For this interval, the current is constant, so the area under the curve is simply a rectangle:
\[ Q_1 = I \times \Delta t = 2 \times (3 - 1) = 4 \, \text{C} \]
For this interval, the current-time graph forms a triangle. The area of a triangle is given by:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values (base = 2 s, height = 2 A):
\[ Q_2 = \frac{1}{2} \times (6 - 4) \times 2 = 2 \, \text{C} \]
From the calculations above:
Thus, we have: \[ Q_1 > Q_2 \]