Problem 92 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int 4^{-x} d x $$

Step-by-Step Solution

Answer

\(-\frac{4^{-x}}{\ln(4)} + C\)

Step 1: Identify the Integral Form

The given integral is \( \int 4^{-x} \, dx \). This is in the form of \( \int a^{u} \, du \), where \( a = 4 \) and \( u = -x \).

Step 2: Apply Substitution

Let \( u = -x \), which gives \( du = -dx \) or \( dx = -du \). Substitute these into the integral to get \( \int 4^{u} (-du) \).

Step 3: Simplify the Expression

Rewrite the integral as \( -\int 4^{u} \, du \). This expression is now ready for integration using the formula for exponential functions.

Step 4: Integrate Using the Exponential Formula

The integral of \( a^{u} \) with respect to \( u \) is \( \frac{a^{u}}{\ln(a)} \). Thus, we have \( -\int 4^{u} \, du = -\frac{4^{u}}{\ln(4)} + C \).

Step 5: Substitute Back for x

Substitute \( u = -x \) back into the integrated result to get \( -\frac{4^{-x}}{\ln(4)} + C \) as the final result.

Key Concepts

Substitution Method

The substitution method is a powerful tool for solving integrals. It simplifies the integration process by transforming the original variable into a new one. This method is particularly useful when dealing with functions that involve composite operations.

In our exercise, the expression to integrate is \( \int 4^{-x} \, dx \). Recognizing that \(-x\) complicates integration, we make a substitution. Let \( u = -x \). This gives us \( du = -dx \).

What does this change mean? By substituting, we're effectively altering the differential \( dx \) to \( -du \). This will deliver a new integral: \( \int 4^{u} (-du) \). Notice how the integral now appears more straightforward and ready for integration.

One essential point to keep in mind with any substitution is to remember to revert your substituted variable back to the original variable at the end, ensuring the answer relates to the initial variable of integration.

Exponential Integration

Exponential integration centers on integrating functions of the form \( a^{u} \), where \( a \) is a constant, and \( u \) is a variable. This technique requires specific formulas, especially because exponential functions grow or decay so rapidly.

In this context, once we substitute \( u \) for \(-x\), the integral \( \int 4^{u} (-du) \) surfaces. The negative sign in front of \( du \) simply pulls out of the integral to give \(-\int 4^{u} \ du\).

Now, employ the formula \( \int a^{u} \, du = \frac{a^{u}}{\ln(a)} \). For our problem, it becomes \( -\frac{4^{u}}{\ln(4)} \). This result captures the integration of the constant base raised to the power of \( u \). The ln function here helps adjust for the unique growth rate of exponential functions.

Integration Techniques

Integration techniques are diverse and picking the right one depends on the form of the function. In this example of \( \int 4^{-x} \, dx \), both substitution and exponential integration were crucial.

Each integration problem might need different strategies:

• Substitution: Useful when a variable change simplifies the integral.
• Integration by parts: Ideal for products of functions with different integration ease.
• Partial fraction decomposition: Applies to rational functions to break them into simpler parts.

Utilizing these techniques properly can turn even the most daunting integrals into manageable tasks. Practicing variety is key to mastering when and how to employ each method effectively.