Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-34, use the method of substitution to find each of the following indefinite integrals. \(\int x \sqrt{x^{2}+4} d x\)

Step-by-Step Solution

Answer

\(\int x \sqrt{x^2+4} \, dx = \frac{(x^2+4)^{3/2}}{3} + C\).

Step 1: Identify the Substitution Variable

We look at the expression inside the square root, \(x^2 + 4\), and decide to let \(u = x^2 + 4\). This simplifies the integral by substituting \(u\) for \(x^2 + 4\).

Step 2: Calculate the Differential

Differentiate \(u = x^2 + 4\) with respect to \(x\): \(\frac{du}{dx} = 2x\). This gives us \(du = 2x \, dx\) or \(\frac{1}{2} du = x \, dx\).

Step 3: Substitute in the Integral

Substitute \(u\) and \(\frac{1}{2} du\) into the integral: \(\int x \sqrt{x^2 + 4} \, dx = \int \sqrt{u} \cdot \frac{1}{2} \, du\). Factor out the \(\frac{1}{2}\) to simplify: \(\frac{1}{2} \int \sqrt{u} \, du\).

Step 4: Integrate with respect to u

Integrate \(\sqrt{u} = u^{1/2}\) with respect to \(u\): \(\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2u^{3/2}}{3}\). Multiply by the \(\frac{1}{2}\) we factored out earlier: \(\frac{1}{2} \cdot \frac{2u^{3/2}}{3} = \frac{u^{3/2}}{3}\).

Step 5: Back-Substitute x and Simplify

Replace \(u\) with \(x^2 + 4\) to convert back to the variable \(x\): \(\frac{(x^2 + 4)^{3/2}}{3} + C\), where \(C\) is the constant of integration.

Key Concepts

Integration by Substitution

Integration by substitution is a key technique in calculus that simplifies integrals by changing variables. The main idea is to substitute parts of the integral with a new variable, often making the integral easier to evaluate.
This is particularly helpful when dealing with composite functions, like the one in the problem \(\int x \sqrt{x^{2}+4} \, dx\). By using substitution, we aim to transform the integrand into a simpler form, like a basic power of \(u\).
Common steps for substitution include:

• Choose a substitution (a function and its derivative that appear in the integral).
• Replace the chosen part of the integral with the new variable \(u\).
• Write the differential \(dx\) in terms of \(du\).
• Substitute these into the integral, converting it into an expression in terms of \(u\).

This method essentially transforms the integral into a new problem that is often easier to solve. Remember, the ultimate goal is to integrate with respect to \(u\), and then revert back to the original variable.

Calculus

Calculus is a branch of mathematics focused on change and motion. It is broadly divided into two areas: differentiation and integration.
Integration, which this exercise focuses on, is about finding the whole from understanding the parts. Specifically, indefinite integrals involve finding a general function whose derivative is the given function.
In this exercise:

• We used integration by substitution.
• We converted a more complex expression into an easier form.
• Once in the simpler form, we applied basic integration rules.

Calculus allows us to deal with changing quantities rigorously, and integrating effectively undoes differentiation. Understanding how these parts connect will essential in mastering problems like this one.

Differentials

Differentials serve as a bridge between atoms and integrable functions. They help in systematically changing variables during substitution. In calculus, when we say we differentiate \(u = f(x)\), we find \(\frac{du}{dx}\), which shows how \(u\) changes with respect to \(x\).
For this exercise:

• We had \(u = x^2 + 4\), and found \(\frac{du}{dx} = 2x\).
• By rearranging, we expressed \(du\) in terms of \(dx\), resulting in \( du = 2x \, dx \).
• This differential form is crucial because it allows us to substitute \(du\) directly for a part of the integral, helping transform the original problem into one we can solve.

Once the differentials are set, they help maintain the integrity of the integral during substitution.

Back-Substitution

Back-substitution is the final step in the substitution method, bringing us back to the original variable. Once the integral is solved in terms of \(u\), we must revert \(u\) to its original form, expressing in terms of the original variable \(x\).
In the concluding part of the exercise:

• We solved the integral \(\int \sqrt{u} \, du\) in terms of \(u\), obtaining \(\frac{u^{3/2}}{3} + C\).
• To complete the solution, we substituted back \(u = x^2 + 4\).
• The final answer was \(\frac{(x^2 + 4)^{3/2}}{3} + C\), negating the back-substitution effort became the crucial step in finding the full, correct answer incorporating the initial variable.

Back-substitution closes the circle of substitution, connecting the transformed integral back to its roots.