To solve the problem, we need to find the electric charge that flows through a section of the conductor between \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \) given the current \( I(t) = 3t + 4t^2 \).
### Step-by-Step Solution:
1. **Understand the relationship between current and charge**:
The current \( I \) is defined as the rate of flow of charge \( Q \) with respect to time \( t \). Mathematically, this is expressed as:
\[
I = \frac{dQ}{dt}
\]
From this, we can express the charge \( Q \) as:
\[
dQ = I \, dt
\]
2. **Set up the integral for charge**:
To find the total charge \( Q \) that flows through the conductor from \( t = 1 \, \text{s} \) to \( t = 3 \, \text{s} \), we integrate the current function over this time interval:
\[
Q = \int_{t=1}^{t=3} I(t) \, dt
\]
Substituting the expression for \( I(t) \):
\[
Q = \int_{1}^{3} (3t + 4t^2) \, dt
\]
3. **Perform the integration**:
We can split the integral into two parts:
\[
Q = \int_{1}^{3} 3t \, dt + \int_{1}^{3} 4t^2 \, dt
\]
Now, we calculate each integral separately.
- For the first integral:
\[
\int 3t \, dt = \frac{3t^2}{2}
\]
Evaluating from 1 to 3:
\[
\left[ \frac{3(3)^2}{2} - \frac{3(1)^2}{2} \right] = \left[ \frac{27}{2} - \frac{3}{2} \right] = \frac{24}{2} = 12
\]
- For the second integral:
\[
\int 4t^2 \, dt = \frac{4t^3}{3}
\]
Evaluating from 1 to 3:
\[
\left[ \frac{4(3)^3}{3} - \frac{4(1)^3}{3} \right] = \left[ \frac{108}{3} - \frac{4}{3} \right] = \frac{104}{3}
\]
4. **Combine the results**:
Now, we add the results of the two integrals:
\[
Q = 12 + \frac{104}{3}
\]
To combine these, we convert 12 into a fraction:
\[
12 = \frac{36}{3}
\]
Thus,
\[
Q = \frac{36}{3} + \frac{104}{3} = \frac{140}{3}
\]
5. **Final answer**:
The total charge that flows through the section of the conductor between \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \) is:
\[
Q = \frac{140}{3} \, \text{C}
\]
