91Ó°ÊÓ

When dealing with integrals, especially those like \( \int u^n \, du \), the power rule for integration becomes extremely useful. This rule states:

• \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \)

Where \( n \) is any real number except -1, and \( C \) represents the constant of integration. The power rule is derived from the reverse process of differentiation and allows us to handle polynomials and other expressions raised to a power.
In our example, the expression \( \frac{4}{u^3} \) is rewritten as \( 4u^{-3} \), making it a candidate for the power rule. Applying the rule, we integrated \( u^{-3} \) to get \( \frac{u^{-2}}{-2} = -\frac{1}{2u^{2}} \). So, \( 4 \times -\frac{1}{2u^{2}} \) simplifies to \( -\frac{2}{u^2} \). This straightforward application of the power rule demonstrates the efficiency in simplifying the integration process.

To evaluate a definite integral like \( \int_{1}^{2} f(u) \, du \), applying the limits of integration is crucial. These limits indicate the interval over which the function \( f(u) \) is integrated.
In essence, they transform an indefinite integral into a specific numerical value.

• Substitute the upper limit into the antiderivative: \( F(b) \)
• Substitute the lower limit into the antiderivative: \( F(a) \)
• Find the difference \( F(b) - F(a) \)

In our integral, we determined the antiderivative, \( -\frac{2}{u^2} + \ln|u| \), and then applied the limits from 1 to 2:
Evaluate at \( u=2 \) yields \( -\frac{1}{2} + \ln(2) \), and at \( u=1 \) gives \( -2 + 0 \).
Taking their difference, the expression simplifies to \[ -\frac{1}{2} + \ln(2) + 2 \], ultimately helping us to arrive at the final answer.

Integration involving the natural logarithm often emerges when dealing with the reciprocal function \( \frac{1}{u} \). Its integration yields the natural logarithm function. Specifically:

• \( \int \frac{1}{u} \, du = \ln|u| + C \)

This is because the derivative of \( \ln|u| \) is \( \frac{1}{u} \), which aligns with our integral. It works under the assumption that \( u eq 0 \), as the natural logarithm requires positivity or a negative value for absolute value reasoning.
In the provided exercise, the term \( \frac{1}{u} \) naturally integrates to \( \ln|u| \). Here, we recognize the fundamental connection between the function and its antiderivative across continuous intervals, blending seamlessly with other terms we calculated using different techniques in the problem.