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 $$t = 1$$ s to $$t = 2$$ s is ____________ C.

The relation between current $$I$$ and charge $$Q$$ is given by the formula $$I = \frac{dQ}{dt}$$ Integrating both sides with respect to time, we get $$Q = \int I\,dt \quad\quad -(1)$$

Here, the current is given as $$I = 3t^2 + 4t^3 \quad\quad -(2)$$

Substitute equation $$(2)$$ into equation $$(1)$$: $$Q = \int (3t^2 + 4t^3)\,dt$$

We split the integral into two parts: $$Q = \int 3t^2\,dt \;+\; \int 4t^3\,dt$$

Now we integrate each term separately. Using the power rule $$\int t^n\,dt = \frac{t^{n+1}}{n+1}$$, we have:

$$\int 3t^2\,dt = 3 \cdot \frac{t^{3}}{3} = t^3$$ $$\int 4t^3\,dt = 4 \cdot \frac{t^{4}}{4} = t^4$$

Therefore, the antiderivative is $$Q(t) = t^3 + t^4 + C_1$$ where $$C_1$$ is the constant of integration. Since we are calculating the definite charge flow between two times, the constant cancels out.

The amount of charge flowing from $$t = 1\text{ s}$$ to $$t = 2\text{ s}$$ is given by the definite integral: $$Q_{1 \to 2} = \Bigl[t^3 + t^4\Bigr]_{1}^{2} \quad\quad -(3)$$

Evaluate at the upper limit $$t = 2$$: $$2^3 + 2^4 = 8 + 16 = 24$$

Evaluate at the lower limit $$t = 1$$: $$1^3 + 1^4 = 1 + 1 = 2$$

Subtract to find the net charge: $$Q_{1 \to 2} = 24 - 2 = 22\text{ C}$$

Final Answer: The amount of electric charge that flows during $$t=1\text{ s}$$ to $$t=2\text{ s}$$ is $$22\text{ C}$$.