Problem 20 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(13-48\) . $$ \int \frac{4 y d y}{\sqrt{2 y^{2}+1}} $$

Step-by-Step Solution

Answer

\( 2\sqrt{2y^2 + 1} + C \)

Step 1: Identify the Integral Form

We have the integral \( \int \frac{4y \, dy}{\sqrt{2y^2 + 1}} \). Notice that the integrand has the form \( \frac{f'(y)}{\sqrt{f(y)}} \), which suggests substitution can be applied.

Step 2: Choose an Appropriate Substitution

Let \( u = 2y^2 + 1 \). Then, differentiate to find \( du = 4y \, dy \). This substitution will simplify the integral.

Step 3: Substitute and Simplify the Integral

Using the substitution, substitute into the integral: \( \int \frac{4y \cdot dy}{\sqrt{2y^2 + 1}} = \int \frac{1}{\sqrt{u}} \, du \).

Step 4: Integrate the Simplified Form

The integral is now \( \int u^{-1/2} \, du \). Use the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = -1/2 \). This gives \( 2u^{1/2} + C \).

Step 5: Substitute Back to Original Variable

Substitute \( u = 2y^2 + 1 \) back into the expression to obtain the final answer: \( 2\sqrt{2y^2 + 1} + C \).

Key Concepts

Substitution Method

The substitution method in integral calculus is a technique used to simplify integrals, making them easier to evaluate. The main idea is to change the variable of integration to transform the integral into a simpler form. It is similar to the reverse process of the chain rule in differentiation. To use the substitution method:

• Identify a part of the integrand (the function inside the integral) that can be substituted with a new variable. This is typically something like an inner function composition or an expression inside a root or power.
• Express this part as a new variable, say \( u \). Differentiate \( u \) to find \( du \), replacing all differential parts of the integrand.
• Substitute \( u \) and \( du \) into the integral to simplify it into a more basic form.
• Integrate the simplified expression, and then substitute back the original variables to get back to the original terms of the problem.

In the original example, the substitution method involved setting \( u = 2y^2 + 1 \), simplifying the integral into a basic form that was more straightforward to solve.

Power Rule

The power rule is a fundamental tool in both differentiation and integration. For integration, the power rule states that if you have an integral of the form \( \int x^n \, dx \), it can be solved using the formula \( \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). This rule provides a simple way of integrating polynomial expressions. In the example problem:

• After substitution, the integral \( \int u^{-1/2} \, du \) is tackled using the power rule.
• Recognizing that \( n = -1/2 \), the integrated form becomes \( \frac{u^{1/2}}{1/2} + C \), which simplifies to \( 2u^{1/2} + C \).

This demonstrates how the power rule can simplify complicated integrands into more manageable expressions.

Definite Integrals

Definite integrals are used to compute the area under a curve between two specific points or bounds. While indefinite integrals provide a general family of antiderivatives, definite integrals result in a specific numerical value.For definite integrals:

• Bounds are specified, usually written as \( \int_{a}^{b} f(x) \, dx \).
• The result is dependent on both the function and the limits \( a \) and \( b \).
• Fundamental Theorem of Calculus is used, which states: if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

Definite integrals are common in real-world applications where calculation of area, volume, or other quantities is required.

Indefinite Integrals

Indefinite integrals, unlike definite integrals, do not have specified limits or bounds. They represent a general form of antiderivatives of a function, denoted by \( \int f(x) \, dx = F(x) + C \). Here, \( C \) is a constant of integration that accounts for the general solution since differentiation erases any constant.Indefinite integrals have several key considerations:

• The constant \( C \) is crucial because it symbolizes all potential vertical shifts of the antiderivative.
• An indefinite integral answers: "What function differentiates into this one?"
• Conceptually, it is akin to reversing differentiation.

In the original example, the indefinite integral approach results in the solution \( 2\sqrt{2y^2 + 1} + C \), indicating an entire family of curves related by their constant \( C \).