Question

Задание на фотографии прикрепленной

Answer 1

1.

$F(x) = \sqrt{x} + c$

A(1;0)

$\sqrt{1} + c = 0 \\ 1 + c = 0 \\ c = - 1$

Первообразная, проходящая через точку A

$F(x) = \sqrt{x} - 1$

2.

$\int \frac{dx}{ \sin^{2} (2 - 3x) } = - \frac{1}{3} \times ( - \cot(2 -3x)) = \frac{ \cot(2 - 3x) }{3} + c$

3.

$\int _{1}^{2} (3 {x}^{2} - 4x - \frac{2}{ {x}^{2} } )dx = (3 \times \frac{ {x}^{3} }{3} - 4 \times \frac{ {x}^{2} }{2} + \frac{2}{x} )|_{1}^{2} = ( {x}^{3} - 2 {x}^{2} + \frac{2}{x} )|_{1}^{2} = {2}^{3} - 2 \times {2}^{2} + \frac{2}{2} - ( {1}^{3} - 2 \times {1}^{2} + \frac{2}{1} ) = 8 - 8 + 1 - (1 - 2 + 2) = 1 - 1 = 0$

4.

$- {x}^{2} - 4x = 4 + x \\ - {x}^{2} - 5x - 4 = 0 \\ {x}^{2} + 5x + 4 = 0 \\ (x + 4)(x + 1) = 0 \\ x _{1} = - 4 \: \: \: \: \: \: \: \: \: \: x_{2} = - 1$

$\int _{ - 4} ^{ - 1} (- 4 - {x}^{2} - (4 + x)) dx = \int _{ - 4} ^{ - 1} (- 4 - {x}^{2} - 4 - x)dx = \int _{ - 4} ^{ - 1}( - {x}^{2} - x - 8)dx = ( - \frac{ {x}^{3} }{3} - \frac{ {x}^{2} }{2} - 8x)| _{ - 4} ^{ - 1} = - \frac{ {( - 1)}^{3} }{3} - \frac{ {( - 1)}^{2} }{2} - 8 \times ( - 1) - ( - \frac{ {( - 4)}^{3} }{3} - \frac{ {( - 4)}^{2} }{2} - 8 \times ( - 4)) = \frac{1}{3} - \frac{1}{2} + 8 - ( \frac{64}{3} - \frac{16}{2} + 32 )= \frac{1}{3} - \frac{1}{2} + 8 - \frac{64}{3} + 8 - 32 = - \frac{63}{3} - \frac{1}{2} - 16 = - 21 - \frac{1}{2} - 16 = - 37 - \frac{1}{2} = - 37.5$