Question

Буду дуже вдячна за допомогу

Answer 1

Ответ:

$∫^{4}_{ - 4} \sqrt{ {4}^{2} - {x}^{2} }dx = \frac{1}{2} S_{круга} = \frac{1}{2}\pi R^{2} = \frac{1}{2} \pi {4}^{2} = 8\pi$

.....

$∫_{0}^{ \frac{\pi}{6} } 2cosxdx = 2sinx | ^{ \frac{\pi}{6} }_{0} = \\ = 2sin \frac{\pi}{6} - 2sin0 = 2 \times \frac{1}{2} - 0 = 1$