Problem 94 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int\left(x^{-3}+3^{-x}\right) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(-\frac{1}{2}x^{-2} - \frac{1}{\ln(3)}3^{-x} + C\).

Step 1: Break Down the Integral

First, we identify the individual terms in the integrand. The expression can be written as two separate integrals: \[\int x^{-3} \, dx + \int 3^{-x} \, dx.\] Each term can be integrated independently.

Step 2: Integrate the First Term

The first term is \(x^{-3}\). To integrate \(x^{-n}\), we use the formula:\[\int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} + C_1,\] where \(n eq 1\). For \(n = 3\), this becomes:\[\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2}x^{-2} + C_1.\]

Step 3: Integrate the Second Term

The second term is \(3^{-x}\). We use the formula for integrating \(a^{-x}\), which is:\[\int a^{-x} \, dx = -\frac{1}{\ln(a)} a^{-x} + C_2.\]Here, \(a = 3\) and thus the integral becomes:\[\int 3^{-x} \, dx = -\frac{1}{\ln(3)} 3^{-x} + C_2.\]

Step 4: Combine Results

Now, combine the integrals of the two separate terms to form the indefinite integral of the entire function:\[\int \left(x^{-3} + 3^{-x}\right) \, dx = -\frac{1}{2}x^{-2} - \frac{1}{\ln(3)} 3^{-x} + C.\]Here, \(C = C_1 + C_2\) represents the constant of integration for the entire function.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to integrate products of functions. However, in the exercise given, this method is not directly applied because the integral involves functions that can be more easily integrated by other methods.
However, understanding integration by parts will help you in future exercises where you do encounter products of more complicated functions.
This technique is based on the product rule for differentiation and is particularly useful when dealing with products of polynomial and exponential functions.

• The formula of integration by parts is: \( \int u \, dv = uv - \int v \, du \)
• Here, \( u \) and \( dv \) are parts of the integrand, chosen in such a way that \( du \) and \( v \) can be easily determined.

While the term \( x^{-3} \) in our exercise might initially seem fitting for integration by parts, it's actually more straightforward to use the power rule. This emphasizes the importance of recognizing when to use integration by parts versus other techniques, highlighting the foresight involved in choosing the most effective integration method.

Exponential Decay

Exponential decay refers to situations where quantities decrease at a rate proportional to their current value. In integration, an expression like \( 3^{-x} \) is common and represents an exponential decay.
The integral of such functions involves recognizing the base and the negative exponent.
This is different from regular exponential growth because of the negative sign in the exponent, which flips the growth into decay.

• For integrals involving bases other than \( e \) (like our base \( 3 \)), we adjust using the natural logarithm to find the antiderivative.
• Here, the formula becomes \( \int a^{-x} \, dx = -\frac{1}{\ln(a)} a^{-x} + C \).

The understanding of exponential decay is essential as it's ubiquitous in real-world applications like radioactive decay, cooling processes, and depreciation in value.

Power Rule for Integration

The power rule for integration is one of the most fundamental and simplest techniques. It simplifies the process of integrating functions of the form \( x^n \).
The rule states that for any real number \( n eq -1 \), the integral of \( x^n \) is:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

• It's crucial to remember that this rule does not apply when \( n = -1 \), as this case leads to the natural logarithm function.
• In this exercise, we applied the power rule to integrate \( x^{-3} \), where \( n = -3 \).

The power rule allows us to quickly and accurately find antiderivatives of many functions, making it a go-to technique for initial integration steps. By applying it correctly, we can streamline solving complex integrals and make the process more efficient.