Problem 42 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral. $$ \int_{1}^{2} x^{2} \ln x d x $$

Step-by-Step Solution

Answer

The exact value of the integral is \(\frac{8}{3} \ln 2 - \frac{7}{3}\).

Step 1: Identify 'u' and 'dv'

For integration by parts, a function is broken down into two parts 'u' and 'dv' such that its integral is easier to solve. Having \(x^{2}\) and \(\ln x\) in the integrand, one strategy is to differentiate the logarithm and to integrate the other part. So, we take: \(u = \ln x\) and \(dv = x^{2} dx.\)

Step 2: Find 'du' and 'v'

Now find 'du' by differentiating 'u' and find 'v' by integrating 'dv'. Differentiating \(u = \ln x\), we get: \(du = 1/x dx.\) Integrating \(dv = x^{2} dx\), we get: \(v = x^{3}/3.\)

Step 3: Apply the Integration by Parts Formula

The formula for integration by parts is: \(\int u dv = uv - \int v du.\) Substitute 'u', 'v', 'du', and 'dv' into the formula to obtain: \(\int_{1}^{2} x^{2} \ln x dx = [(\ln x)(x^{3}/3)]_{1}^{2} - \int_{1}^{2} (x^{3}/3)(1/x) dx = (2^3/3 \ln 2 - 1/3 * \ln 1) - 1/3 \int_{1}^{2} x^{2} dx.\)

Step 4: Solve the Integral

Now solve the new simple integral: \(\int_{1}^{2} x^{2} dx\) and perform the other calculations. For the integral, we get 7/3. Simplifying the full expression yields: \(\frac{8}{3} \ln 2 - 1/3 * 0 - 7/3 = \frac{8}{3} \ln 2 - \frac{7}{3}.\)

Step 5: Write the Answer as Exact Value

Keep the \(\ln 2\) term as it is, do not try to approximate it. The exact value of the integral is then: \(\frac{8}{3} \ln 2 - \frac{7}{3}.\)

Key Concepts

Definite Integrals

Definite integrals calculate the net area between the curve of a function and the x-axis, within a specific interval. In this problem, we evaluated the definite integral of \( \int_{1}^{2} x^{2} \ln x \, dx \). The limits of integration \(x = 1\) and \(x = 2\) denote the interval over which the area is calculated.

Definite integrals have properties that make them particularly useful:

• They provide the total accumulation of quantities, such as area or volume.
• The process uses the fundamental theorem of calculus, connecting differentiation and integration.
• They result in a specific value, often exact, rather than a general antiderivative.

The calculation begins with identifying the appropriate technique, such as integration by parts, to simplify the integrand, which is essential in solving complex integrals effectively.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions involving logarithms, which appear frequently in calculus problems. The function \(\ln x\) has distinct properties that make it a good candidate for the 'u' part in integration by parts. When differentiating \(\ln x\), the result is \(du = \frac{1}{x} \, dx\).

This step was a crucial part of our integral evaluation because:

• The derivative of the logarithmic function reduces the complexity of integrating the rest of the expression.
• It transforms a product of functions into a simpler form manageable by integration techniques such as integration by parts.
• The properties of logarithms, like simplifying multiplication and division, are advantageous when handling complex integrands.

By using logarithmic differentiation, the integral \(\int u \, dv\) can be dissected into manageable pieces, making the overall solution more straightforward.

Integration Techniques

Integration techniques are strategies to simplify and solve integrals. Various methods, like substitution and integration by parts, are chosen based on the form of the integrand. In this exercise, integration by parts was used to solve the definite integral \(\int_{1}^{2} x^{2} \ln x \, dx\).

Integration by parts is driven by the formula \(\int u \, dv = uv - \int v \, du\) and requires:

• Choosing \(u\) and \(dv\) wisely to simplify the integral.
• Finding derivatives \(du\) and integrals \(v\) of the chosen parts.
• Substituting these into the formula to transform the original integral into a more solvable form.

This method is particularly useful when integrating products of functions that don't easily lend themselves to direct integration. By breaking down the integrand, you simplify the process into smaller, more manageable integrations, as seen when solving the integral resulting in \(\frac{8}{3} \ln 2 - \frac{7}{3}\).