Problem 46 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral. \( \displaystyle \int x^2 \sqrt{2 + x} \, dx \)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2}{7}(2+x)^{3.5} - \frac{8}{5}(2+x)^{2.5} + \frac{8}{3}(2+x)^{1.5} + C \).

Step 1: Choose a Substitution

For the integral \( \int x^2 \sqrt{2 + x} \, dx \), let's use the substitution \( u = 2 + x \). This choice simplifies the square root expression. Then, compute the differential: \( du = dx \) and \( x = u - 2 \).

Step 2: Rewrite the Integral

Substitute \( u = 2 + x \), \( du = dx \), and \( x = u - 2 \) into the integral. The integral becomes \( \int (u - 2)^2 \sqrt{u} \, du \).

Step 3: Expand the Expression

Expand \( (u - 2)^2 \) to get \( u^2 - 4u + 4 \). The integral now is \( \int (u^2 - 4u + 4) \sqrt{u} \, du \).

Step 4: Simplify and Separate

Simplify by distributing the \( \sqrt{u} \): \( \int (u^{2.5} - 4u^{1.5} + 4u^{0.5}) \, du \). This breaks down into three separate integrals: \( \int u^{2.5} \, du - 4 \int u^{1.5} \, du + 4 \int u^{0.5} \, du \).

Step 5: Integrate Term by Term

Use the power rule for integration on each term:1. \( \int u^{2.5} \, du = \frac{u^{3.5}}{3.5} + C_1 \)2. \( \int u^{1.5} \, du = \frac{u^{2.5}}{2.5} + C_2 \)3. \( \int u^{0.5} \, du = \frac{u^{1.5}}{1.5} + C_3 \)

Step 6: Combine and Simplify the Result

Combine the integrals:\[ \frac{u^{3.5}}{3.5} - 4 \times \frac{u^{2.5}}{2.5} + 4 \times \frac{u^{1.5}}{1.5} + C \]= \[ \frac{2}{7}u^{3.5} - \frac{8}{5}u^{2.5} + \frac{8}{3}u^{1.5} + C \].

Step 7: Resubstitute Back in Terms of x

Convert back to the variable \( x \) by substituting \( u = 2 + x \):\[ \frac{2}{7}(2+x)^{3.5} - \frac{8}{5}(2+x)^{2.5} + \frac{8}{3}(2+x)^{1.5} + C \].

Key Concepts

Substitution Method

The substitution method is particularly useful when dealing with integrals that contain composite functions, like the one with a square root term here. The idea is to simplify the expression by replacing a part of the integral with a single variable, typically denoted as "u." In our example, we chose the substitution \( u = 2 + x \), which simplifies the square root expression \( \sqrt{2 + x} \) to \( \sqrt{u} \). This changes the form of the integral into something more manageable.

When using substitution, don't forget to also change the differential \( dx \) to \( du \). In this step, the relationship \( du = dx \) directly shows that swapping \( u \) for \( x \) is straightforward. We also expressed \( x \) in terms of \( u \), from the equation \( u = 2 + x \), so we have \( x = u - 2 \).

By converting all terms in the integrand to functions of \( u \), you can rewrite the whole integral in terms of \( u \) and \( du \). This often results in a simpler expression, as seen with \( (u-2)^2 \sqrt{u} \), making the integration process more manageable.

Power Rule of Integration

The power rule of integration is a fundamental technique that allows us to integrate simple power functions. The general formula of the power rule, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), is applied where \( n eq -1 \). This rule is especially efficient for polynomial terms and is utilized in step 5 of our example.

Once the integral was expressed fully in terms of \( u \), we had terms like \( u^{2.5} \), \( u^{1.5} \), and \( u^{0.5} \). Using the power rule, each term was integrated:

• For \( u^{2.5} \), the integral becomes \( \frac{u^{3.5}}{3.5} + C_1 \).
• For \( u^{1.5} \), it is \( \frac{u^{2.5}}{2.5} + C_2 \).
• For \( u^{0.5} \), it results in \( \frac{u^{1.5}}{1.5} + C_3 \).

Each of these expressions follows from adding one to the power and dividing by the new power, illustrating the simplicity and effectiveness of the power rule in calculating indefinite integrals. Remember that the constant \( C \) represents the integration constant, which accounts for any constant term that might have initially been present in the derivative.

Integration of Polynomials

Integrating polynomials involves breaking down the polynomial expression into simpler components that are easier to integrate, primarily using the power rule. In our example, after the substitution, \((u-2)^2\) was expanded into \(u^2 - 4u + 4\), providing a clear path to applying the power rule.

Upon expansion, each polynomial term such as \(u^2\), \(-4u\), and constant \(4\) is associated with a \(\sqrt{u}\) term. This results in the powers \(u^{2.5}\), \(u^{1.5}\), and \(u^{0.5}\). By isolating each term, we were able to apply the integration process step-by-step.

The approach of integration by parts breaks the integral into manageable sections. Each term separately corresponds to a straightforward application of the power rule, leading to the final result. It’s crucial in polynomial integrations to express all terms in a form ready for the power rule, ensuring clear integration of each part of the polynomial.