Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(G^{\prime}(x)\). $$ G(x)=\int_{x}^{1} 2 t d t $$

Step-by-Step Solution

Answer

The derivative is \( G'(x) = -2x \).

Step 1: Recognize the Type of Problem

We are asked to find the derivative of a function defined as an integral with a variable limit of integration. This type of problem is solved using the Fundamental Theorem of Calculus, Part 1.

Step 2: Identify the Fundamental Theorem of Calculus, Part 1

The Fundamental Theorem of Calculus, Part 1, states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). However, if the upper limit is a constant and the lower limit is \(x\), the derivative involves a negative sign: \( F'(x) = -f(x) \).

Step 3: Apply the Fundamental Theorem of Calculus

Our function is \( G(x) = \int_{x}^{1} 2t \, dt \). To differentiate, apply the theorem: the derivative will be \( G'(x) = -f(x) \), where \( f(t) = 2t \).

Step 4: Determine the Value of f(x)

Since \( f(t) = 2t \), we substitute \( t = x \) to find \( f(x) = 2x \).

Step 5: Compute G'(x)

Using Step 2 and Step 4, substitute \( f(x) \) into the formula from Step 3: \( G'(x) = -2x \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. When we differentiate a function, we are essentially determining how it changes at any given point. The derivative provides us with the slope of the function at a particular point, thereby helping us understand how the function behaves.

• The derivative of a function is represented as \( f'(x) \) or \( \frac{df}{dx} \).
• It tells us the rate at which \( f \) is changing with respect to \( x \).
• Differentiation is used extensively in various fields such as physics, economics, and engineering to model and analyze real-world situations.

Understanding how to differentiate functions, especially when they are expressed in integral form, is crucial to mastering calculus. In cases where we have functions defined as integrals with variable limits, we use specific theorems to differentiate correctly.

Variable Limits of Integration

When dealing with integrals, the limits of integration define the interval over which the function is being integrated. Typically, these are fixed numbers; however, when the limits are variable, it introduces an additional layer of consideration, especially when finding derivatives.

• In the expression \( \int_{x}^{1} 2t \, dt \), \(x\) is a variable limit, implying the integration range is dependent on the variable \(x\).
• This can make problems more complex because the computation of integrals with variable limits involves the function changing as \(x\) changes.
• When differentiating such integrals, the Fundamental Theorem of Calculus helps bridge the gap between integration and differentiation even with variable limits.
• Understanding how to handle variable limits of integration is key in solving problems involving rates of change and accumulation in dynamic situations.

Derivative of an Integral

Finding the derivative of an integral, especially when the integral involves variable limits, requires us to use a specific part of the Fundamental Theorem of Calculus.

Application of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus acts as a link between differentiation and integration. Particularly, it provides a way to differentiate an integral calculated with variable limits. This process involves reversing the action of integration by finding the derivative.

• This theorem states that if you have an integral \( G(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( G'(x) = f(x) \).
• In the exercise, the function \( G(x) = \int_{x}^{1} 2t \, dt \) has its variable in the lower limit, and the result involves a negative sign due to this arrangement, leading to \( G'(x) = -f(x) \).
• To find \( G'(x) \), substitute \( t = x \) in \( f(t) \) to get \( f(x) = 2x \), resulting in \( G'(x) = -2x \).

This is how we leverage both the integration principles and differentiation rules to evaluate the derivative effectively, providing a strong training ground for further applications in calculus.