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 integral is \( \int (x^{-3} + 3^{-x}) \, dx = -\frac{1}{2x^2} - \frac{3^{-x}}{\ln(3)} + C \).

Step 1: Identify the Integral Expression

The integral we need to compute is given by:\[ \int (x^{-3} + 3^{-x}) \, dx \] This expression consists of two functions that need to be integrated separately: \( x^{-3} \) and \( 3^{-x} \). You can use the properties of integrals to address them one at a time.

Step 2: Integrate \( x^{-3} \)

To integrate \( x^{-3} \), recall the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).Apply the power rule:\[ \int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C_1 = \frac{x^{-2}}{-2} + C_1 = -\frac{1}{2x^2} + C_1 \]

Step 3: Integrate \( 3^{-x} \)

For \( 3^{-x} \), recognize it as an exponential function and use the formula \( \int a^x \, dx = \frac{a^x}{\ln(a)} + C \), where \( a = 3 \).Therefore:\[ \int 3^{-x} \, dx = \int (\frac{1}{3})^x \, dx = -\frac{3^{-x}}{\ln(3)} + C_2 \] This uses the substitution \( a^{-x} = (\frac{1}{a})^x \) and accounts for the negative exponent.

Step 4: Combine the Indefinite Integrals

Combine the results from Steps 2 and 3 to finalize the integral of the entire function:\[ \int (x^{-3} + 3^{-x}) \, dx = -\frac{1}{2x^2} - \frac{3^{-x}}{\ln(3)} + C \] Here, \( C = C_1 + C_2 \), the arbitrary constant of integration.

Key Concepts

Power Rule for Integration

The power rule for integration is one of the most fundamental tools for solving integrals. It helps us integrate functions of the form \( x^n \), where \( n \) is any real number except \(-1\). The formula is given by:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).

This rule allows us to find the antiderivative of any polynomial term by simply increasing the exponent by one and dividing by the new exponent.
For instance, to integrate \( x^{-3} \), you would:

• Add 1 to the exponent: \(-3 + 1 = -2\).
• Divide by the new exponent: \(\frac{x^{-2}}{-2}\).

Thus, the integral of \( x^{-3} \) becomes \(-\frac{1}{2x^2} + C\).
It's important to note that this rule does not apply when \( n = -1 \); for \( n = -1 \), the integral becomes \( \ln |x| + C \), because of the nature of logarithmic functions.

Exponential Functions

Exponential functions have a unique behavior due to their base being a constant raised to variable exponents. The general exponential function can be written as \( a^x \), where \( a \) is a positive constant. When integrating such functions, we use a special rule:

• \( \int a^x \, dx = \frac{a^x}{\ln(a)} + C \).

This rule helps integrate any exponential function efficiently by dividing the exponential by the natural log of its base.
A common case is integrating expressions like \( 3^{-x} \). Notice, this is equivalent to integrating \( (\frac{1}{3})^x \). By applying the exponential function rule:

• Express \( 3^{-x} \) as \( (\frac{1}{3})^x \).
• Use the formula: \( \int (\frac{1}{3})^x \, dx = -\frac{3^{-x}}{\ln(3)} + C \).

The negative sign arises since we handle it as \( (\frac{1}{3})^x \) and adjust for the negative exponent during integration.
This approach is fundamental when managing exponential terms where the exponent includes a variable part.

Properties of Integrals

The properties of integrals provide essential techniques for solving complex integrals by breaking them into simpler parts. Some key properties are:

• Linearity: \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \).
• Constant Multipliers: \( \int c\,f(x) \, dx = c \int f(x) \, dx \), where \( c \) is a constant.

These properties allow us to decompose integrals into manageable segments, simplifying the computation process significantly.
In the exercise of integrating \( x^{-3} + 3^{-x} \), we apply the linearity of integrals to split the problem:

• Integrate \( x^{-3} \) and \( 3^{-x} \) separately.
• Combine the results to obtain: \( -\frac{1}{2x^2} - \frac{3^{-x}}{\ln(3)} + C \).

This method leverages the integral properties to make the integration process straightforward. Remember, the constant \( C \) represents the general solution, accommodating all possible antiderivative forms.