254. ∫₁⁵ ((x-3)²-4) dx.

Ответ: -28/3

\[\int_1^5 ((x-3)^2 - 4) dx = \int_1^5 (x^2 - 6x + 9 - 4) dx = \int_1^5 (x^2 - 6x + 5) dx = (\frac{x^3}{3} - 3x^2 + 5x) \vert_1^5 = (\frac{5^3}{3} - 3 \cdot 5^2 + 5 \cdot 5) - (\frac{1^3}{3} - 3 \cdot 1^2 + 5 \cdot 1) = (\frac{125}{3} - 75 + 25) - (\frac{1}{3} - 3 + 5) = (\frac{125}{3} - 50) - (\frac{1}{3} + 2) = (\frac{125 - 150}{3}) - (\frac{1 + 6}{3}) = -\frac{25}{3} - \frac{7}{3} = -\frac{32}{3}\]

Ответ: -32/3