Problem 7 - Calculus Single Variable

Question

Use the first part of the Fundamental Theorem of Calculus together with the ideas of Examples 1 and 2 to evaluate the definite integrals in Exercises \(1-10\) $$ \int_{\pi / 2}^{\pi} \sec (x / 3) \tan (x / 3) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 6 - 2\sqrt{3} \).

Step 1: Understanding the Fundamental Theorem of Calculus

The first part of the Fundamental Theorem of Calculus states that if a function \( f \) is continuous over [a, b], and F is an antiderivative of f, then \( \int_a^b f(x) \ dx = F(b) - F(a) \). In this exercise, we aim to find an antiderivative of the function \( \sec(x/3) \tan(x/3) \).

Step 2: Simplifying the Integral

Recognize that \( \frac{d}{dx} \sec(u) = \sec(u)\tan(u) \) where \( u = x/3 \). This indicates that the integral of \( \sec(u)\tan(u) \) with respect to \( u \) is \( \sec(u) \). Thus, \( F(x) = \sec(x/3) \).

Step 3: Change of Variable

Setting \( u = \frac{x}{3} \) implies \( du = \frac{1}{3}dx \). Thus, \( dx = 3du \). When \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{6} \), and when \( x = \pi \), \( u = \frac{\pi}{3} \). The integral in terms of \( u \) becomes \( 3\int_{\pi/6}^{\pi/3} \sec(u)\tan(u) \ du \).

Step 4: Applying the Antiderivative

We know that \( \int \sec(u)\tan(u) \, du = \sec(u) \). Therefore, \( F(u) = \sec(u) \).

Step 5: Evaluating the Definite Integral

Substitute back to evaluate the definite integral: \[ 3[\sec(\pi/3) - \sec(\pi/6)] \].

Step 6: Calculating the Values

Calculate \( \sec(\pi/3) = 2 \) and \( \sec(\pi/6) = \frac{2\sqrt{3}}{3} \).Therefore, the result is \[ 3 \left(2 - \frac{2\sqrt{3}}{3}\right) = 6 - 2\sqrt{3} \].

Key Concepts

Definite Integrals

Definite integrals allow us to find the total accumulation of a quantity, which can be interpreted as the area under a curve within a specific range. Basically, when you perform a definite integral from point \(a\) to \(b\), you are calculating the net area between the function \(f(x)\) and the \(x\)-axis from \(x = a\) to \(x = b\). This is different from an indefinite integral, which produces a general form of the antiderivative plus a constant.

• The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \).
• It is evaluated as \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \).
• The limits of integration, \( a \) and \( b \), are key to determining specific values rather than a general function.

The Fundamental Theorem of Calculus, which connects differentiation and integration, makes it straightforward to evaluate these integrals once you find an antiderivative of the function.

Antiderivative

An antiderivative of a function is simply a function whose derivative gives back the original function. It is an essential concept in integration.

• The process of finding an antiderivative is known as integration.
• For the function \( \sec(u) \tan(u) \), the antiderivative is \( \sec(u) \).
• This means when you differentiate \( \sec(u) \), you obtain \( \sec(u) \tan(u) \), confirming it’s the correct antiderivative.

In problems involving definite integrals, having the correct antiderivative allows us to use the Fundamental Theorem of Calculus efficiently. We plug the limits of integration into the antiderivative to find the solution.

Change of Variable

The change of variable technique is a powerful tool that facilitates simplification of integral expressions. This method is also known as "substitution" because you replace variables to simplify the integration process.

• You set a part of the expression, often a complicated term, as a new variable, typically \( u \).
• For example, with \( u = \frac{x}{3} \), it simplifies the integral with respect to \( u \) rather than \( x \).
• Don't forget to adjust the differential (\( dx = 3du \)), and re-calculate the limits of integration based on the new variable.

The change of variable is especially useful in functions involving composite expressions or trigonometric identities, making them easier to integrate.

Trigonometric Functions

Trigonometric functions like \( \sec(x) \) and \( \tan(x) \) are common in calculus problems, often appearing in integrals and derivatives.

• \( \sec(x) = \frac{1}{\cos(x)} \) is the reciprocal of \( \cos(x) \).
• \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), the ratio of \( \sin(x) \) to \( \cos(x) \).
• These functions are periodic and have specific values at standard angles, like \( \pi/3 \) and \( \pi/6 \).

Working with trigonometric functions often requires understanding their properties and identities, as they are crucial in simplifying expressions and evaluating integrals effectively.