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Chapter 5: Problem 5

Find the average value of the function over the given interval. $$ f(x)=\sin x ;[0, \pi] $$

Short Answer

Expert verified
The average value is \( \frac{2}{\pi} \).

Step by step solution

01

Understand the Average Value Formula

The average value of a continuous function \( f(x) \) over the interval \([a, b]\) is given by the formula: \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] In this case, \( f(x) = \sin x \), \( a = 0 \), and \( b = \pi \).
02

Set Up the Integral

Using the average value formula, set up the integral: \[ \text{Average Value} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin x \, dx \] This simplifies to: \[ \frac{1}{\pi} \int_{0}^{\pi} \sin x \, dx \]
03

Integrate the Function

Integrate \( \sin x \) from \( 0 \) to \( \pi \). The antiderivative of \( \sin x \) is \(-\cos x\). Therefore, \[ \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} \]
04

Evaluate the Definite Integral

Substitute the limits into the antiderivative: \[ -\cos(\pi) - (-\cos(0)) \] Knowing that \( \cos(\pi) = -1 \) and \( \cos(0) = 1 \), we get: \[ -(-1) - (-1) = 1 + 1 = 2 \]
05

Calculate the Average Value

Substitute the result back into the average value formula: \[ \frac{1}{\pi} \times 2 = \frac{2}{\pi} \] Therefore, the average value of the function \( f(x) = \sin x \) over the interval \([0, \pi]\) is \( \frac{2}{\pi} \).

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus, which helps in finding the accumulation of quantities, such as area under curves. When we talk about integrating a function, we refer to finding its antiderivative or integral.
  • An integral can be indefinite, representing a family of functions, or definite, which involves a specific calculation over an interval.
  • In this case, we work with definite integrals as it calculates the average value of a function over an interval.
The process involves determining the integral of a continuous function which, for trigonometric functions like sine, involves understanding the antiderivative. The integral of \( \sin x \) results in an antiderivative of \( -\cos x \), a crucial component for solving such problems.
Definite Integral
A definite integral computes the net area under a curve defined by a function, over a specified interval. This is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the interval's bounds.
  • In our exercise, \( a = 0 \) and \( b = \pi \).
  • The definite integral tells us how much area exists between the function and the x-axis from \( x = 0 \) to \( x = \pi \).
  • To solve it, we first find the antiderivative of \( f(x) = \sin x \).
After computing the antiderivative, we substitute the upper and lower bounds to get the exact area. By doing these calculations for \( \int_{0}^{\pi} \sin x \, dx \), we utilize the values of cosine at the endpoints, which leads to finding the net area of \( 2 \) over the interval. This result is then used in the formula for average value.
Trigonometric Functions
Trigonometric functions relate angles to the sides of a triangle and are foundational in both algebra and calculus. In calculus, we often deal with functions like \( \sin x \), \( \cos x \), and \( \tan x \).
  • These functions are essential in calculating integrals and derivatives.
  • The sine function, \( \sin x \), oscillates between \(-1\) and \(1\), making it vital for modeling wave-like phenomena.
When integrating \( \sin x \), we find the antiderivative, which is \( -\cos x \). Knowing the calculated values at specific points, like \( 0 \) and \( \pi \), allows us to evaluate the integral. Understanding these basics aids in solving problems involving the average value of these periodic functions over a defined interval.

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Most popular questions from this chapter

(a) Over what open interval does the formula $$ F(x)=\int_{1}^{x} \frac{d t}{t} $$ represent an antiderivative of \(f(x)=1 / x ?\) (b) Find a point where the graph of \(F\) crosses the \(x\) -axis.

Find the area of the region enclosed by the graphs of \(y=\sin ^{-1} x, x=0,\) and \(y=\pi / 2\).

Evaluate the integrals by any method. $$ \int_{0}^{\pi / 4} 4 \sin x \cos x d x $$

Sketch the region described and find its area. The region under the curve \(y=x^{2}+1\) and over the interval \([0,3] .\)

Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{\pi / 6}^{\pi / 2}\left(x+\frac{2}{\sin ^{2} x}\right) d x $$
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