91Ó°ÊÓ

In calculus, finding the antiderivative is like doing a reverse operation of differentiation. When you're given a function, like the second derivative of a function, you're tasked with finding its primitive function(s). In the context of our exercise, we start by determining the antiderivative of the second derivative \( f''(t) = \frac{3}{\sqrt{t}} \). To achieve this, we rewrite the function as \( 3t^{-1/2} \) to facilitate simple integration. This rewriting allows us to easily integrate it to find the first derivative \( f'(t) \).

• The antiderivative of \( 3t^{-1/2} \) is found by using the power rule of integration, resulting in \( 6t^{1/2} + C_1 \), where \( C_1 \) is a constant.
• This process continues as we find another antiderivative to reach the function \( f(t) \), starting from \( f'(t) = 6t^{1/2} - 5 \), leading to \( f(t) = 4t^{3/2} - 5t + C_2 \).

Antiderivatives give us a family of solutions, showing us how a function accumulates over time or space. It is important to note that integrating does not produce a single solution but a set of possible functions, each shifted by a constant.

Initial conditions allow us to pin down a specific solution from the family of solutions given by the antiderivative. When you integrate a function, you arrive at a general solution with unknown constants, like \( C_1 \) and \( C_2 \) in our example. Initial conditions are specific points or values that the function must satisfy, which help immensely in finding these constants.

• Consider \( f'(4) = 7 \), an initial condition that determines \( C_1 \). By substituting \( t = 4 \) in \( f'(t) = 6t^{1/2} + C_1 \), you can solve for \( C_1 \).
• Similarly, the condition \( f(4) = 20 \) allows us to determine \( C_2 \) by substituting \( t = 4 \) in \( f(t) = 4t^{3/2} - 5t + C_2 \).

With these initial conditions, we're obtaining particular solutions, anchoring our potential family of solutions to a precise one that meets the problem's requirements.

The constant of integration is an integral component when you solve antiderivatives. Each antiderivative operation introduces a constant (like \( C_1 \) and \( C_2 \) in our solution) because integration is essentially 'losing' derivative information, as differentiation reduces dimensional specifics.

• For each integration performed, a constant is added, representing that the antiderivative could be shifted in the vertical or horizontal space.
• For the problem we're working on, these constants are crucial to finding solutions that meet our initial conditions.

These constants ensure the function remains as flexible as needed until further details (initial conditions) narrow down the choices. Without constants, integrations would yield incomplete, inaccurate representations of potential functions, making problem-solving with integrations ineffective.