Problem 11 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x} .\) $$ y=\int_{-2}^{x} \frac{1}{u+3} d u, x>-2 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{x+3} \) for \( x > -2 \).

Step 1: Understand the Problem

We are asked to find the derivative \( \frac{dy}{dx} \) of the function given as an integral with variables in its limits.

Step 2: Identify Applicable Theorem

Since \( y \) is defined as an integral with a variable upper limit, we'll apply the Fundamental Theorem of Calculus, Part 1, which states that if \( y = \int_{a}^{x} f(u) \, du \), then \( \frac{dy}{dx} = f(x) \).

Step 3: Differentiate Using the Fundamental Theorem

Apply the theorem to differentiate the given integral. Here, \( y = \int_{-2}^{x} \frac{1}{u+3} \ du \). Therefore, by the Fundamental Theorem, \( \frac{dy}{dx} = \frac{1}{x+3} \).

Step 4: Check the Domain

Verify if the domain condition is satisfied. The domain given is \( x > -2 \), which aligns with the derivative \( \frac{1}{x+3} \), as the function is defined in this domain without any discontinuities.

Key Concepts

Derivative of an Integral

When dealing with integrals in calculus, it's common to encounter situations where you need to find the derivative of an integral. This is where the Fundamental Theorem of Calculus comes into play, specifically its first part. This powerful theorem links the concept of differentiation and integration, which are often seen as inverse operations. Here’s the concept broken down:- If you have a function defined as an integral with one of the limits being a variable, the derivative with respect to that upper variable limit is simply the integrand evaluated at that limit. This is given by: \[ y = \int_{a}^{x} f(u) \, du \Rightarrow \frac{dy}{dx} = f(x) \] - In our exercise, where we have \[ y = \int_{-2}^{x} \frac{1}{u+3} \, du, \] using the theorem gives us: \[ \frac{dy}{dx} = \frac{1}{x+3} \] This means that if you know how to differentiate regular expressions, differentiating an integral becomes straightforward once you understand the theorem.

Variable Limits of Integration

In calculus, integrals with variable limits arise frequently in applications where a quantity depends on an upper limit that can change. These limits usually involve a variable, like the upper limit of our integral represented by \(x\) in the problem.Here’s how it works:

• The variable limit of integration means that as your upper limit changes, it can dynamically affect the outcome of your integral expression.
• In the exercise, we have the integral from \(-2\) to \(x\) of \(\frac{1}{u+3} \, du\), meaning the result of the integral is a function of \(x\) rather than a constant.

Upon differentiating such an integral, the variable limit becomes the main focus because it determines the form of the derivative. This is a neat way calculus allows handling dynamic systems where variables are continuously adjusted.

Calculus for Biology and Medicine

In fields like biology and medicine, understanding calculus concepts like the derivative of an integral is crucial. Here's why: - Calculus helps model and interpret biological systems, where biological quantities often change with respect to one another. - Consider a situation in biology where you need to model how a substance diffuses in a cell over time. The rate of change in concentration is often expressed as a derivative of an integral function. Such mathematical models in calculus allow scientists and medical professionals to predict complex biological processes and their interactions with greater accuracy. Understanding variable limits of integration and their derivatives provides the tools to represent how biological processes evolve over time, making calculus indispensable in fields beyond mathematics itself.