Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int \frac{3 y}{\sqrt{2 y^{2}+5}} d y $$

Step-by-Step Solution

Answer

\( \frac{3}{4} \sqrt{2y^2 + 5} + C \)

Step 1: Identify Substitution

We need to simplify the integral \( \int \frac{3 y}{\sqrt{2 y^{2}+5}} dy \). Notice the expression \( 2y^2 + 5 \) inside the square root. To simplify, use substitution. Let \( u = 2y^2 + 5 \).

Step 2: Differentiate and Substitute

Differentiate the substitution \( u = 2y^2 + 5 \) to find \( du \). We get \( du = 4y \, dy \). Therefore, \( y \, dy = \frac{1}{4} du \). Substitute into the integral: \( \int \frac{3}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{1}{4} \, du \).

Step 3: Simplify and Integrate

Simplify the integral to \( \frac{3}{8} \int u^{-1/2} du \). Integrate \( u^{-1/2} \) to obtain \( \frac{3}{8} (2 u^{1/2}) + C \), which simplifies to \( \frac{3}{4} u^{1/2} + C \).

Step 4: Back Substitute

Replace \( u \) with \( 2y^2 + 5 \). The final solution is \( \frac{3}{4} \sqrt{2y^2 + 5} + C \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept when dealing with integration, especially in relation to calculating areas under curves within specific limits. A definite integral, such as \( \int_{a}^{b} f(x) \, dx \), represents the net area between the function \( f(x) \) and the x-axis, from x = a to x = b. This is unlike indefinite integrals, which do not have limits and represent a family of functions.

To evaluate a definite integral, we apply the Fundamental Theorem of Calculus. This theorem states that if a function \( F \) is an antiderivative of \( f \), then:

• \( \int_{a}^{b} f(x) \ dx = F(b) - F(a) \)

Definite integrals are used throughout various fields like physics for calculating quantities such as work done by a force, or in economics for evaluating total cost or revenue over a certain period.

In our exercise, however, we are working with an indefinite integral since there are no specified limits.

Indefinite Integrals

Indefinite integrals, sometimes referred to as antiderivatives, involve finding a function whose derivative gives the original function. In mathematical terms, finding the indefinite integral of a function \( f(x) \) means finding the function \( F(x) \) such that \( F'(x) = f(x) \). The solution includes a constant \( C \), representing any constant value that could be part of the family of solutions.

The indefinite integral is denoted as \( \int f(x) \, dx = F(x) + C \). It does not include upper and lower bounds like a definite integral.

In our example problem, \( \int \frac{3y}{\sqrt{2y^2+5}} \, dy \), we find the indefinite integral using substitution. The substitution technique helps simplify the integral into a form easier to solve by reducing complexity. Ultimately, we use antiderivatives to get to a function that, when differentiated, returns our original integrand.

The final result of the indefinite integral in our example is an expression \( \frac{3}{4} \sqrt{2y^2 + 5} + C \), indicating we successfully integrated the function related to the expression.

Techniques of Integration

Integration by substitution is a key technique used within calculus to solve complex integrals. It simplifies the integral into a manageable form, similar to the chain rule in differentiation. The idea is to transform the variable of integration into another variable that makes integration more straightforward.

In integration by substitution, we follow these steps:

• Identify and choose a substitution \( u = g(x) \) that simplifies the integral.
• Differentiation follows to express \( dx \) in terms of \( du \).
• Rewrite the integral in terms of \( u \) and solve.
• Finally, substitute back to the original variable.

In the original step-by-step solution provided for \( \int \frac{3y}{\sqrt{2y^2+5}} \, dy \), we let \( u = 2y^2 + 5 \), which simplifies the square root and creates an integrable form. This approach highlights how substitution can transform complex tasks into something more manageable by focusing on transforming parts of the integrand effectively.

Substitution is especially useful when the integrand includes composite functions, as seen in our problem, where direct integration is not immediately possible.