Problem 11 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle F(x) = \int^0_x \sqrt{1 + \sec t} \,dt \) $$ \biggl[ \textit{Hint:} \int^0_x \sqrt{1 + \sec t} \,dt = - \int^x_0 \sqrt{1 + \sec t} \,dt \biggr] $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) = -\sqrt{1 + \sec x} \).

Step 1: Understand the Function and the Hint

The function given is a definite integral: \[ F(x) = \int^0_x \sqrt{1 + \sec t} \,dt. \]Using the hint provided, we rewrite this integral:\[ F(x) = - \int^x_0 \sqrt{1 + \sec t} \,dt. \]This transformation is possible because reversing the limits of integration introduces a negative sign.

Step 2: Apply Part 1 of the Fundamental Theorem of Calculus

Part 1 of the Fundamental Theorem of Calculus states that if \[ G(x) = \int_a^x g(t) \, dt, \]then the derivative \[ G'(x) = g(x). \]In our case, \[ G(x) = \int_0^x \sqrt{1 + \sec t} \, dt \]is the integral from 0 to x. Therefore, the derivative is \[ G'(x) = \sqrt{1 + \sec x}. \]

Step 3: Adjust the Sign due to the Negative Integral

Due to the negative sign introduced in Step 1, we need to adjust the derivative. Thus, the derivative of \[ F(x) = - \int^x_0 \sqrt{1 + \sec t} \,dt \]is\[ F'(x) = -\sqrt{1 + \sec x}. \]

Key Concepts

Derivative Calculation

Calculating the derivative of a function involves finding its rate of change. When faced with finding the derivative of a function defined by an integral, the Fundamental Theorem of Calculus becomes a powerful tool.

• The theorem states that if you have an integral of a function from a constant to a variable, then the derivative of the integral is essentially the original function evaluated at the variable.
• This simplification allows us to convert the integral-based function back into a simple algebraic expression for differentiation.

For example, if we have a function defined by an integral like \[ F(x) = \int_a^x g(t) \,dt, \]the derivative would naturally be \[ F'(x) = g(x). \]This transformation bypasses a lot of intermediate calculus steps, making it much simpler to process the integral function directly.

Definite Integral

A definite integral is a fundamental concept in calculus that represents the area under a curve within specified bounds. This area gives us valuable information about the cumulative value of a function over an interval.

• The integral of a function between two points, say \( a \) and \( b \), is denoted as \( \int_a^b f(x) \, dx \).
• The actual value computed from a definite integral is a number, representing this total area or accumulation, rather than another function or expression.
• In the context of our exercise, understanding the role of definite integrals helps visualize how changing the bounds or limits affects the result.

This definite integral changes its sign when its limits are reversed, highlighting the orientation of the area calculation as integral limits dictate the direction from which we integrate.

Integration Limits

Integration limits are boundaries that define the range over which an integration operation is performed. These limits crucially impact the result and interpretation of a definite integral.

• When given an integral like \( \int_a^b f(x) \, dx \), the limits \( a \) and \( b \) set the start and end for our collection of area (or accumulated value).
• Reversing these limits, changing an integral from \( \int_a^b \) to \( \int_b^a \), introduces a negative sign. This sign change occurs because the orientation of the integral and the direction of accumulation are switched.
• Our textbook problem highlights this point by reversing the integration limits and showing how it affects the function's derivative.

Understanding how to manipulate these limits, especially in light of derivative calculations, empowers you to solve more complex calculus problems with accuracy.

Secant Function

The secant function, denoted as \( \sec(x) \), is one of the six fundamental trigonometric functions. Representing it as the reciprocal of the cosine function, \( \sec(x) = \frac{1}{\cos(x)} \), shows how it fits into trigonometric relationships.

• This function is essential in calculus, especially in solving integrals and derivatives involving angle measures.
• Its unique properties and behavior, particularly its undefined values where \( \cos(x) = 0 \), present both challenges and opportunities in calculus problems.
• In our given exercise, the expression \( \sqrt{1 + \sec t} \) illustrates how secant contributes to more complex integrals, which can be tricksome but rewarding to resolve.

The secant function offers a useful perspective on how trigonometric identities interplay with calculus operations, particularly when evaluating integrals involving variable limits.