Problem 16 - University Calculus: Early Transcendentals

Question

Evaluate the indefinite integrals by using the given substitutions to reduce the integrals to standard form. \(\int \frac{d x}{\sqrt{5 x+8}}\) a. Using \(u=5 x+8\) b. Using \(u=\sqrt{5 x+8}\)

Step-by-Step Solution

Answer

Both substitutions give the same integral result: \( \frac{2}{5} \sqrt{5x + 8} + C \).

Step 1: Choose the Substitution for Part a

For part a, we are given the substitution \( u = 5x + 8 \). This means that \( du = 5 \, dx \) or \( dx = \frac{du}{5} \).

Step 2: Substitute and Integrate for Part a

Substitute \( u = 5x + 8 \) and \( dx = \frac{du}{5} \) into the integral to get: \[ \int \frac{dx}{\sqrt{5x+8}} = \int \frac{1}{\sqrt{u}} \frac{du}{5} = \frac{1}{5} \int u^{-1/2} \, du \]Integrate: \[ \frac{1}{5} \int u^{-1/2} \, du = \frac{1}{5} \left( 2u^{1/2} + C \right) = \frac{2}{5} \sqrt{u} + C \]

Step 3: Back-substitute for Part a

Replace \( u \) with \( 5x + 8 \) to get the final answer for part a: \[ \frac{2}{5} \sqrt{5x + 8} + C \]

Step 4: Choose the Substitution for Part b

For part b, we are given the substitution \( u = \sqrt{5x + 8} \). Therefore, \( u^2 = 5x + 8 \) and differentiating both sides, we get: \( 2u \, du = 5 \, dx \), or \( dx = \frac{2u \, du}{5} \).

Step 5: Substitute and Integrate for Part b

Plug \( u = \sqrt{5x + 8} \) and \( dx = \frac{2u \, du}{5} \) into the integral to obtain: \[ \int \frac{dx}{u} = \int \frac{2u \, du}{5u} = \frac{2}{5} \int du \]Integrate:\[ \frac{2}{5} \int du = \frac{2}{5} u + C \]

Step 6: Back-substitute for Part b

Replace \( u \) with \( \sqrt{5x + 8} \) in the integral to finalize the solution for part b: \[ \frac{2}{5} \sqrt{5x + 8} + C \]

Key Concepts

Integration by Substitution

Integration by substitution is a powerful method for solving integrals. It works by changing the variable of integration, making the integral easier to solve. This method is very similar to reversing the chain rule used in differentiation. Here's how it works:

• First, identify a part of the integrand that can be substituted with a single variable, say \( u \).
• Compute the differential of the substitution. For example, if \( u = 5x + 8 \), then \( du = 5 \, dx \).
• Rewrite the integral in terms of \( u \) and \( du \).
• Once the integral is simplified, proceed with the integration.
• Finally, substitute back the original variable to get the result in terms of the original variable.

In our exercise, choosing \( u = 5x + 8 \) transforms the integral into a simpler form. This allows us to integrate with respect to \( u \) more easily.
Notice how transformation of variables helps to simplify otherwise complex integral evaluations.

Change of Variables

The change of variables technique is applied in integration to simplify the computation process. It involves introducing a new variable to make the math more straightforward. Here's why changing variables helps:

• By selecting an appropriate substitution, you can transform complicated expressions into simpler integrable forms.
• The goal is to match the integrand or part of it to a known formula or a simpler form.

When choosing \( u \) as the square root of the expression, as shown in part b of the example, the entire integral obfuscates less complicated functions.
The approach involves equating your chosen \( u \) to a part of the integrand and then rewriting the entire integral accordingly. This simplification process is key in changing complex integrals into a basic form.

Definite and Indefinite Integrals

Integrals are categorized into two types: definite and indefinite integrals. In this exercise, we worked with indefinite integrals.

An indefinite integral represents a family of all antiderivatives of a function. These integrals introduce an integration constant \( C \), recognizing that the original function could have differed by a constant. In the provided solutions, the integration resulted in an expression like \( \frac{2}{5} \sqrt{5x + 8} + C \).

Here are the characteristics:

• Indefinite integrals yield a function plus an arbitrary constant \( C \).
• They might not have an upper or lower limit to evaluate a specific numeric area.

Understanding this difference is essential when dealing with integration challenges as it may affect how you apply boundary conditions or initial values in various problems.