Chapter 7: Problem 5
In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=\sin (x) \quad I=[\pi / 3, \pi] $$
Short Answer
Expert verified
The average value is \( \frac{9}{4\pi} \).
Step by step solution
01
Identify Formula for Average Value
The formula to find the average value of a function \( f(x) \) over the interval \([a, b]\) is given by: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
02
Determine Parameters for the Given Interval
In this problem, the function is \( f(x) = \sin(x) \) and the interval \( I = \left[\frac{\pi}{3}, \pi\right] \). Here, \( a = \frac{\pi}{3} \) and \( b = \pi \).
03
Setup the Integral
Substitute the function and the interval into the formula: \[ \text{Average value} = \frac{1}{\pi - \frac{\pi}{3}} \int_{\frac{\pi}{3}}^{\pi} \sin(x) \, dx \] This simplifies the interval length to \( \frac{2\pi}{3} \).
04
Integrate the Function
Evaluate the integral \( \int_{\frac{\pi}{3}}^{\pi} \sin(x) \, dx \). The antiderivative of \( \sin(x) \) is \( -\cos(x) \): \[ \int \sin(x) \, dx = -\cos(x) + C \] So, \[ \int_{\frac{\pi}{3}}^{\pi} \sin(x) \, dx = \left. -\cos(x) \right|_{\frac{\pi}{3}}^{\pi} = (-\cos(\pi)) - (-\cos(\frac{\pi}{3})) \]
05
Calculate Definite Integral
Substitute the upper and lower limits into the antiderivative: \[ -\cos(\pi) = 1 \] \[ -\cos(\frac{\pi}{3}) = -\left(\frac{1}{2}\right) = -\frac{1}{2} \] Thus, \[ 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]
06
Calculate the Average Value
Substitute the integral result into the average value formula: \[ \text{Average value} = \frac{1}{\frac{2\pi}{3}} \cdot \frac{3}{2} = \frac{3}{2} \times \frac{3}{2\pi} = \frac{9}{4\pi} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is a powerful tool in calculus. It is used to find the accumulated total of a function over a specific interval. The notation \[ \int_a^b f(x) \, dx \]indicates that we are calculating the integral of a function \(f(x)\) from the lower limit \(a\) to the upper limit \(b\). By doing this, we can find the net area between the function curve and the x-axis within this interval.
Here are some key points about definite integrals:
Here are some key points about definite integrals:
- The definite integral gives a precise total value as opposed to an antiderivative, which represents a family of functions.
- To calculate it, you need to find the antiderivative of the function, then apply the fundamental theorem of calculus by plugging in the upper and lower limits.
- After evaluating their results, you subtract the lower limit evaluation from the upper one to get the final integral value.
Trigonometric Functions
Trigonometric functions form a core group of functions in calculus and have applications in various fields including physics and engineering. The sine function \(\sin(x)\) is one such function, characterized by its periodic, wave-like shape.
Here are some important attributes of trigonometric functions, particularly \(\sin(x)\):
Here are some important attributes of trigonometric functions, particularly \(\sin(x)\):
- It varies between -1 and 1, making it a bounded function.
- The function is periodic with a period of \(2\pi\), meaning \(\sin(x) = \sin(x + 2\pi)\) for any \(x\).
- Its derivative is \(\cos(x)\) and its antiderivative is \(-\cos(x)\)\, which comes in handy when dealing with integrals such as the one in our exercise.
Calculus Problem-Solving
Solving problems in calculus involves a step-by-step approach, which includes understanding the problem, setting up equations, and carrying out computations.
For instance, solving a calculus problem to find the average value of a trigonometric function over a given interval involves several steps:
For instance, solving a calculus problem to find the average value of a trigonometric function over a given interval involves several steps:
- Understand the formula: For average value, use \[ \frac{1}{b-a} \int_a^b f(x) \, dx \].
- Set the parameters: Identify your function \(f(x)\) and the correct limits \([a, b]\).
- Compute the integral: Find the antiderivative of \(f(x)\) and use the bounds of the interval to evaluate the integral.
- Perform arithmetic: Use the evaluated integral to calculate the average, ensuring each mathematical operation is correct.