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Chapter 14: Problem 10

Evaluate the integrals using integration by parts where possible. $$\int(3 x-2) 4^{x} d x$$

Short Answer

Expert verified
The short answer to the integral \(\int(3x-2)4^x dx\) using integration by parts is: \((3x-2)\frac{4^x}{\ln(4)} - \frac{3\cdot 4^x}{\ln^2(4)} + C\)

Step by step solution

01

Choose u and dv

First, let's choose the functions \(u\) and \(dv\), according to the given integral: \[u = 3x - 2\] \[dv = 4^x dx\]
02

Calculate du and v

Now, let's compute derivatives for \(u\) and integrals for \(dv\): \[du = (3) dx\] \[v = \int 4^x dx\] Recall that to integrate \(4^x\), we use the formula: \[\int a^x dx = \frac{a^x}{\ln(a)} + C\] In this case, \(a = 4\), so: \[v = \frac{4^x}{\ln(4)}\]
03

Apply integration by parts formula

Now that we have calculated \(u, du, v\), and \(dv\), we will apply the integration by parts formula: \[\int u dv = uv - \int v du\] Substituting our values, we get: \[\int (3x - 2)4^x dx = (3x - 2)\frac{4^x}{\ln(4)} - \int \frac{4^x}{\ln(4)}(3) dx\]
04

Solve the remaining integral

Now we need to solve the remaining integral: \[\int \frac{4^x}{\ln(4)}(3) dx\] This is simply an exponential integral, which can easily be evaluated: \[\int 3\frac{4^x}{\ln(4)} dx = \frac{3}{\ln(4)}\int 4^x dx\] Using the formula for exponential integrals again, we get: \[\frac{3}{\ln(4)}\int 4^x dx = \frac{3\cdot 4^x}{\ln^2(4)} + C\]
05

Combine terms

Finally, we will combine all terms: \[\int (3x-2)4^x dx = (3x-2)\frac{4^x}{\ln(4)} - \frac{3\cdot 4^x}{\ln^2(4)} + C\] Thus, the evaluated integral is: \((3x-2)\frac{4^x}{\ln(4)} - \frac{3\cdot 4^x}{\ln^2(4)} + C\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integral
In calculus, an indefinite integral represents a family of functions that gives rise to the original function when differentiated. It's essentially finding the antiderivative of a function. For instance, consider the indefinite integral \[ \int f(x) dx \], which is the antiderivative of the function \( f(x) \). The resulting function includes a constant of integration, often denoted by \(C\), which accounts for the fact that differentiation eliminates constants.

The process of calculating an indefinite integral is known as integration, and it comes with a variety of techniques to handle different types of functions. In our exercise, the indefinite integral we are looking to find is \[ \int(3x-2)4^x dx \], which requires the application of integration techniques to solve.
Exponential Functions
An exponential function is a mathematical function of the form \( f(x) = a^x \), where \(a\) is a constant base and \(x\) is the exponent. These functions exhibit the property that the rate of change (or derivative) of the function is proportional to the value of the function itself. This makes them particularly important in compound interest, population growth models, and other natural phenomena.

When integrating exponential functions, a key formula is \[ \int a^x dx = \frac{a^x}{\ln(a)} + C \], where \(a\) is greater than 0 and not equal to 1, and \(C\) is the integration constant. In the context of our exercise, we integrate the function \(4^x\) using this knowledge, resulting in \[ \frac{4^x}{\ln(4)} \] to further apply integration by parts.
Integration Techniques
There are several integration techniques used when the standard formula for integration is not directly applicable. One of the most versatile methods is integration by parts, which is based on the product rule for differentiation. The formula is \[ \int u dv = uv - \int v du \], where \(u\) and \(dv\) are parts of the original integral we choose to differentiate and integrate, respectively. The goal is to transform the original integral into a simpler form that can be easily integrated.

In our exercise, we've used this technique to integrate \((3x-2)4^x\). By choosing \(u = 3x - 2\) and \(dv = 4^x dx\), we make the integration process more manageable. The subsequent steps involve calculating \(du\) and \(v\), then applying the integration by parts formula to reach a solution. This method is pivotal when dealing with products of algebraic expressions and exponential functions, as in the problem we're exploring.

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