Решение:
а) \( \int_{-1}^{2} (3x^2 + 2x) dx \)
= \( [x^3 + x^2]_{-1}^{2} \)
= \( (2^3 + 2^2) - ((-1)^3 + (-1)^2) \)
= \( (8 + 4) - (-1 + 1) \)
= \( 12 - 0 = 12 \)
б) \( \int_{1}^{4} (4x^3 + 6x) dx \)
= \( [x^4 + 3x^2]_{1}^{4} \)
= \( (4^4 + 3 \cdot 4^2) - (1^4 + 3 \cdot 1^2) \)
= \( (256 + 48) - (1 + 3) \)
= \( 304 - 4 = 300 \)