Question

ДАЮ 40 БАЛОВ СРОЧНО РЕШИТЕ!!!!!​

Answer 1

1.

$\int\limits^{ 4 } _ {1}(2x - \frac{1}{2 \sqrt{x} })dx = \int\limits^{ 4} _ {1}(2x - \frac{1}{2} {x}^{ - \frac{1}{2} }) dx = \\ = ( \frac{2 {x}^{2} }{2} - \frac{1}{2} \times \frac{ {x}^{ \frac{1}{2} } }{ \frac{1}{2} } )|^{ 4 } _ {1} = ( {x}^{2} - \sqrt{x} ) |^{ 4 } _ {1} = \\ = 16 - 2 - (1 - 1) = 14$

2.

$\int\limits^{ \frac{\pi}{3} } _ { \frac{}{6} } \frac{dx}{ \sin {\pi}^{2} (x) } = - ctgx|^{ \frac{\pi}{3} } _ { \frac{\pi}{6} } = - ctg \frac{\pi}{3} + ctg \frac{\pi}{6} = \\ = - \frac{ \sqrt{3} }{3} + \sqrt{3} = \frac{2 \sqrt{3} }{3}$

3.

$\int\limits^{ 1 } _ {0}(2 - \frac{5}{ \sqrt{5x + 4} } )dx = 2\int\limits^{ 1} _ {0}dx - \int\limits^{ 1 } _ {0} \frac{d(5x + 4)}{ {(5x + 4)}^{ \frac{1}{2} } } = \\ = (2x - \frac{ {(5x + 4)}^{ \frac{1}{2} } }{ \frac{1}{2} } ) |^{ 1 } _ {0}= (2x - 2 \sqrt{5x + 4}) |^{ 1 } _ {0} = \\ = 2 - 2 \times 3 - (0 - 4) = 2 - 6 + 4 = 0$