The electric charge \(q\) is given by:
\[ q = \int_{t_1}^{t_2} I(t) \, dt \]
Substituting \(I(t) = 3t^2 + 4t\) and integrating from \(t = 1 \, \text{s}\) to \(t = 2 \, \text{s}\):
\[ q = \int_{1}^{2} (3t^2 + 4t) \, dt \]
Calculating the integral:
\[ q = \left[ t^3 + 2t^2 \right]_{1}^{2} = \left( 2^3 + 2 \times 2^2 \right) - \left( 1^3 + 2 \times 1^2 \right) \]
\[ q = (8 + 8) - (1 + 2) = 16 - 3 = 22 \, \text{C} \]
A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: