Problem 14 - Calculus

Question

If \(f(x)=\int_{-2}^{x} \frac{1}{t+3} d t,-2 \leq x,\) find \(f^{\prime}(7)\).

Step-by-Step Solution

Answer

The derivative \( f'(7) = \frac{1}{10} \).

Step 1: Understanding the Problem

We are given a function defined by an integral, \( f(x) = \int_{-2}^{x} \frac{1}{t+3} \, dt \), and we need to find the derivative \( f'(x) \) at \( x = 7 \).

Step 2: Using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a}^{x} g(t) \, dt \), then \( F'(x) = g(x) \). Here, \( g(t) = \frac{1}{t+3} \), so \( f'(x) = \frac{1}{x+3} \).

Step 3: Evaluating the Derivative at a Specific Point

Now that we have \( f'(x) = \frac{1}{x+3} \), we substitute \( x = 7 \) to find \( f'(7) \).

Step 4: Calculating \( f'(7) \)

Substitute \( x = 7 \) into \( f'(x) = \frac{1}{x+3} \) to get \( f'(7) = \frac{1}{7+3} = \frac{1}{10} \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities. It allows us to compute areas under curves, volumes, and other quantities that accumulate continuously. In this problem, we have a function defined by an integral: \( f(x) = \int_{-2}^{x} \frac{1}{t+3} \, dt \). This integral represents an accumulation function, which is essentially a sum of the values of \( \frac{1}{t+3} \) from \( t = -2 \) to \( t = x \).
The fundamental concept is that integral calculus can relate an accumulation function to its original function, providing a powerful tool for analyzing changes over an interval. The Fundamental Theorem of Calculus, which connects differentiation and integration, is pivotal here. It states that the process of differentiation and integration are inverses of each other, allowing us to evaluate derivatives of integral-based functions easily.
This makes integral calculus not just about accumulation, but a tool for evaluating rates of change in dynamic systems, which is why it's crucial for understanding the world that involves continuous processes.

Derivative

A derivative represents the rate of change of a function concerning its variable. It is often interpreted as the slope of a curve at any given point. In this exercise, we want the derivative of an integral function, \( f(x) \), which is \( f'(x) \). The Fundamental Theorem of Calculus is utilized here, as it tells us how to find this derivative simply.
To put it simply, the theorem provides a formula: if \( F(x) = \int_{a}^{x} g(t) \, dt \), then the derivative \( F'(x) \) is just \( g(x) \). This simplifies the entire process, as it bypasses direct computation of limits. By using this theorem, we determine that \( f'(x) = \frac{1}{x+3} \).
This indicates that at every point \( x \), the rate of change of our function \( f(x) \) is \( \frac{1}{x+3} \), offering a straightforward way to find the change without recalculating from scratch. Derivatives are essential in various fields like physics, engineering, and economics because they help us understand how a small change in one variable affects another.

Evaluation of Functions

Evaluation of functions is the process of determining the output of a function for a given input. In this exercise, we are asked to find the derivative at a specific point, \( x = 7 \). First, we found the general derivative \( f'(x) = \frac{1}{x+3} \), and to evaluate this at \( x=7 \), we simply substitute 7 into the derivative:

• Calculate \( f'(7) = \frac{1}{7+3} \)
• Simplify to get \( f'(7) = \frac{1}{10} \)

Function evaluation is fundamental because it allows us to find tangible results that can be interpreted or used for further analysis. In simpler terms, it's like testing a formula with specific numbers to see what results it produces. This step is crucial in problems dealing with real-world applications where precise values are needed.
By evaluating functions at particular points, we gain insights into their behavior over specific domains, enabling better predictive models in sciences and enhancing design and control in engineering scenarios.