The current in a conductor is expressed as `I = 3t^2 + 4t^3`, where I is in Ampere and t is in second. The amount of electric charge that flows through a section of the conductor during t = 1s to t = 2s is ____________ C.

Answer

To find the amount of electric charge that flows through a section of the conductor from \( t = 1 \, \text{s} \) to \( t = 2 \, \text{s} \), we start with the given expression for current:

\[ I = 3t^2 + 4t^3 \]

### Step 1: Relate current to charge
The relationship between current \( I \) and charge \( Q \) is given by:

\[ I = \frac{dQ}{dt} \]

This means that the charge \( Q \) can be found by integrating the current with respect to time \( t \):

\[ Q = \int I \, dt \]

### Step 2: Set up the integral
We need to find the charge that flows from \( t = 1 \, \text{s} \) to \( t = 2 \, \text{s} \). Therefore, we set up the integral:

\[ Q = \int_{1}^{2} (3t^2 + 4t^3) \, dt \]

### Step 3: Integrate the current expression
Now we will integrate the expression \( 3t^2 + 4t^3 \):

\[ \int (3t^2 + 4t^3) \, dt = \int 3t^2 \, dt + \int 4t^3 \, dt \]

Calculating the integrals separately:

1. \( \int 3t^2 \, dt = 3 \cdot \frac{t^3}{3} = t^3 \)
2. \( \int 4t^3 \, dt = 4 \cdot \frac{t^4}{4} = t^4 \)

Combining these results, we have:

\[ \int (3t^2 + 4t^3) \, dt = t^3 + t^4 \]

### Step 4: Evaluate the definite integral
Now we evaluate this from \( t = 1 \) to \( t = 2 \):

\[ Q = \left[ t^3 + t^4 \right]_{1}^{2} = \left( 2^3 + 2^4 \right) - \left( 1^3 + 1^4 \right) \]

Calculating the limits:

- Upper limit: \( 2^3 + 2^4 = 8 + 16 = 24 \)
- Lower limit: \( 1^3 + 1^4 = 1 + 1 = 2 \)

Thus, we have:

\[ Q = 24 - 2 = 22 \, \text{C} \]

### Final Answer
The amount of electric charge that flows through the section of the conductor during the time interval from \( t = 1 \, \text{s} \) to \( t = 2 \, \text{s} \) is:

\[ \boxed{22 \, \text{C}} \]