{8}.\)
07
Conclusion: Result of Area Calculation
After evaluating both definite integrals, the area of the region bounded by the curves \(y = x \sin x\) and \(y = x \cos x\) from \(x = 0\) to \(x = \pi / 4\) is \(\frac{\pi}{8}.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area Between Curves
When we discuss finding the area between curves, we are essentially talking about the space that lies between two functions over a certain interval. This is a common application of definite integrals in calculus. To find this area, we need to determine which function is the upper curve (top one) and which is the lower curve (bottom one). Once identified, the strategy is to subtract the lower curve's function from the upper curve's function and integrate this difference over the given interval. In our case, from the original problem, we have:
- Upper curve: \( y = x \cos x \)
- Lower curve: \( y = x \sin x \)
Integration by Parts
Integration by Parts is a powerful technique in calculus used to integrate products of functions. It works on the principle of the product rule for differentiation. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Here, you choose one function as \(u\), which you will differentiate, and \(dv\), which you will integrate. In the original step-by-step solution, both functions \(x \cos x\) and \(x \sin x\) were integrated using this method.For example, to integrate \(x \cos x\), we set:
- \( u = x \)
- \( dv = \cos x \, dx \)
- \( du = dx \)
- \( v = \sin x \)
Points of Intersection
Before calculating the area, we need to know where the two curves intersect. These points of intersection define our limits for integration. In the problem, we set the two functions equal to find these points:\[x \sin x = x \cos x\]Simplifying, we arrive at \(\tan x = 1\). Solving this within the interval \([0, \pi/4]\), we find that the curves intersect at \(x = 0\) and \(x = \pi/4\).Identifying these intersection points is crucial because they mark the bounds of integration, ensuring the area calculation is accurately performed between the right segments of the curves.
