Problem 91 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int 3^{-2 x} d x $$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Form

The integral to compute is given by \( \int 3^{-2x} \, dx \). Recognize that this is an exponential function with a base of 3. The form we have is of an exponential function where the variable exponent is linear, \( -2x \).

Step 2: Use the Integration Formula for Exponentials

The formula for integrating an exponential function \( a^{bx} \) is \( \frac{a^{bx}}{b \ln a} + C \). In this situation, \( a = 3 \) and \( b = -2 \).

Step 3: Apply the Formula and Simplify

Plug the values into the formula: \( \int 3^{-2x} \, dx = \frac{3^{-2x}}{-2 \ln 3} + C \). Simplify the expression to achieve the final integral result.

Key Concepts

Exponential Functions

Exponential functions are an important class of mathematical functions that appear frequently in many areas such as biology, economics, and physics. These functions have the form \( f(x) = a^{bx} \), where \( a \) is the base and \( b \) is the exponent multiplier. They are characterized by their constant growth rate, and properties such as multiplicative increase and continuous compounding. Exponential functions can represent rapid growth or decay, depending on whether the exponent multiplier \( b \) is positive or negative.- For \( b > 0 \), the function grows as \( x \) increases.- For \( b < 0 \), the function decreases as \( x \) increases.Understanding this behavior helps us handle exponential functions in integration, as seen in the example \( 3^{-2x} \) where the function represents an exponential decay due to the negative exponent.

Integration Formula

In calculus, integrating exponential functions involves recognizing specific patterns and applying standard formulas. The key formula used for integrating an exponential function of the form \( a^{bx} \) is:\[ \int a^{bx} \, dx = \frac{a^{bx}}{b \ln a} + C \]Here, \( a \) is the base of the exponential, \( b \) is part of the exponent, and \( \ln \) is the natural logarithm. The formula helps us directly find the indefinite integral, which in simple terms is the collection of all antiderivatives of the function.It's important to understand each component:- **\( a^{bx} \)**: The original function within the integral.- **\( b \ln a \)**: Forms the denominator due to the integration process.- **\( + C \)**: The constant of integration, which accounts for all possible antiderivatives.

Variable Exponent

A variable exponent in exponential functions means that the exponent contains the variable we are interested in, such as \(-2x\) in \(3^{-2x}\). The presence of the variable exponent introduces variable rates of change, which makes integrating these functions different from polynomial functions.- **In our example:** - The variable exponent is \(-2x\), indicating a decay (since it's negative).When the exponent is variable, it significantly affects how the graph of the function behaves. Moreover, handling these expressions during integration requires special techniques, like using the aforementioned integration formula for exponentials.- We identify \(-2x\) as the exponent affecting the rate of decay.- The formula balances this variability with the log of the base and the multiplier \(-2\) in this case.

Base of Exponential

The base in an exponential function, such as the 3 in \(3^{-2x}\), is a crucial element of the function's behavior and outcome after integration. It determines the growth factor of the function.- **Significance of Base: 3** - While integrating, knowing the base allows us to use the integration formula. - It influences the scale of the graph of the function.In our example, the base 3 indicates that if it were growth (had the exponent been positive), each increment in \( x \) would multiply the outcome by 3. However, with the negative exponent in \(3^{-2x}\), it refers to a decay process rather than growth.- During integration, divide by \( \ln 3 \) as part of the formula \( \int a^{bx} \, dx = \frac{a^{bx}}{b \ln a} + C \).- Recognizing the base and its logarithm is essential to correctly applying the formula.