Problem 24 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the integral. In many cases it will be advantageous to begin by doing a substitution. For example, in Problem 19, let \(w=\sqrt{x}, w^{2}=x ;\) then \(2 w d w=d x .\) This eliminates \(\sqrt{x}\) by replacing \(x\) with a perfect square. $$ \int \sqrt{x} \ln x d x $$

Step-by-Step Solution

Answer

The evaluation of the integral \( \int \sqrt{x} \ln x \, dx \) results in \( \frac{2x^{3/2} \ln x}{3} - \frac{x^2}{3} + C\).

Step 1: Apply the given substitution

Start by applying the substitution \(w = \sqrt{x}\), which implies \(x = w^2\) and \(dx = 2w dw\). Substitute this into the original integral to get \(\int w \ln (w^2) \cdot 2w \, dw\). This simplifies to \(2 \int w^2 \ln (w^2) \, dw\).

Step 2: Use integration by parts

Now, we apply the method of integration by parts. The integration by parts formula is \(\int u dv = uv - \int v du\). Choose \(u = \ln (w^2)\) and \(dv = w^2 dw\). Then, \(du = \frac{2}{w} dw\) and \(v = w^3/3\). Substituting these values into the integration by parts formula, we obtain \(2 \left[ \frac{w^3 \ln (w^2)}{3} - \int \frac{2 w^3}{3} dw \right]\).

Step 3: Simplify integral and compute

The remaining integral can be easily computed to get \( \frac{2w^4}{12} = \frac{w^4}{6}\). Substituting back, the simplified integral becomes \(2 \left[ \frac{w^3 \ln (w^2)}{3} - \frac{w^4}{6} \right] + C\).

Step 4: Substitute the original variable

Finally, resubstitute \(w = \sqrt{x}\) to get the solution in terms of the original variable \(x\). The integral is then \(2 \left[ \frac{(\sqrt{x})^3 \ln ((\sqrt{x})^2)}{3} - \frac{(\sqrt{x})^4}{6} \right] + C\), which simplifies to \( \frac{2x^{3/2} \ln x}{3} - \frac{x^2}{3} + C\).

Key Concepts

Integration by Parts

Integration by parts is a crucial technique derived from the product rule of differentiation. It simplifies the integration of products of functions. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here, \(u\) and \(dv\) are chosen from the original integral such that they simplify the problem when differentiated and integrated, respectively.

During the computation of \( \int \sqrt{x} \ln x \, dx \), we first use substitution to simplify our integrand to something more manageable with standard integration techniques. Once the integral becomes \( \int w^2 \ln (w^2) \ dw \), we apply integration by parts:

• Choose \(u = \ln(w^2)\), which differentiates to \(du = \frac{2}{w} \, dw\).
• Choose \(dv = w^2 \, dw\), which integrates to \(v = \frac{w^3}{3}\).

By substituting these into the integration by parts formula, we split our integral into a simpler form that can be solved and evaluated.

Definite and Indefinite Integrals

Integral calculus extends into definite and indefinite integrals. While indefinite integrals provide antiderivatives of functions, definite integrals evaluate the net area under a curve within specific bounds.

In the original problem, we are dealing with an indefinite integral \( \int \sqrt{x} \ln x \, dx \). This means we are seeking a general formula for the antiderivative, denoted by adding a constant \(C\) to account for the infinitely many antiderivatives differing by a constant. Through substitution and integration by parts, the solution was simplified and evaluated to:

• \( \frac{2x^{3/2} \ln x}{3} - \frac{x^2}{3} + C \)

Remember, indefinite integrals are open-ended, covering the entire family of antiderivatives until specific boundaries are set for definite integrals.

Integration Techniques

In calculus, integration is a powerful tool with several techniques to tackle different types of integrals. These techniques often transform complex integrals into manageable forms.

**Common integration techniques include:**

• Substitution Method: Used when an integral contains a function and its derivative. By substituting one variable in terms of another, it simplifies complicated functions into recognizable forms.
• Integration by Parts: Especially useful for integrating products of two functions, as described above.
• Partial Fraction Decomposition: Applicable for rational functions, breaking them into simpler parts that are easier to integrate.

In this particular problem, the substitution \(w = \sqrt{x}\) was essential in transforming the integrand \( \sqrt{x} \ln x \) into a form \(w^2 \ln(w^2)\) that allowed for easier application of integration by parts.

Knowing when and how to apply these techniques is key to mastering calculus and solving complex integrals efficiently.