Question

Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value. $$f(x)=\cos x \text { on }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

Short answer

Answer: The average value of the function $$f(x) = \cos{x}$$ on the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ is $$\frac{2}{\pi}$$. This can be represented on a graph by drawing a horizontal line at the height of $$\frac{2}{\pi}$$ that spans across the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$.

Step-by-step solution

Setup the Average Value Formula

To find the average value of the function, we will use the average value formula: $$f_{avg} = \frac{1}{b - a}\int_{a}^{b} f(x)dx$$ with $$a = -\frac{\pi}{2}, b = \frac{\pi}{2},$$ and $$f(x) = \cos{x}$$. So our formula becomes: $$f_{avg} = \frac{1}{\frac{\pi}{2} - (-\frac{\pi}{2})}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{x} dx$$

Evaluate the Integral

Now, we need to evaluate the integral in the formula: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{x} dx$$ We know that the integral of $$\cos{x}$$ is $$\sin{x}$$. So, $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{x} dx= \sin{x}\Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin{\frac{\pi}{2}} - \sin{-\frac{\pi}{2}} = 1 - (-1) = 2$$

Calculate the Average Value

Now that we have evaluated the integral, we can plug it back into the average value formula: $$f_{avg} = \frac{1}{\frac{\pi}{2} - (-\frac{\pi}{2})}\cdot 2$$ The denominator simplifies to: $$\frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$$ So our average value is: $$f_{avg} = \frac{1}{\pi}\cdot 2 = \frac{2}{\pi}$$

Draw the Graph and Indicate the Average Value

Based on the calculations, the average value of the function $$f(x) = \cos{x}$$ on the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ is $$\frac{2}{\pi}$$. To draw the graph of the function and indicate the average value, plot the function $$\cos{x}$$ and draw a horizontal line at the height of $$\frac{2}{\pi}$$ that spans across the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$. This line represents the average value of the function on the given interval.

Key concepts

Definite Integral

In integral calculus, the \textbf{definite integral} of a function is a fundamental concept representing the area under the curve of the function between two specified points on the x-axis. Specifically, to find the definite integral of a function, say \( f(x) \) from \( a \) to \( b \) on the x-axis, we use the notation \[ \int_{a}^{b} f(x)dx \. \]

This can be intuitively understood as the total accumulation of the quantity represented by the function, from one point to another. It is essential not just in mathematics but also in physics, engineering, and economics to calculate areas, volumes, displacement, work, or any quantity where accumulation over an interval is necessary. In the context of finding average values, the definite integral helps us understand how a function behaves on average over the interval from \( a \) to \( b \).

Visualize the definite integral as literally adding up infinitely small rectangles under the curve from \( a \) to \( b \) to find this exact total accumulation. This visualization plays a significant role in understanding how the average value of a function is computed over an interval.

Integral Calculus

Integral calculus is the branch of mathematics that deals with integrals and their properties. It is essentially about addition—more accurately, the 'continuous addition' of infinitely many infinitesimally small quantities. The sister branch to integral calculus is differential calculus which focuses on the concept of the derivative, or the rate at which quantities change.

Integrals can be divided into two major types: indefinite integrals and definite integrals. The indefinite integral, also known as the antiderivative, is the reverse of taking a derivative and includes an arbitrary constant, \( C \). On the other hand, the definite integral, which has already been introduced, computes a specific numerical value and represents a precise quantity.

Integral calculus is fundamental in calculating areas under curves, solving the problem of finding functions given their rates of change, modeling physical phenomena, and is deeply intertwined with the idea of inverting the process of differentiation, tying back to the Fundamental Theorem of Calculus.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions, such as \( \sin(x) \) and \( \cos(x) \). These are common in calculus due to the periodic nature of trigonometric functions, which model many natural phenomena such as waves and oscillations.

Common Trigonometric Integrals

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• \( \int \sin(x) dx = -\cos(x) + C \)
• \( \int \cos(x) dx = \sin(x) + C \)
• \( \int \tan(x) dx = -\ln(|\cos(x)|) + C \)
• \( \int \sec(x) dx = \ln(|\sec(x) + \tan(x)|) + C \)
\

In the textbook exercise provided, we dealt specifically with the integral of \( \cos(x) \), which has a straightforward antiderivative, \( \sin(x) \). The mastery of trigonometric identities and integrals of trigonometric functions is quintessential in solving more complex calculus problems as well as modeling scenarios in various scientific fields.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is one of the most powerful principles in mathematics, bridging the seemingly separate concepts of differentiation and integration together. It states that if a function \( f \) is continuous over the interval \( [a,b] \), and \( F \) is an antiderivative of \( f \) on that interval, then: \[ \int_{a}^{b} f(x)dx = F(b) - F(a) \. \]

In simpler terms, it means the definite integral of a function over an interval can be found using its antiderivatives. This theorem not only has central importance in theoretical mathematics but also profound practical applications. It allows us to compute definite integrals quickly, without having to manually sum infinitesimal quantities.

Relating to the average value exercise, the Fundamental Theorem of Calculus eases the process of finding the average value of a function, by linking the area under the curve (integral) to the antiderivatives of the function. It reassures us that by finding the antiderivative, we can then evaluate at the boundaries to compute the integral, which then plays a vital role in calculating the average value.