Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int z a^{z} d z $$

Step-by-Step Solution

Answer

\( \int z a^z \, dz = \frac{z a^z}{\ln a} - \frac{a^z}{(\ln a)^2} + C \)

Step 1: Identify Parts for Integration by Parts

Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \). Let \( u = z \) and \( dv = a^{z} dz \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u = z \) to get \( du = dz \). Integrate \( dv = a^z dz \) to find \( v \). The integration of \( a^z \) with respect to \( z \) is \( \frac{a^z}{\ln a} \), so \( v = \frac{a^z}{\ln a} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula. We have:\[ \int z a^z \, dz = z \cdot \frac{a^z}{\ln a} - \int \frac{a^z}{\ln a} \, dz \]

Step 4: Simplify and Solve Remaining Integral

The remaining integral is \( \int \frac{a^z}{\ln a} \, dz = \frac{1}{\ln a} \int a^z \, dz \). Integrate \( a^z \) to get \( \frac{a^z}{\ln a} \). Thus, the integral becomes \( \frac{a^z}{(\ln a)^2} \).

Step 5: Write the Final Solution

Substitute back to get:\[ \int z a^z \, dz = \frac{z a^z}{\ln a} - \frac{a^z}{(\ln a)^2} + C \]where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

Integration by parts is a powerful technique to solve integrals of products of functions. It can be thought of as the reverse process of the product rule for differentiation. The formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, we pick "u" as the function which simplifies upon differentiation and "dv" as the part that remains manageable when integrated. For the given integral \( \int z a^z \, dz \), choosing \( u = z \) allows easy differentiation, while \( dv = a^z dz \) is straightforward to integrate using logarithmic properties.
The strategy is about turning a complex integral into simpler parts that are easier to compute. This particular integration technique is vital especially when direct integration methods are ineffective or cumbersome.

Definite Integral

Definite integrals are used to find the area under a curve between two specific limits. Represented by \( \int_{a}^{b} f(x) \, dx \), it calculates the net area, taking into account areas above the x-axis as positive and below as negative.
To apply integration by parts to definite integrals, the formula slightly modifies to include the limits of integration:

• \( \int_{a}^{b} u \, dv = \left[u v\right]_{a}^{b} - \int_{a}^{b} v \, du \)

In the context of the integral \( \int z a^z \, dz \), if evaluated as a definite integral, one would substitute the limits after performing integration by parts. The definite integral offers precise numerical values for tangible applications like calculating total accumulated quantities.

Indefinite Integral

An indefinite integral represents a family of functions and is expressed without upper and lower bounds, typically involving a constant of integration, denoted as \( C \). When integrating \( \int z a^z \, dz \) using integration by parts, the outcome is:

• \( \frac{z a^z}{\ln a} - \frac{a^z}{(\ln a)^2} + C \)

The "C" accounts for any constant that may be present since differentiation of a constant results in zero, ensuring no loss of information about the original function.
Indefinite integrals are essential in defining the general form of antiderivatives and are fundamental in calculus, serving as the basis for solving many real-world problems.

Exponential Functions

Exponential functions are characterized by the constant base raised to a variable exponent, such as \( a^z \). Calculus involving exponentials frequently requires properties of logarithms, such as \( \ln(a^b) = b \ln a \), to aid computation.
In our integration example, the function \( a^z \) is integrated by recognizing the derivative form,

• \( \int a^z \, dz = \frac{a^z}{\ln a} + C \)

This stems from the rule that the integral of an exponential function \( a^x \) involves dividing by the natural logarithm of the base. This rule is derived because exponential growth or decay is consistently proportional to the function's current value.
Mastering these operations is crucial, as exponential functions appear in many disciplines including population dynamics, economics, and physics, often describing growth processes or decay models.