Problem 13 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find the average value of \(y=\tan x\) from \(x=\frac{\pi}{4}\) to \(x=\frac{\pi}{3}\).

Step-by-Step Solution

Answer

Answer: The average value of the function \(y=\tan x\) from \(x=\frac{\pi}{4}\) to \(x=\frac{\pi}{3}\) is \(\frac{12}{\pi} \ln|\sqrt{2}|\).

Step 1: Analyze the interval and the range of the function

Our interval goes from \(x=\frac{\pi}{4}\) to \(x=\frac{\pi}{3}\), which is inside the first quadrant where the function \(y=\tan x\) is defined and positive. Given that \(y=\tan x\) is continuous and differentiable in this interval, we can proceed with the average value formula.

Step 2: Find the integral of the function

Now we have to find the integral of \(f(x)=\tan x\) with respect to \(x\). We know that the integral of \(\tan x\) is: \(\int \tan x dx = \ln|\sec x| + C\) We will be using the definite integral from \(x= \frac{\pi}{4}\) to \(x= \frac{\pi}{3}\): \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x dx = \left[ \ln|\sec x| \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\)

Step 3: Evaluate the integral at the limits

Now, we have to evaluate this integral at the upper and lower limits: \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x dx = \ln|\sec \frac{\pi}{3}| - \ln|\sec \frac{\pi}{4}|\) To simplify it further, we can use the properties of logarithms: \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x dx = \ln\left|\frac{\sec \frac{\pi}{3}}{\sec \frac{\pi}{4}}\right|\) Now, we will find the values of \(\sec \frac{\pi}{3}\) and \(\sec \frac{\pi}{4}\): \(\sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = 2\) \(\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \sqrt{2}\) Thus: \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x dx = \ln\left|\frac{2}{\sqrt{2}}\right| = \ln\left|\sqrt{2}\right|\)

Step 4: Apply the average value formula

Now that we found the definite integral of the function, we can use the average value formula: \(\text{Average value}=\frac{1}{\frac{\pi}{3}-\frac{\pi}{4}} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \tan x dx = \frac{1}{\frac{\pi}{12}} \ln|\sqrt{2}|\) Finally, we simplify the expression: \(\text{Average value}= \frac{12}{\pi} \ln|\sqrt{2}|\) The average value of the function \(y=\tan x\) from \(x=\frac{\pi}{4}\) to \(x=\frac{\pi}{3}\) is \(\frac{12}{\pi} \ln|\sqrt{2}|\).

Key Concepts

Definite Integral

When we want to understand the accumulation of a quantity, or the total area under a curve, the concept we turn to in calculus is that of the definite integral. In simple terms, the definite integral computes the area bounded by the graph of a function and the x-axis between two given points. The notation used is \[\int_{a}^{b} f(x) \, dx\] where \(a\) and \(b\) are the limits of integration representing the interval on the x-axis, and \(f(x)\) is the function we're integrating over that interval.

For the student tackling problems in AP Calculus AB, mastering definite integrals is crucial because they lay the foundation for understanding other concepts in calculus, such as antiderivatives and the calculations of volumes and average values of functions. Moreover, for functions that are continuous over the interval \[a, b\], the Fundamental Theorem of Calculus asserts that we can compute the definite integral using antiderivatives, making the process much more efficient and analytically solvable.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, allowing us to describe angles and their relationships to lengths and positions of lines or shapes. These functions, which include sine, cosine, tangent, and their reciprocals, cosecant, secant, and cotangent, emerge from the study of right triangles and are extended to the unit circle definition for all angles.

Understanding trig functions is essential in AP Calculus, particularly in performing integration and differentiation of these functions. In our example, the function \(y = \tan x\) is one such trigonometric function. The tangent function, which represents the ratio of the opposite side to the adjacent side in a right-angled triangle, exhibits periodic behavior and has its own set of properties and identities that simplify calculus operations.

AP Calculus AB

AP Calculus AB is an advanced placement course and examination offered in the United States to high school students. The curriculum covers a range of topics in differential and integral calculus, including limits, derivatives, definite integrals, and the Fundamental Theorem of Calculus. This course emphasizes a multi-representational approach to calculus, with concepts, results, and problems expressed graphically, numerically, analytically, and verbally.

In terms of our example problem, students in AP Calculus AB learn to calculate the average value of a function over an interval, which involves finding the definite integral of the function. The ability to connect the concept of the integral to the physical idea of area and the calculation of averages is a significant learning outcome of the course.

Function Continuity

Function continuity is a core concept in calculus that deals with how functions behave at points and over intervals. A function is said to be continuous at a point if there is no break, jump, or hole in the graph of the function at that point. Specifically, for a function to be continuous at point \(x=a\), the function must be defined at \(a\), the limit as \(x\) approaches \(a\) from both sides must exist, and the limit must equal the function's value at \(a\).

Continuity over an interval means the function is continuous at every point within the interval. In the context of the average value calculation of \(y=\tan x\), analyzing the function's continuity over the interval \[\frac{\pi}{4}, \frac{\pi}{3}\] is crucial. The function must be continuous and differentiable (except possibly at the endpoints) to apply the average value formula correctly. This ensures that there are no discontinuities that could disrupt the calculation of the area under the curve.

Antiderivative

An antiderivative, also known as an indefinite integral, represents a function whose derivative is the original function. In simpler terms, it's the reverse process of differentiation. When we know the derivative of a function and we want to recover the original function, we look for an antiderivative. The notation for an antiderivative is \(\int f(x) \, dx\), which reads as the integral of \(f(x)\) with respect to \(x\).

For example, if we have the function \(f(x) = \tan x\), its antiderivative is \(\int \tan x \, dx = \ln|\sec x| + C\), where \(C\) represents the constant of integration. This antiderivative is useful when calculating definite integrals, as seen in our example problem. It allows us to find a function's average value by integrating over a closed interval. Understanding the relationship between a function and its antiderivative is fundamental in calculus, as it forms the basis for solving complex problems involving areas, volumes, and average values.