Problem 123 - Precalculus

Question

In calculus, the value \(F(b)-F(a)\) of a function \(F(x)\) at \(x=a\) and \(x=b\) plays an important role in the calculation of definite integrals. Find the exact value of \(F(b)-F(a)\) $$F(x)=\sec ^{2} x+1, a=\frac{5 \pi}{6}, b=\frac{4 \pi}{3}$$

Step-by-Step Solution

Answer

The exact value of \(F(b) - F(a)\) is \(\frac{8}{3}\).

Step 1: Evaluate the Function at x=b

Substitute the value of \(x = b\) into the function \(F(x)\).\[ F \left( \frac{4\pi}{3} \right) = \sec^2 \left(\frac{4\pi}{3}\right) + 1 \]Recall that \(\sec x = \frac{1}{\cos x}\). First, find \(\cos \left( \frac{4\pi}{3} \right)\), which is \(-\frac{1}{2}\). Thus, \(\sec \left( \frac{4\pi}{3} \right) = -2\). So, \[ \sec^2 \left( \frac{4\pi}{3} \right) = (-2)^2 = 4 \]Therefore, \[ F \left( \frac{4\pi}{3} \right) = 4 + 1 = 5 \]

Step 2: Evaluate the Function at x=a

Substitute the value of \(x = a\) into the function \(F(x)\).\[ F \left( \frac{5\pi}{6} \right) = \sec^2 \left(\frac{5\pi}{6}\right) + 1 \]For \(\frac{5\pi}{6}\), \(\cos \left( \frac{5\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2}\).Thus, \(\sec \left( \frac{5\pi}{6} \right) = -\frac{2}{\sqrt{3}}\).So, \[ \sec^2 \left( \frac{5\pi}{6} \right) = \left(-\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]Therefore, \[ F \left( \frac{5\pi}{6} \right) = \frac{4}{3} + 1 = \frac{4}{3} + \frac{3}{3} = \frac{7}{3} \]

Step 3: Calculate F(b) - F(a)

Now that both values \(F(b)\) and \(F(a)\) have been calculated, find \(F(b) - F(a)\).Given that:\[ F \left( \frac{4\pi}{3} \right) = 5 \quad \text{and} \quad F \left( \frac{5\pi}{6} \right) = \frac{7}{3} \]So, \[ F(b) - F(a) = 5 - \frac{7}{3} \]To subtract, rewrite 5 as \(\frac{15}{3}\):\[ F(b) - F(a) = \frac{15}{3} - \frac{7}{3} = \frac{8}{3} \]

Key Concepts

Trigonometric Functions

Trigonometric functions are the building blocks for many concepts in calculus and mathematics in general. They relate the angles of a triangle to the lengths of its sides. Common trigonometric functions include sine (\(\sin\theta\)), cosine (\(\cos\theta\)), and tangent (\(\tan\theta\)). These functions are periodic and often found in mathematical contexts involving waves, circles, and oscillations.

The study of trigonometric functions is crucial for understanding phenomena in various fields such as physics and engineering. They provide necessary function transformations, including stretches and compressions that modify the basic wave shapes.

These functions have inverses and reciprocals, which are also widely used. Particularly, the secant function, which is the focus here, is the reciprocal of the cosine function. Understanding these relationships allows mathematicians and scientists to model and solve real-world problems effectively.

Secant Function

The secant function (\(\sec x\)) is less commonly encountered than sine or cosine, but it is still essential. It is defined as the reciprocal of the cosine function:

• Formula: \(\sec x = \frac{1}{\cos x}\)

This function has important properties related to its domain and range. The secant function is undefined wherever the cosine function is zero because division by zero is undefined. Therefore, \(\sec x\) is undefined at values where \(x=\frac{\pi}{2} + n\pi\), where \(n\) is any integer.

The range of the secant function is:

• \(( -\infty, -1 ] \cup [ 1, \infty )\)

While evaluating \(\sec x\) for specific angles, understanding the unit circle can be very helpful as it simplifies the process. In the exercise, the evaluation of \(\sec(\frac{4\pi}{3})\) resulted in \(-2\), which was calculated by finding \(\cos(\frac{4\pi}{3})\) as \(-\frac{1}{2}\). Similarly, for \(\sec(\frac{5\pi}{6})\), the reciprocal value was \(-\frac{2}{\sqrt{3}}\). By squaring these secant values, we found \(\sec^2 x\) needed for integrating certain functions in calculus.

Function Evaluation

Function evaluation is simply the process of determining the output of a function for a specific input. In the context of integrating or using calculus, this often involves substituting a specific value into a given formula.

In the exercise presented, we evaluated the function \(F(x) = \sec^2 x + 1\) at certain inputs to find needed outputs. Specifically, evaluating at \(x=b\) and \(x=a\) tasks us to substitute these values into the function to determine \(F(b)\) and \(F(a)\) respectively.

To conduct a function evaluation, follow these steps:

• Substitute the given value of \(x\) into the function equation.
• Compute the resulting expression by simplifying it.
• Use trigonometric identities or tables as necessary to find exact values, especially for trigonometric functions.

In this exercise, calculating \(F(b)\) and \(F(a)\) involved using the secant function, calculating its value, squaring it, and then adding 1. The final step was to find \(F(b) - F(a)\) which requires basic subtraction of the outcomes from these evaluations. This process helps in understanding the role of definite integrals as it involves finding the net change over a period of time, which is fundamental in calculus.