Problem 58
Question
Find the average value of the function on the given interval. \(f(x)=\sin x,[0, \pi / 2]\)
Step-by-Step Solution
Verified Answer
The average value of the function \(f(x)=\sin x\) on the interval \([0, \pi / 2]\) is \(\frac{2}{\pi}\).
1Step 1 : Identify a, b and \(f(x)\)
The function is \(f(x)=\sin x\), and the interval is [0, \(\pi / 2]\). So, \(a=0\), \(b = \pi / 2\), and \(f(x) = \sin x\).
2Step 2: Calculate the integral
The integral of \(f(x) = \sin x\) from 0 to \(\pi / 2\) is given by: \(\int_{0}^{\pi / 2}\sin x dx=-\cos x \Big|_0^{\pi / 2} =-(\cos(\pi / 2) -\cos(0)) = - (0 - 1) = 1.\)
3Step 3: Apply the formula for average value of a function
The average value of the function is given by \(f_{av}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). Substituting \(a=0, b = \pi / 2\), and \(\int_{a}^{b}f(x) dx = 1\), you get: \(f_{av}=\frac{1}{\pi / 2 - 0}*1 = \frac{2}{\pi}\)
Key Concepts
Definite IntegralTrigonometric FunctionsAverage Value FormulaFundamental Theorem of Calculus
Definite Integral
When we talk about the definite integral of a function, we're referring to the process of calculating the area under the curve of that function between two specific points on the x-axis. In mathematical terms, a definite integral is expressed as \(\int_{a}^{b} f(x) dx\), where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function we're integrating.
To visualize this, imagine plotting the function \(f(x)\) on a graph. The area between the curve, the x-axis, and the vertical lines at \(x=a\) and \(x=b\) is what we're interested in. For the function \(f(x)=\sin x\), this integral computes the total area under the sine curve from 0 to \(\pi/2\). As you can see from the solution, this area is exactly 1 square unit.
To visualize this, imagine plotting the function \(f(x)\) on a graph. The area between the curve, the x-axis, and the vertical lines at \(x=a\) and \(x=b\) is what we're interested in. For the function \(f(x)=\sin x\), this integral computes the total area under the sine curve from 0 to \(\pi/2\). As you can see from the solution, this area is exactly 1 square unit.
Trigonometric Functions
Trigonometric functions, such as \(\sin x\), \(\cos x\), and \(\tan x\), are fundamental in mathematics, especially in fields like geometry, engineering, and physics. These functions relate the angles of a right triangle to the lengths of its sides and extend these ratios to describe circular motion and periodic phenomena.
The sine function, \(\sin x\), represents the y-coordinate of a point on the unit circle, which is a circle with a radius of 1 centered at the origin of a coordinate system, as the angle x (measured in radians) sweeps from 0 to \(2\pi\). In this exercise, we specifically look at \(\sin x\) on the interval [0, \(\pi/2\)], which corresponds to the first quadrant of the unit circle where \(\sin x\) values range from 0 to 1.
The sine function, \(\sin x\), represents the y-coordinate of a point on the unit circle, which is a circle with a radius of 1 centered at the origin of a coordinate system, as the angle x (measured in radians) sweeps from 0 to \(2\pi\). In this exercise, we specifically look at \(\sin x\) on the interval [0, \(\pi/2\)], which corresponds to the first quadrant of the unit circle where \(\sin x\) values range from 0 to 1.
Average Value Formula
The average value formula is a way to find the average height of the function's graph over a certain interval \[a, b\]. Think of it like averaging out all the function's values over [a, b] to find the overall 'mean' value. The formula is given by \( f_{av} = \frac{1}{b-a} \int_{a}^{b} f(x) dx \).
Applying this formula requires you to integrate the function over the interval, which gives us the total area under the curve, and then divide by the length of the interval, which normalizes this area into an average height. For the sine function, the average value from 0 to \(\pi/2\) is found to be \(\frac{2}{\pi}\), which intuitively means that if you were to smooth out the sine wave over this range into a flat line, it would sit at the height of \(\frac{2}{\pi}\) on the graph.
Applying this formula requires you to integrate the function over the interval, which gives us the total area under the curve, and then divide by the length of the interval, which normalizes this area into an average height. For the sine function, the average value from 0 to \(\pi/2\) is found to be \(\frac{2}{\pi}\), which intuitively means that if you were to smooth out the sine wave over this range into a flat line, it would sit at the height of \(\frac{2}{\pi}\) on the graph.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects two central concepts in calculus: differentiation and integration. It tells us that the definite integral of a function over an interval can be calculated by finding the antiderivative of the function and evaluating it at the endpoints of the interval.
This theorem is essentially what makes it possible to compute integrals without having to actually measure the area under the curve piece by piece. It has two parts: the first part provides the method for evaluating a definite integral, while the second part tells us that the derivative of an integral function is the original function. In solving for the average value of \(\sin x\) on the interval [0, \(\pi/2\)], we effectively use the first part of this theorem by taking the antiderivative of \(\sin x\), and then we subtract its values at the interval's endpoints.
This theorem is essentially what makes it possible to compute integrals without having to actually measure the area under the curve piece by piece. It has two parts: the first part provides the method for evaluating a definite integral, while the second part tells us that the derivative of an integral function is the original function. In solving for the average value of \(\sin x\) on the interval [0, \(\pi/2\)], we effectively use the first part of this theorem by taking the antiderivative of \(\sin x\), and then we subtract its values at the interval's endpoints.
Other exercises in this chapter
Problem 57
The following table shows the velocity of a falling object at different times. For each time interval, estimate the distance fallen and the acceleration. $$\beg
View solution Problem 58
Generalize exercises 56 and 57 to \(\int \frac{1}{x^{(p+1 / / q}+x^{p / q}} d x\) for positive integers \(p\) and \(q\)
View solution Problem 58
Use a geometric formula to compute the integral. $$\int_{-3}^{0} \sqrt{9-x^{2}} d x$$
View solution Problem 58
The following table shows the velocity of a falling object at different times. For each time interval, estimate the distance fallen and the acceleration. $$\beg
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