Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int \sqrt{3 x+2} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{2}{9} (3x + 2)^{3/2} + C \).

Step 1: Choose a substitution

We begin by identifying a substitution that will simplify the integral. We can let \( u = 3x + 2 \) since the expression \( \sqrt{3x + 2} \) appears under the square root. Differentiating both sides with respect to \( x \), we obtain \( \frac{du}{dx} = 3 \) or \( dx = \frac{1}{3} du \).

Step 2: Rewrite the integral in terms of \( u \)

Substitute \( u = 3x + 2 \) and \( dx = \frac{1}{3} du \) into the integral. The integral becomes: \[ \int \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int u^{1/2} \, du. \]

Step 3: Perform the integration

To compute \( \int u^{1/2} \, du \), we use the power rule for integration \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Applying this, we have: \[ \frac{1}{3} \cdot \left( \frac{u^{3/2}}{3/2} \right) + C. \] Simplifying: \( \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2} + C. \)

Step 4: Substitute back in terms of \( x \)

Replace \( u \) with the original expression \( 3x + 2 \) to revert back to the variable \( x \). So, the integral becomes: \[ \frac{2}{9} (3x + 2)^{3/2} + C. \]

Key Concepts

Indefinite Integral

Integrals can either be definite or indefinite. An indefinite integral does not have limits of integration and represents a family of functions. It includes a constant of integration, usually denoted as "C." This is important because when differentiating a constant, it becomes zero, meaning the original function could have contained a constant that isn't evident in the derivative.

If we denote the indefinite integral of a function, it looks like this: \[ \int f(x) \, dx = F(x) + C \] Here, \( f(x) \) is the function to be integrated, \( F(x) \) is the antiderivative of \( f(x) \), and \( C \) is the constant of integration. Indefinite integrals are a fundamental concept in calculus because they allow us to reverse the process of differentiation, giving us back the original function or at least the family of such functions.

Remember, when you compute an indefinite integral, you might get different functions due to this constant of integration. Hence, specifying \( C \) is crucial for completeness in solutions.

Integration by Substitution

Integration by substitution is a common technique used to simplify an integral. It involves changing the variable of integration to make the integral easier to solve. It's akin to the reverse chain rule used in differentiation.

The key steps in substitution include:

• Selecting a substitution, often denoted as \( u \), that simplifies part of the integral.
• Replacing all instances of the original variable and its differential with the new variable and its corresponding differential.
• Calculating the integral in terms of \( u \).
• Finally, substituting back the original variables to get the answer in the initial terms.

In the exercise, we chose \( u = 3x + 2 \) because it appears inside the square root, making it a natural choice. Differentiating gives \( du = 3 \, dx \) or \( dx = \frac{1}{3} du \), which allows us to further simplify our integral. This technique, while simple in concept, becomes incredibly powerful for more complex integrals, making it a cornerstone method in integral calculus.

Power Rule for Integration

The power rule for integration is straightforward and similar to the power rule for differentiation, but reversed. It allows us to integrate functions of the form \( x^n \) easily.

The rule states:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \]where \( n eq -1 \). The constraint \( n eq -1 \) exists because \( \int x^{-1} \, dx \) is a separate case, specifically yielding \( \ln |x| + C \).

In our solution, once we converted our integral using a substitution, we ended up with an expression that required applying the power rule: \[ \int u^{1/2} \, du. \]By applying the power rule, the integral became \( \frac{u^{3/2}}{3/2} + C \). However, since we had a constant outside the integral \( \frac{1}{3} \), we multiplied the result by this constant to get our final answer. This simple tool of the power rule is indispensable for handling polynomials within integrals.