91Ó°ÊÓ

When we talk about the movement of charge through a wire, it's important to connect the concept of current with charge. Current, denoted by \( I(t) \), is essentially the rate at which charge flows through a point. To find the total charge that has passed, we need the integral of the current over a specific time period. Think of integration as a way to sum up tiny amounts of current over each instant between two times. By integrating current, we effectively combine all these tiny amounts of charge to find the total charge that has moved through the wire.

So, when you see a function like \( I = 3t^2 + 2t + 5 \), this function tells you how the current changes over time. To know how much charge has passed from time \( t = 0 \) to \( t = 2 \), you integrate this function over the interval \( [0, 2] \). That’s how you move from a rate of flow to an accumulated quantity.

Calculating charge from current involves clear steps:

• Identify the function for current, \( I(t) \).
• Set up the integral over the time interval of interest.
• Find the antiderivative of the current function.
• Evaluate the definite integral to find the total charge.

In the problem, our current function was \( I(t) = 3t^2 + 2t + 5 \). We want to convert this rate (current) into a total amount (charge) from time \( t = 0 \) to \( t = 2 \) seconds.

Breaking it down, we set up our integral:\[ Q = \int_{0}^{2} (3t^2 + 2t + 5) \, dt \]To do this, we first need the antiderivative, ensuring each term is correctly integrated. So, we obtain \( t^3 + t^2 + 5t \), which combines terms showing the cumulative effect of current over the interval. Using the limits 0 and 2, we substitute to compute charge \( Q \). Remember to evaluate at both bounds and subtract, giving us the net charge over the time period.

Evaluating a definite integral in the context of this problem means determining the total charge using precise boundaries, here between \( t = 0 \) and \( t = 2 \). After finding the antiderivative, which was \( t^3 + t^2 + 5t \), we apply these limits to get the exact result.

When evaluating:

• Substitute the upper limit \( t=2 \) into the antiderivative.
• Substitute the lower limit \( t=0 \) into the antiderivative.
• Subtract the lower limit result from the upper limit result.

This calculation process looks like:\[ Q = \left[(2^3 + 2^2 + 5 \times 2) - (0^3 + 0^2 + 5 \times 0)\right] \]Simplifying part by part, it flows as:\[ Q = [8 + 4 + 10] - [0] = 22 \, \text{C} \]The clear decision points in calculation illustrate how charges accumulate over time due to changing current guided by the integral boundaries.