Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int z a^{z} d z $$

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 the Components for Integration by Parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) from the integral \( \int z a^z \, dz \). A common strategy is to let \( u \) be the algebraic part and \( dv \) be the exponential part. Thus, let \( u = z \) and \( dv = a^z \, dz \).

Step 2: Differentiate and Integrate the Components

First, differentiate \( u = z \) to find \( du \, \). Therefore, \( du = dz \). Next, integrate \( dv = a^z \, dz \) to find \( v \, \). Since the integral of \( a^z \) is \( \frac{a^z}{\ln(a)} \), we have \( v = \frac{a^z}{\ln(a)} \).

Step 3: Apply the Integration by Parts Formula

Plug \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula \( \int u \, dv = uv - \int v \, du \). This gives us: \( \int z a^z \, dz = z \cdot \frac{a^z}{\ln(a)} - \int \frac{a^z}{\ln(a)} \, dz \).

Step 4: Simplify and Integrate the Remaining Integral

The expression becomes \( \frac{z a^z}{\ln(a)} - \frac{1}{\ln(a)} \int a^z \, dz \). We already know \( \int a^z \, dz = \frac{a^z}{\ln(a)} \), so this part becomes \( \frac{1}{\ln(a)} \cdot \frac{a^z}{\ln(a)} \) or \( \frac{a^z}{(\ln(a))^2} \).

Step 5: Write the Final Expression

Combine the terms: \( \frac{z a^z}{\ln(a)} - \frac{a^z}{(\ln(a))^2} + C \), where \( C \) is the constant of integration.

Key Concepts

Exponential Functions

Exponential functions are a key topic in calculus and appear frequently in various mathematical contexts. They are functions of the form \( f(x) = a^x \), where \( a \) is a positive constant (the base) and \( x \) is the exponent or power. These functions are unique because their rate of growth or decay is proportional to their current value.

• When \( a > 1 \), the function represents exponential growth.
• When \( 0 < a < 1 \), the function represents exponential decay.

Exponential functions are notable for their distinct curve shape and constant relative variability. They are widely used in modeling real-world phenomena such as population growth, radioactive decay, and interest calculations. In the context of integration by parts for the integral \( \int z a^z \, dz \), the exponential function \( a^z \) requires special techniques to integrate effectively, which highlights the importance of understanding these types of functions in calculus.

Differentiation

Differentiation is one of the core concepts in calculus, focusing on how functions change. It deals with finding the derivative, which measures the rate at which a quantity changes. The process involves finding an expression for the slope of the function's graph at any given point.

• The derivative of a power function \( f(x) = x^n \) is \( f'(x) = nx^{n-1} \).
• The derivative of an exponential function \( f(x) = a^x \) involves the natural logarithm: \( f'(x) = a^x \ln(a) \).

In solving the integral \( \int z a^z \, dz \), differentiation is used to determine \( du \) from \( u \), where \( u = z \). By differentiating \( z \), we obtain \( du = dz \), which is essential in applying the integration by parts technique. Differentiation seamlessly ties into integration by parts, as it helps transform and simplify the integral into a form that is easier to integrate.

Indefinite Integration

Indefinite integration, often referred to as finding an antiderivative, is the process of determining a function whose derivative matches the original function. Unlike definite integration, it doesn't involve calculated limits; it produces a family of functions represented with a constant of integration, \( C \).

• The integral \( \int f(x) \, dx = F(x) + C \), where \( F'(x) = f(x) \), signifies indefinite integration.
• For exponential functions, \( \int a^z \, dz = \frac{a^z}{\ln(a)} + C \).

In the problem \( \int z a^z \, dz \), indefinite integration plays a key role after utilizing integration by parts. Once we isolate the integral \( \int a^z \, dz \), we use indefinite integration to solve it, considering the properties of exponential functions. This step highlights the importance of understanding how to reverse differentiation to simplify complex integrals effectively.

Calculus Techniques

Calculus offers various techniques for solving integration and differentiation problems, providing a rich toolbox for addressing mathematical challenges. Integration by parts is one such technique, particularly useful when direct integration is complicated or not possible.

• The formula is \( \int u \, dv = uv - \int v \, du \), derived from the product rule of differentiation.
• Choosing appropriate \( u \) and \( dv \) is crucial; a common approach is to let \( u \) be a polynomial and \( dv \) involve an exponential or trigonometric function.

In the exercise \( \int z a^z \, dz \), integration by parts simplifies the problem to manageable parts, enabling the solution of integrals that might otherwise be too complex to handle directly. This demonstrates the power and flexibility of calculus techniques, which are designed to tackle difficult problems by breaking them down into simpler steps.