Problem 32 - Calculus and its applications

Question

Evaluate using integration by parts. $$ \int_{1}^{2} x^{3} \ln x d x $$

Step-by-Step Solution

Answer

The value of the integral is \(4 \ln 2 - \frac{15}{16}\).

Step 1: Choose Functions for Integration by Parts

Integration by parts is given by the formula \[ \int u \, dv = uv - \int v \, du \]For the integral \( \int x^3 \ln x \, dx \), choose \( u = \ln x \) (so that its derivative, \( du = \frac{1}{x} dx \) is simpler) and \( dv = x^3 \, dx \) (which can be integrated easily).

Step 2: Differentiate and Integrate the Chosen Functions

Now differentiate \( u = \ln x \) to get \( du = \frac{1}{x} dx \).Integrate \( dv = x^3 \, dx \) to obtain \( v = \frac{x^4}{4} \).

Step 3: Apply the Integration by Parts Formula

Apply the integration by parts formula:\[ \int u \, dv = uv - \int v \, du \]Substitute \( u = \ln x \), \( du = \frac{1}{x} dx \), \( v = \frac{x^4}{4} \), and \( dv = x^3 \, dx \):\[ \int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx \].

Step 4: Simplify the Integral and Solve

Simplify the integral:\[ \frac{x^4}{4} \ln x - \int \frac{x^4}{4x} \, dx = \frac{x^4}{4} \ln x - \int \frac{x^3}{4} \, dx \]This becomes:\[ \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx \].

Step 5: Integrate the Remaining Part

To integrate \( \int x^3 \, dx \), compute:\[ \int x^3 \, dx = \frac{x^4}{4} \]Substituting this back:\[ \frac{x^4}{4} \ln x - \frac{1}{4} \cdot \frac{x^4}{4} \].

Step 6: Evaluate from 1 to 2

Plug in the limits from 1 to 2:\[ \left[ \frac{x^4}{4} \ln x - \frac{x^4}{16} \right]_1^2 = \left( \frac{2^4}{4} \ln 2 - \frac{2^4}{16} \right) - \left( \frac{1^4}{4} \ln 1 - \frac{1^4}{16} \right) \]Since \( \ln 1 = 0 \), the second part cancels out:\[ = \left( \frac{16}{4} \ln 2 - \frac{16}{16} \right) - (0 - \frac{1}{16}) \]Simplify:\[ = 4 \ln 2 - 1 + \frac{1}{16} \]\[ = 4 \ln 2 - \frac{15}{16} \].

Key Concepts

Integral Calculus

Integral calculus is a crucial part of calculus that primarily deals with integrals, which represent accumulation or area under a curve. In this exercise, we're focusing on evaluating the integral \( \int_{1}^{2} x^3 \ln x \, dx \) using integration by parts, a technique often used when direct integration is challenging. The fundamental theorem of calculus connects differentiation and integration, providing a powerful tool for finding areas and solving real-world problems. By understanding integral calculus, you can solve various problems involving rates of change and accumulation over an interval.

Definite Integrals

Definite integrals calculate the total area under a curve between two specific points, known as limits. Here, we're working with \( \int_{1}^{2} x^3 \ln x \, dx \), where the limits are 1 and 2. When you evaluate a definite integral, you're essentially summing up infinitely many infinitesimally small quantities between these two points.

You find the value of the integral's antiderivative at the upper limit and subtract its value at the lower limit. In our exercise, after applying integration by parts and simplifying, the integral is evaluated from 1 to 2 to find the exact total area under the curve between these points.

Logarithmic Integration

Logarithmic integration involves integrals that contain a logarithmic function like \( \ln x \). In such cases, you may need to employ techniques like integration by parts to simplify the process.

• Choosing \( u = \ln x \) ensures that its derivative, \( du = \frac{1}{x} \, dx \), is straightforward.
• Logarithmic functions can make integrals tricky, but when properly managed, they unravel nicely.

In this exercise, \( ln x \) is differentiated, enabling us to separate the problem into more manageable parts, greatly helping with the evaluation.

Integration Techniques

Various integration techniques, like substitution, partial fractions, and integration by parts, allow you to solve complex integrals. Integration by parts is particularly helpful when dealing with products of functions, specifically when one component is easily differentiated while the other is easily integrated.

• The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).
• This method transforms the integral into a simpler problem, as seen when \( x^3 \ln x \) is split into parts—\( u = \ln x \) and \( dv = x^3 \, dx \).

After differentiating and integrating the chosen parts, the new integral is often simpler, making it easier to solve and evaluate. Mastery of techniques like integration by parts significantly broadens your ability to tackle diverse integral calculus problems.