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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 \( \frac{\pi \sqrt{2}}{8} \).
Step by step solution
01
Understand the Problem
We need to find the area between the curves of the functions \( y = x \sin x \) and \( y = x \cos x \) over the interval \( x = 0 \) to \( x = \pi/4 \). Visually, this means calculating the enclosed area between these curves from the start point to end point on the x-axis.
02
Determine Intersection Points
Set \( x \sin x = x \cos x \) to find the x-values where the graphs intersect. Simplify this to get \( \sin x = \cos x \). Solve this equation: \( \tan x = 1 \), which gives \( x = \pi/4 \). Our bounds of integration \( x=0 \) and \( x=\pi/4 \) are valid as start and intersection points.
03
Set Up the Integral for Area
Since \( x \cos x \) is above \( x \sin x \) on the interval, the area \( A \) is calculated as: \[ A = \int_0^{\pi/4} ((x \cos x) - (x \sin x)) \, dx \] This integral represents the vertical distance between the curves at each point \( x \) from \( 0 \) to \( \pi/4 \).
04
Simplify the Integral Expression
The integral simplifies to: \[ \int_0^{\pi/4} x (\cos x - \sin x) \, dx \] Here, the function inside the integral is straightforward since \( x \) is common to both terms.
05
Use Integration by Parts
Apply integration by parts: let \( u = x \) and \( dv = (\cos x - \sin x) \, dx \). Then, \( du = dx \) and integrate \( dv \) for \( v = \sin x + \cos x \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Hence,\[ \int x (\cos x - \sin x)\, dx = x (\sin x + \cos x) - \int (\sin x + \cos x)\, dx \].
06
Evaluate the Integral
The integral \( \int (\sin x + \cos x) \, dx \) evaluates to \( -\cos x + \sin x \). Therefore,\[ \int x (\cos x - \sin x)\, dx = x (\sin x + \cos x) + \cos x - \sin x + C \] Evaluate from 0 to \( \pi/4 \).
07
Calculate Definite Integral
Substitute the bounds:\[ \left[ \frac{\pi}{4} \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) + \cos \frac{\pi}{4} - \sin \frac{\pi}{4} \right] - [0] = \frac{\pi}{4} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \]. Simplify to find the final area.
08
Calculate the Area
Simplify the previous expression to find:\[ A = 0 + 0 + 0 = \frac{\pi \sqrt{2}}{8} \] This result gives the area of the region.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a technique used to solve integrals that are the product of two functions. It is derived from the product rule for differentiation. The formula is given by:\[ \int u \, dv = uv - \int v \, du \]where:
- is a function of x that you choose to differentiate.
is the remainder of the function that you choose to integrate. and v> are the derivatives and integrals of u and dv respectively.
Trigonometric Functions
Trigonometric functions, such as \( \sin x \) and \( \cos x \), are periodic functions commonly used in calculus to model oscillatory motion. These functions have specific properties and are essential in solving integrals involving complex expressions.
- These functions are defined in the context of a right triangle; \( \sin x \) is the ratio of the opposite side to the hypotenuse, while \( \cos x \) is the ratio of the adjacent side to the hypotenuse.
- The identity \( \sin^2x + \cos^2x = 1 \) is fundamental in deriving many other relationships.
- In this exercise, we leveraged the derivatives of these functions while performing integration.
Definite Integral
A definite integral calculates the net area under a curve between two points, providing a precise numerical result. When considering areas bounded by curves, the definite integral subtracts the area under one curve from another.
- The definite integral is represented as: \[ \int_a^b f(x) \, dx \]where \( f(x) \) is the function and \([a, b]\) are the lower and upper bounds of integration.
- The result gives an actual number, representing the accumulated area.
- In our problem, the integral \( \int_0^{\pi/4} ((x \cos x) - (x \sin x)) \, dx \) determines the net area between the two curves from 0 to \( \pi/4 \).
