a) ∫ 3 cos xdx π 3. 2 2 b)∫(1-) dx 0

a) ∫ 3 cos x dx

• Шаг 1: Вычисляем интеграл.

\[ ∫_{0}^{π/2} 3 cos(x) dx = 3 ∫_{0}^{π/2} cos(x) dx \] \[ = 3 [sin(x)]_{0}^{π/2} \] \[ = 3 (sin(π/2) - sin(0)) \] \[ = 3 (1 - 0) \] \[ = 3 \]

b) ∫ (1 - x/2)⁴ dx

• Шаг 1: Вычисляем интеграл.

\[ ∫_{0}^{2} (1 - \frac{x}{2})^4 dx \] Сделаем замену переменных: \(u = 1 - \frac{x}{2}\), тогда \(du = -\frac{1}{2} dx\) и \(dx = -2 du\). Когда \(x = 0\), \(u = 1 - \frac{0}{2} = 1\). Когда \(x = 2\), \(u = 1 - \frac{2}{2} = 0\). Новый интеграл: \[ ∫_{1}^{0} u^4 (-2) du = -2 ∫_{1}^{0} u^4 du \] \[ = 2 ∫_{0}^{1} u^4 du \] \[ = 2 \left[ \frac{u^5}{5} \right]_{0}^{1} \] \[ = 2 \left( \frac{1^5}{5} - \frac{0^5}{5} \right) \] \[ = 2 \left( \frac{1}{5} - 0 \right) \] \[ = \frac{2}{5} \cdot 8 = \frac{16}{5} \]