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Chapter 6: Problem 16

Sketch each region (if a figure is not given) and then find its total area. The regions between \(y=\sin x\) and \(y=\sin 2 x,\) for \(0 \leq x \leq \pi\)

Short Answer

Expert verified
Based on the given solution, the total area between the curves y = sin x and y = sin 2x in the interval 0 鈮� x 鈮� 蟺 is 2.

Step by step solution

01

Sketch the Region

To begin, we must sketch the region between the curves \(y = \sin x\) and \(y = \sin 2x\) in the given interval \(0 \leq x \leq \pi\). Let's plot the two functions on the same graph, with the x-axis limited to the given range. After plotting, we can observe that there are two regions enclosed by the curves. We will calculate the area of each region and sum their areas to get the total area.
02

Determine Points of Intersection

We must find the points where the two functions intersect within the given interval. To do this, we set the functions equal to each other: \(\sin x = \sin 2x\) Now we need to solve this equation for x in the range of \([0, \pi]\). Using the double angle identity, we can rewrite \(\sin 2x\) as \(2 \sin x \cos x\). The equation becomes: \(\sin x = 2 \sin x \cos x\) Divide both sides by \(\sin x\) (excluding the case \(x = 0\) or \(x = \pi\), as \(\sin x \neq 0\)): \(1 = 2 \cos x\) Now solve for x: \(\cos x = \frac{1}{2}\) \(x = \frac{\pi}{3}\) or \(x=\frac{5\pi}{3}\) However, only \(\frac{\pi}{3}\) is inside the interval \([0, \pi]\). So, we have two intersection points at \(x=0\) and \(x=\frac{\pi}{3}\).
03

Integrate the Difference between Functions

We will now integrate the difference between the given functions over the two regions to find the total area. First, we will find the area of the region between \(x = 0\) and \(x = \frac{\pi}{3}\). In this region, \(y = \sin x\) is above \(y = \sin 2x\). The area A1 of this region can be found by integrating the difference between the functions: \(A1 = \int_{0}^{\frac{\pi}{3}} (\sin x - \sin 2x) dx\) Next, we will find the area of the region between \(x = \frac{\pi}{3}\) and \(x = \pi\). In this region, \(y = \sin 2x\) is above \(y = \sin x\). The area A2 of this region can be found by integrating the difference between the functions: \(A2 = \int_{\frac{\pi}{3}}^{\pi} (\sin 2x - \sin x) dx\) Now we can add these two areas to find the total area: \(Total Area = A1 + A2 = \int_{0}^{\frac{\pi}{3}} (\sin x - \sin 2x) dx + \int_{\frac{\pi}{3}}^{\pi} (\sin 2x - \sin x) dx\) Evaluate each integral separately: \(A1 = \left[-\cos x - \frac{1}{2}\cos 2x\right]_{0}^{\frac{\pi}{3}} = \frac{3}{2}\) \(A2 = \left[\frac{1}{2}\cos 2x + \cos x\right]_{\frac{\pi}{3}}^{\pi} = \frac{1}{2}\) Now sum the areas: \(Total Area = \frac{3}{2} + \frac{1}{2} = 2\) The total area between the two functions is 2.

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Key Concepts

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

Integration
Integration is a core concept in calculus that involves finding the accumulation of quantities, which can be represented as the area under curves. In mathematics, we often encounter integrals in the form of definite integrals, where you calculate the net area between the x-axis and the curve over a specific interval.
There are different methods to perform integration, such as the substitution method, integration by parts, and using standard integral formulas. The indefinite integral, or antiderivative, represents a family of functions whose derivative is the original function, whereas the definite integral represents a specific numerical value or area.
  • For example, if you have a function like \( (x) = \sin x\)\, its integral over the interval \[0, \frac{\pi}{3}\]\
Trigonometric Functions
Trigonometric functions such as sine and cosine play a crucial role in calculus, particularly in periodic functions. These functions describe the relationships in right-angled triangles but extend to periodic, oscillating patterns, such as waves. They often appear in integration problems and are practical for modeling scenarios involving circles or other periodic phenomena.
Key trigonometric identities simplify integration involving these functions.
  • For example, the double angle identity used in calculus is \(\sin 2x = 2 \sin x \cos x\), which can simplify solving and integrating equations involving trigonometric functions.
  • These identities help break down complex equations into more manageable parts, making the calculation of areas between curves and other integration problems easier.
Areas Between Curves
The concept of finding the area between curves is an important application of integration. This process involves calculating the integral of the difference between two functions over a specified interval. In simpler terms, it determines how much space (area) lies between two curves on a graph over a given range on the x-axis.
To find this area, one must:
  • Start by sketching the curves and identifying any intersection points because they divide the domain into smaller, manageable intervals for integration.
  • Then, determine which function is on top (or above) for each interval, as the area between curves is derived from subtracting the lower function from the upper function over the selected interval.
  • Finally, integrate this difference over each interval to get the total area.
This application is especially useful for physical problems involving velocity, position, or even physics and engineering scenarios where areas between curves can represent various real-world quantities.
Definite Integrals
Definite integrals form the backbone for calculating areas, particularly in finding areas between curves. A definite integral computes the net area from a function across an interval \([a, b]\), often looked at as the signed area.
Definite integrals are vital because they give us a precise value, unlike indefinite integrals, which result in a function + \(C\) where \(C\) is an unknown constant. This method makes solving for exact areas in calculus very effective.
  • For instance, the definite integral goes beyond just providing a cumulative total; it also accounts for areas above and below the x-axis, subtracting the latter to reflect the net accumulated area.
  • This is crucial in problems where we look at areas between two curves because the domains will have different integral calculations depending on intersection points.
Techniques like the Fundamental Theorem of Calculus bridge the integral to derivatives, creating a powerful toolset for many types of mathematical and scientific calculations.

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Most popular questions from this chapter

A simple model (with different parameters for different people) for the flow of air in and out of the lungs is $$V^{\prime}(t)=-\frac{\pi}{2} \sin \frac{\pi t}{2}$$ where \(V(t)\) (measured in liters) is the volume of air in the lungs at time \(t \geq 0, t\) is measured in seconds, and \(t=0\) corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs in exchanged with each breath. The amount that is exchanged is called the tidal volume. a. Find and graph the volume function \(V\) assuming that $$ V(0)=6 \mathrm{L} $$ b. What is the breathing rate in breaths/min? c. What is the tidal volume and what is the total capacity of the lungs?

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