Problem 31 - Calculus Single Variable

Question

Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{0}^{\pi / 3} \sec (x) \tan (x) d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to 1.

Step 1: Identify the Antiderivative

First, we recognize that the integral involves the function \( \sec(x) \tan(x) \), and we need to find the antiderivative. Notice that the derivative of \( \sec(x) \) is \( \sec(x) \tan(x) \). Therefore, the antiderivative of \( \sec(x) \tan(x) \) is \( \sec(x) \).

Step 2: Apply the Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \), then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]Here, \( F(x) = \sec(x) \). Thus, we evaluate \( F(x) \) from 0 to \( \pi/3 \).

Step 3: Evaluate the Antiderivative at the Bounds

Calculate \( F(\pi/3) \) and \( F(0) \):- \( F(\pi/3) = \sec(\pi/3) = \frac{1}{\cos(\pi/3)} = 2 \).- \( F(0) = \sec(0) = \frac{1}{\cos(0)} = 1 \).

Step 4: Compute the Definite Integral

Subtract the values obtained:\[ \int_{0}^{\pi/3} \sec(x) \tan(x) \, dx = F(\pi/3) - F(0) = 2 - 1 = 1 \].

Key Concepts

Antiderivatives

Antiderivatives are an integral concept in calculus that can help us solve definite integrals. They are functions whose derivative is the given function. To understand this, imagine a road and the antiderivative as a traveler retracing their steps from a known path back to its origin. For any function such as \( \sec(x) \tan(x) \), finding its antiderivative means discovering a function that would have \( \sec(x) \tan(x) \) as its derivative.

In our given problem, we identified that the derivative of \( \sec(x) \) is \( \sec(x) \tan(x) \). Thus, the antiderivative of \( \sec(x) \tan(x) \) is \( \sec(x) \). This reverse process of differentiation is essential since it allows us to handle integral calculations, which is crucial for solving areas under curves or solving for accumulated values.

Through understanding antiderivatives, you gain an essential tool for mastering calculus. You learn to see the bigger picture and connect the dots between derivatives and the integrals they relate to.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between derivatives and integrals, acting as the foundation for integral calculus. This theorem is divided into two parts; the first part guarantees the existence of an antiderivative for continuous functions, while the second part provides a powerful method for evaluating definite integrals.

In the exercise, we used the second part of the theorem, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral from \( a \) to \( b \) is given by \( F(b) - F(a) \).

• This means you evaluate the antiderivative at the upper limit, \( b \), and subtract the evaluation at the lower limit, \( a \).
• For our problem, this step translated to calculating \( \sec(\pi/3) \) and \( \sec(0) \).

Overall, the Fundamental Theorem allows us to compute the definite integral efficiently without needing to find areas by summation. It saves time and effort while affording precision in evaluating integral bounds.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are periodic functions frequently encountered in calculus. They have a vast array of applications, from calculating angles to modeling periodic behaviors.

In the context of integrals, understanding these functions is vital. The specific function we dealt with, \( \sec(x) \tan(x) \), consists of the secant \( \sec(x) \) and tangent \( \tan(x) \):

• Secant, \( \sec(x) \), is the reciprocal of cosine, \( \frac{1}{\cos(x)} \).
• Tangent, \( \tan(x) \), is sine divided by cosine, or \( \frac{\sin(x)}{\cos(x)} \).

In the exercise, understanding these relationships is crucial for recognizing that \( \sec(x) \tan(x) \) is the derivative of \( \sec(x) \). This realization was pivotal in applying the antiderivative concept effectively. By deepening your knowledge of trigonometric functions, you enhance your calculus skills, handling more complex problems with confidence.