Problem 20 - Calculus for Biology and Medicine

Question

Find the general antiderivative of the given function. $$ f(x)=e^{-x / 2}-e^{-2 x} $$

Step-by-Step Solution

Answer

The general antiderivative is \( -2 e^{-x/2} - \frac{1}{2} e^{-2x} + C \).

Step 1: Identify the Function Components

The function given is \( f(x) = e^{-x/2} - e^{-2x} \). It is composed of two separate exponential functions: \( e^{-x/2} \) and \( e^{-2x} \). To find the general antiderivative, we need to find the antiderivative of each component individually.

Step 2: Antiderivative of the First Component

We start by finding the antiderivative of the first component, \( e^{-x/2} \). The antiderivative of \( e^{ax} \) is \( \frac{1}{a} e^{ax} \). Here, \( a = -1/2 \), so the antiderivative is \( \frac{1}{-1/2} e^{-x/2} = -2 e^{-x/2} \).

Step 3: Antiderivative of the Second Component

Next, we find the antiderivative of the second component, \( e^{-2x} \). Using the same rule, here \( a = -2 \), thus the antiderivative is \( \frac{1}{-2} e^{-2x} = -\frac{1}{2} e^{-2x} \).

Step 4: Combine Results and Add Constant of Integration

Now, we combine the antiderivatives from Steps 2 and 3 to find the general antiderivative of the whole function. This gives \( -2 e^{-x/2} - \frac{1}{2} e^{-2x} + C \), where \( C \) is the constant of integration.

Key Concepts

Exponential Functions

Exponential functions are a key concept in calculus and mathematical analysis. These functions are defined by expressions of the form \(f(x) = e^{ax}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828, and \(a\) is an exponent that can greatly alter the behavior of the function.
Exponential functions can either grow or decay extremely fast depending on the sign of the exponent \(a\):

• If \(a > 0\), the function \(f(x)\) is increasing, representing exponential growth.
• If \(a < 0\), the function \(f(x)\) is decreasing, representing exponential decay.

In the exercise given, the exponential functions \(e^{-x/2}\) and \(e^{-2x}\) represent decay because their exponents are negative. Understanding these functions is crucial because they frequently appear in many fields, like biology for population models, finance for interest calculations, and physics for radioactive decay.

Constant of Integration

When finding an antiderivative, it is essential to consider the constant of integration, often denoted as \(C\). The reason for this constant stems from the idea that the process of taking a derivative removes any information about constant values added to the original function.
For instance, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative of \( f(x) \) for any constant \( C \). The constant of integration is:

• Acknowledging the infinite set of functions that could be antiderivatives, all differing only by a constant.
• Vital for representing all possible solutions to an indefinite integral.

In the solved example, after finding the antiderivatives of \(e^{-x/2}\) and \(e^{-2x}\), the combined result is \(-2e^{-x/2} - \frac{1}{2}e^{-2x} + C\), where \(C\) accounts for any constant term in the antiderivative.

Calculus

Calculus is the branch of mathematics that studies change and motion. It consists of two main parts: differential calculus and integral calculus. Differential calculus focuses on finding the rate at which quantities change, which is quantified as the derivative.
Integral calculus, on the other hand, concerns itself with the concept of the antiderivative. It answers the question: what was the original function if given its rate of change? This process is known as integration. There are two types of integrals:

• Definite Integrals: Represent the accumulated change with definite boundaries, giving a numerical result.
• Indefinite Integrals: Finding the antiderivative of a function without specific limits, expressed with a constant of integration \(C\).

The exercise provided is an example of an indefinite integral, where the antiderivative of the function \( f(x) = e^{-x/2} - e^{-2x} \) was determined to be \(-2e^{-x/2} - \frac{1}{2}e^{-2x} + C\). This process not only highlights the power of calculus but also its utility in finding a function given its rate of change.