Problem 121 - 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 \(F(b)-F(a)\) $$F(x)=2 \tan x+\cos x, a=-\frac{\pi}{6}, b=\frac{\pi}{4}$$

Step-by-Step Solution

Answer

The exact value of \(F(b)-F(a)\) is \(2 + \frac{\sqrt{2}}{2} + \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2}\).

Step 1: Evaluate F(a)

First, we need to find the value of the function at the point \(a\), which is \(a=-\frac{\pi}{6}\). Substitute \(-\frac{\pi}{6}\) into the function: \(F(-\frac{\pi}{6}) = 2 \tan(-\frac{\pi}{6}) + \cos(-\frac{\pi}{6})\). Compute \(\tan(-\frac{\pi}{6})\) and \(\cos(-\frac{\pi}{6})\): \(\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}\) and \(\cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\). Hence, \(F(-\frac{\pi}{6}) = 2(-\frac{1}{\sqrt{3}}) + \frac{\sqrt{3}}{2}\). Simplify this expression to get: \(-\frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{2}\).

Step 2: Evaluate F(b)

Next, we evaluate the function at the point \(b\), which is \(b=\frac{\pi}{4}\). Substitute \(\frac{\pi}{4}\) into the function: \(F(\frac{\pi}{4}) = 2 \tan(\frac{\pi}{4}) + \cos(\frac{\pi}{4})\). Calculate \(\tan(\frac{\pi}{4})\) and \(\cos(\frac{\pi}{4})\): \(\tan(\frac{\pi}{4}) = 1\) and \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). Thus, \(F(\frac{\pi}{4}) = 2(1) + \frac{\sqrt{2}}{2}\). Simplify this expression to get: \(2 + \frac{\sqrt{2}}{2}\).

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

With the values from the previous steps, calculate \(F(b) - F(a)\). Compute \(F(\frac{\pi}{4}) - F(-\frac{\pi}{6}) = \left(2 + \frac{\sqrt{2}}{2}\right) - \left(-\frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{2}\right)\). Simplify the expression: \(2 + \frac{\sqrt{2}}{2} + \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2}\). Combine like terms to get the exact value.

Key Concepts

Function Evaluation

Function evaluation involves calculating the output of a function at given input values. To evaluate the function, substitute the specified values into the function's formula. In this problem, we have a function given by \( F(x) = 2 \tan x + \cos x \). We first evaluate the function at \( x = a \), where \( a = -\frac{\pi}{6} \). This involves:

• Substituting \( -\frac{\pi}{6} \) into \( F(x) \)
• Calculating \( F(-\frac{\pi}{6}) = 2 \, \tan(-\frac{\pi}{6}) + \cos(-\frac{\pi}{6}) \)

Next, similarly, we evaluate \( F(x) \) at \( x = b \), where \( b = \frac{\pi}{4} \), and find \( F(\frac{\pi}{4}) \). The process helps find the value of \( F(b) - F(a) \), which is a critical component in definite integrals, offering the net area between the function and the x-axis over an interval.

Trigonometric Functions

Trigonometric functions like tangent and cosine are crucial in evaluating the function \( F(x) \) in this context. These functions relate the angles of a right triangle to the lengths of its sides and are fundamental in many areas of calculus.**How Trigonometric Functions are Used:**- **Tangent Function:** For \( \tan(-\frac{\pi}{6}) \), the tangent of a negative angle can be found using the symmetry property of the tangent function, \( \tan(-x) = -\tan(x) \). This gives \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \).- **Cosine Function:** For \( \cos(-\frac{\pi}{6}) \), cosine is an even function, meaning \( \cos(-x) = \cos(x) \), resulting in \( \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).Knowing these properties allows us to simplify expressions and make calculations accurate, aiding in the overall evaluation of the function.

Limit of Integration

The limits of integration, \( a \) and \( b \), define the interval over which we evaluate the definite integral of a function. In this exercise, \( a = -\frac{\pi}{6} \) and \( b = \frac{\pi}{4} \) are the endpoints for evaluating \( F(b) - F(a) \). They represent the points on the x-axis over which the function is considered. **Understanding Limits in Context:**- These points decide where to start and stop summing up or evaluating the area under the curve of the function \( F(x) \).- - Here, \( F(x) \) transforms inputs into outputs across this interval, and \( F(b) - F(a) \) gives the accumulated change in the function's values over \( [a, b] \).Mastering limits of integration is essential for understanding and applying definite integrals in various mathematical problems.