Problem 17 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$ f(t)=\sec (t) \tan (t) \quad a=-\pi / 4 \quad(-\pi / 2

Step-by-Step Solution

Answer

\( F(x) = \sec(x) - \sqrt{2} \).

Step 1: Understand the Function for Integration

The problem asks us to calculate \( F(x) = \int_{a}^{x} f(t) \ dt \) where \( f(t) = \sec(t) \tan(t) \). Given the integral interval from \( a = -\pi/4 \) to \( x \), we will evaluate this integral.

Step 2: Identify the Antiderivative

Recognize that the function \( f(t) = \sec(t) \tan(t) \) is the derivative of \( \sec(t) \). Therefore, the antiderivative of \( f(t) \) is \( \sec(t) \). This implies that the indefinite integral of \( \sec(t) \tan(t) \) is \( \sec(t) + C \), where \( C \) is a constant.

Step 3: Compute the Definite Integral

As an antiderivative is known (\( \sec(t) \)), apply the Fundamental Theorem of Calculus for the definite integral:\[ F(x) = \int_{-\pi/4}^{x} \sec(t) \tan(t) \ dt = \sec(x) - \sec(-\pi/4) \]

Step 4: Evaluate the Constant Term

Calculate \( \sec(-\pi/4) \). Since \( \sec(t) = 1/\cos(t) \), we have:\( \cos(-\pi/4) = \cos(\pi/4) = \sqrt{2}/2 \), therefore, \[ \sec(-\pi/4) = 2/\sqrt{2} = \sqrt{2} \]

Step 5: Write the Final Expression

We have previously calculated:\[ \sec(-\pi/4) = \sqrt{2} \] Therefore, the final expression for \( F(x) \) is:\[ F(x) = \sec(x) - \sqrt{2} \].

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a crucial pillar in understanding how differentiation and integration interact. It connects the concept of the derivative with the definite integral.

• The **First Part** of the theorem states that if you have a function that is continuous over an interval and you take its antiderivative, then integrating the derivative over that interval will give you the change in the antiderivative.
• The **Second Part** says if a function is integrable on the interval \([a, b]\), then the integral of the function from \(a\) to \(b\) can be found using its antiderivative. Mathematically, this is expressed as \(\int_{a}^{b} f(t) \ dt = F(b) - F(a)\), where \(F \) is any antiderivative of \(f\).

In our exercise, we applied this theorem to compute the definite integral. We started with the function \( f(t) = \sec(t) \tan(t) \), found its antiderivative, and used that to find \( F(x) = \int_{-\pi/4}^{x} \sec(t) \tan(t) \ dt = \sec(x) - \sec(-\pi/4) \). This shows the power of the Fundamental Theorem of Calculus in allowing us to find the area under a curve from just the antiderivative.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that are combinations of trigonometric functions. These types of integrals are often complex but also quite common in calculus.

• A **trigonometric integral** may involve basic functions like sine, cosine, tangent or secant, as well as more complex combinations of these functions.
• The goal is often to use identities or known derivatives to simplify or directly find these integrals.

In our particular problem, \( f(t) = \sec(t) \tan(t) \) is a trigonometric function. We knew from trigonometric identities and derivatives that the derivative of \( \sec(t) \) is exactly \( \sec(t) \tan(t) \). This direct relationship helped us identify the antiderivative quickly, which was essential in applying the Fundamental Theorem of Calculus.

Antiderivatives

Antiderivatives are functions that reverse the process of differentiation. Simply put, an antiderivative is a function whose derivative yields a specific function.

• Expressed generally, if \( F(t) \) is an antiderivative of \( f(t) \), then \( F'(t) = f(t) \).
• Antiderivatives are not unique; they are defined up to a constant \( C \), so the general form is \( F(t) + C \).

In the given problem, recognizing that \( f(t) = \sec(t) \tan(t) \) is the derivative of \( \sec(t) \) allowed us to quickly identify the antiderivative as \( \sec(t) + C \). This determination was vital to solving the integral from \(-\pi/4\) to \(x\). The antiderivative, coupled with the bounds of integration, helped us calculate the definite integral.