91Ó°ÊÓ

In calculus, u-substitution is a technique used for simplifying the process of integration. It can be particularly useful when dealing with complex integrals. Instead of integrating the function directly, you select part of the expression within the integrant to substitute with a new variable 'u'. This substitution converts the problem into an easier form.
For example, in our problem, we have the function \( g'(t) = \frac{5.45218}{(1+1.09t)^{0.9}} \). Here, we set \( u = (1+1.09t) \). This substitution makes differentiation straightforward: \( \frac{du}{dt} = 1.09 \) so \( dt = \frac{du}{1.09} \). After these substitutions, the integral simplifies to a more workable form, allowing us to focus on integrating with respect to 'u' instead of 't'. This step transforms the integral into \( \int \frac{5.45218}{u^{0.9}} \frac{du}{1.09} \). This makes the whole solving process much simpler and clearer.

The power rule for integration is an essential concept in calculus that helps in solving integrals involving polynomials. It is a key formula that derives from the rule used in differentiation, adjusted for integration.
The basic formula is \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). This formula was applied after performing the u-substitution. We reached the point where we needed to integrate \( u^{-0.9} \). According to the power rule, the integration of \( u^{-0.9} \) becomes: \( \frac{u^{(-0.9 + 1)}}{-0.9 + 1} \), which simplifies to \( \frac{u^{0.1}}{0.1} \).
This computation leads us to find the antiderivative required to solve the given integral. The power rule thus turns an intimidating problem into a series of simple calculations.

Life expectancy models use mathematical equations to predict the average time individuals are expected to live, using factors like time or demographics. In this exercise, we are dealing with a life expectancy model that involves integration over time.
The model provides a function \( g(t) \), representing the life expectancy at birth of a female, given a rate of change \( g'(t) \). To gain insight into life expectancy at different years, the function \( g(t) = C + \frac{5.45218}{1.09} \frac{u^{0.1}}{0.1} \) was derived by integrating the rate function \( g'(t) \).
This model allows us to predict life expectancy at any given year, measured in this context since the year 1900. By substituting \( t \) with specific years, like 2000 (\( t = 100 \)), we use the model to determine the expected life spans and study demographic trends.

In integration, the constant of integration is denoted as 'C' and represents the family of solutions for indefinite integrals. Every time we perform an integration without limits, there might be an added constant, because when you differentiate back, constants disappear.
In the problem, once we found the antiderivative, we needed to solve for the constant of integration. Initial conditions often determine the value of 'C'. We used the given information that life expectancy at the start of 1900 (when \( t = 0 \)) was 50.02 years to find \( C \). Plugging this into the equation allowed us to solve for \( C \) as shown: \( C = 50.02 - \frac{5.45218}{1.09 \times 0.1} \).
This step ensures that the function \( g(t) \) correctly models the life expectancy over time, reflecting the given historical data.