91Ó°ÊÓ

Definite integrals allow us to calculate the accumulation of quantities, like area under curves. Unlike indefinite integrals, which include a constant of integration, definite integrals provide a specific numeric result. In this exercise, the definite integral is evaluated between the limits 1 and 2.

This process involves several steps:

• First, compute the antiderivative of the integrand function (the function you're integrating).
• Apply the limits by substituting the upper limit value, then the lower limit value, and finally subtract the two results.
• Using these steps gives a numerical answer, providing the exact accumulated value of the function over the specified interval.

For example, the integral \( \int_{1}^{2} x^2 \ln x \, dx \) was calculated by finding an antiderivative and evaluating it from 1 to 2, resulting in a precise value: \( \frac{8}{3} \ln 2 - \frac{7}{9} \). This tells us the accumulated area under the curve from \(x = 1\) to \(x = 2\). Understanding definite integrals in this context helps us grasp various real-world applications, like calculating distances or volumes.

Logarithmic functions often appear in calculus, especially when dealing with integrals and derivatives. The natural logarithm function, denoted by \( \ln x \), is particularly common because it elegantly simplifies many calculus tasks.

The derivative of a logarithmic function, like \( \ln x \), is \( \frac{1}{x} \). This property helps when integrating by parts as it makes differentiation straightforward. However, integrating logarithmic functions directly can be challenging. That's where techniques like integration by parts become essential.

• Logarithmic functions are paired well with polynomial expressions using integration by parts due to their simple derivatives.
• Using these functions requires careful handling, as they behave differently near their asymptotes and zero points.

In this exercise, \( \ln x \) was chosen as \(u\) because its derivative \(du\) \(= \frac{1}{x} dx\) is manageable, allowing integration by parts to simplify the original integral effectively.

Integration by parts is a powerful technique in calculus useful for integrating the product of two functions. The formula used is \( \int u \, dv = uv - \int v \, du \).

To effectively use this method:

• Identify two parts of the integrand: choose \(u\) and \(dv\). Typically, choose \(u\) as the part that becomes simpler when differentiated, like \( \ln x \) in this exercise.
• Differentially derive \(u\), and integrate \(dv\) to find \(v\).
• Apply the formula by multiplying \(u\) and \(v\), then subtracting the integral of \(v \, du\).

In this example, choosing \(u = \ln x\) and \(dv = x^2 dx\), we derived \(du = \frac{1}{x} dx\) and found \(v = \frac{x^3}{3}\). Substituting these into the formula, the integral was split into manageable parts, allowing successful computation of the definite integral. This is a testament to the versatility and importance of integration by parts in tackling complex integral problems.