Problem 31 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

\( \frac{8}{3} \ln 2 - \frac{7}{9} \)

Step 1: Identify Parts

Identify the parts of the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln x \) and \( dv = x^2 \, dx \).

Step 2: Differentiate and Integrate

Differentiate \( u = \ln x \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv = x^2 \, dx \) to find \( v = \frac{x^3}{3} \).

Step 3: Substitute into Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula to get:\[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x \bigg|_{1}^{2} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \bigg|_{1}^{2}. \]

Step 4: Simplify the Integral

Simplify the remaining integral: \( \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \int \frac{x^2}{3} \, dx = \frac{1}{3} \int x^2 \, dx \). Perform the integration: \( \frac{1}{3} \times \frac{x^3}{3} = \frac{x^3}{9}. \)

Step 5: Evaluate the Integrals and Solve

Evaluate \( \frac{x^3}{3} \ln x \) and \( \frac{x^3}{9} \) from 1 to 2:\[ \left( \frac{8}{3} \ln 2 - \frac{8}{9} \right) - \left( 0 - \frac{1}{9} \right) = \frac{8}{3} \ln 2 - \frac{7}{9}. \]

Step 6: Final Answer

The final answer is the evaluation of the integral, which is \( \frac{8}{3} \ln 2 - \frac{7}{9} \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus used to find the area under a curve between two specific points on a graph, often referred to as the limits of integration. In this case, the definite integral of the function \( x^2 \ln x \) from 1 to 2 is being evaluated. Definite integrals provide a precise value rather than just the anti-derivative of a function.
When solving a definite integral, it's crucial to follow the specific steps: determine the anti-derivative of the function, substitute the upper and lower limits into this anti-derivative, and then subtract the result of the lower limit from that of the upper limit. This is why step-by-step calculations are essential because each part involves careful algebraic manipulations and understanding of calculus principles. Integrating between known limits is what helps turn these abstract calculations into concrete answers.

Logarithmic Integration

Logarithmic integration comes into play when the function you're integrating involves a logarithm, such as \( \ln x \), as is the case in this problem. The presence of a logarithm in integration generally indicates the use of integration by parts due to the specific properties of logarithms and their derivatives.
To tackle logarithmic integration, remember that the derivative of \( \ln x \) is \( \frac{1}{x} \), a crucial piece of information used within integration by parts. This scenario showcases how understanding the basic properties of logarithms and their behavior within integrals can empower you to solve more complex calculus problems. Each part of the expression— the logarithm and the polynomial— requires distinct steps to simplify it before the integration can proceed smoothly.

Polynomial Integration

Polynomial integration refers to finding the integral of polynomial expressions, such as \( x^2 \). These are generally straightforward problems in integral calculus, and they often serve as a building block for more complicated integrals like the one in this problem.
For polynomial integration, the key principle is that each term of the polynomial is integrated separately. With the power rule of integration, it's simple to find the integral of any polynomial term: increase the exponent by one and divide by the new exponent. For instance, integrating \( x^2 \) involves elevating the power to 3 and dividing by 3, resulting in \( \frac{x^3}{3} \).

• Step 1: Increase the exponent by 1.
• Step 2: Divide by the new exponent.

These steps are applied repeatedly in calculus to integrate polynomial functions quickly and efficiently. Understanding the routine nature of polynomial integration can help demystify more complex problems by breaking them down into simpler parts.