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Chapter 6: Problem 21

Sketch the region enclosed by the given curves and ind its area. $$ y=\tan x, \quad y=2 \sin x, \quad-\pi / 3 \leqslant x \leqslant \pi / 3 $$

Short Answer

Expert verified
The enclosed area is 0.

Step by step solution

01

Understand the Curves

Examine the given functions: \(y = \tan x\) and \(y = 2 \sin x\). These represent the tangent and a transformed sine curve, respectively. Both functions are defined over the interval \(-\pi/3 \leq x \leq \pi/3\).
02

Find Points of Intersection

Set the two expressions for \(y\) equal to each other to find the intersection points: \(\tan x = 2 \sin x\). Rearrange the equation to \(\frac{\tan x}{\sin x} = 2\) or \(\frac{\sin x}{\cos x \cdot \sin x} = 2\) which simplifies to \(\frac{1}{\cos x} = 2\), giving \(\cos x = \frac{1}{2}\). Solve \(\cos x = \frac{1}{2}\) in the interval \(-\pi/3 \leq x \leq \pi/3\). The solutions are \(x = \pi/3\) and \(x = -\pi/3\).
03

Setup the Integral for Area

Select the area between the function curves. The area \(A\) between two curves from \(a\) to \(b\) is calculated by \(A = \int_{a}^{b} |f(x) - g(x)| \, dx\) where \(f(x)\) and \(g(x)\) are the two functions. Since \(y = \tan x\) is above \(y = 2 \sin x\) in the interval, the area is given by the integral \(A = \int_{-\pi/3}^{\pi/3} (\tan x - 2 \sin x) \, dx\).
04

Solve the Integral

Compute the integral: \(\int_{-\pi/3}^{\pi/3} (\tan x - 2 \sin x) \, dx\). This can be split into two integrals: \(\int_{-\pi/3}^{\pi/3} \tan x \, dx - 2 \int_{-\pi/3}^{\pi/3} \sin x \, dx\). The integral of \(\tan x\) is \(-\ln |\cos x|\) and the integral of \(\sin x\) is \(-\cos x\). Evaluating these from \(-\pi/3\) to \(\pi/3\), simplify accordingly to find the area.
05

Evaluate and Simplify

Evaluate: \[-\ln |\cos(\pi/3)| + \ln |\cos(-\pi/3)| - 2(-\cos(\pi/3) + \cos(-\pi/3))\]. For \(\cos(\pi/3) = 1/2\), we have:\[-\ln |1/2| + \ln |1/2| - 2(-1/2 + 1/2)\] which results in \(0\), thus the full evaluation results in the integral of zero, showing no enclosed area.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Between Curves
When dealing with calculus problems, finding the area between curves is a common and essential task. In this exercise, we have two functions: one represented by the tangent function and another by a transformed sine function. The region of interest is defined between these two curves within the specified interval.To find the area between two curves, you must:
  • Identify the interval over which you need to calculate the area. In this case, it was from \(-\pi/3\) to \(\pi/3\).
  • Determine which function is above the other within that range. For this example, \(y = \tan x\) is above \(y = 2 \sin x\) throughout the interval.
  • Set up the definite integral to calculate the area. The basic formula used is: \( A = \int_{a}^{b} |f(x) - g(x)| \, dx \), where \(f(x)\) and \(g(x)\) are the two functions in question.
The step-by-step solution evaluates this integral, leading to zero, indicating no enclosed area between the curves over the given interval.
Integration
Integration is a fundamental concept in calculus used to calculate areas under curves, among other applications. When you integrate a function, you are essentially summing up an infinite number of infinitesimally small areas to find the total area within specified bounds.In this problem:
  • The integral \( A = \int_{-\pi/3}^{\pi/3} (\tan x - 2 \sin x) \, dx \) is evaluated.
  • This integral can be split into two separate integrals: \( \int_{-\pi/3}^{\pi/3} \tan x \, dx \) and \( - 2 \int_{-\pi/3}^{\pi/3} \sin x \, dx \).
The integral of \(\tan x\) is given by \( -\ln |\cos x|\), and the integral of \(\sin x\) is \( -\cos x\). Calculating these integrals over the interval \([-\pi/3, \pi/3]\) reveals that the resulting area is zero.This reinforces the importance of correctly setting up and evaluating integrals to determine the area between curves.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are crucial in analyzing oscillatory problems and calculating areas in calculus.For this exercise:
  • The function \(y = \tan x\) represents a tangent curve that rises drastically as \(x\) approaches \(\pm \pi/2\), but within our defined interval it remains manageable.
  • On the other hand, \(y = 2 \sin x\) is a scaled sine wave, peaking at 2 within its range.
In order to find where these curves intersect, setting \( \tan x = 2 \sin x \) led to solving \( \cos x = 1/2 \), which provided critical points for evaluating the area.Understanding how these functions interact is key to setting up integrals accurately and interpreting geometric relationships between curves.

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Most popular questions from this chapter

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places. \(y=e^{-x^{2}}, y=0, x=-1, x=1\) $$ \text { (a) About the } x \text { -axis } \quad \text { (b) About } y=-1 $$

Find the volume of the described solid \(S\). $$ \begin{array}{l}{\text { A frustum of a pyramid with square base of side } b, \text { square top }} \\ {\text { of side } a, \text { and height } h}\end{array} $$ $$ \text { What happens if } a=b ? \text { What happens if } a=0 ? $$

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to \(x\) or \(y .\) Draw a typical approximating rectangle and label its height and width. Then find the area of the region. $$ 4 x+y^{2}=12, \quad x=y $$

Sketch the region enclosed by the given curves and ind its area. $$ y=\frac{x}{1+x^{2}}, \quad y=\frac{x^{2}}{1+x^{3}} $$

Sketch the region enclosed by the given curves and ind its area. $$ y=\cos x, \quad y=1-\cos x, \quad 0 \leqslant x \leqslant \pi $$
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