Question

1 и 2 пожалуйста....

Answer 1

Ответ:

1.

$F(x) = \int\limits(3 {x}^{2} - 2x + 3)dx = \\ = \frac{3 {x}^{3} }{3} - \frac{2 {x}^{2} }{2} + 3x +C = \\ = {x}^{3} - {x}^{2} + 3x + C$

- общий вид

В точке М:

$- 3 = {1}^{3} - {1}^{2} + 3 \times 1 + C \\ C= - 3 - 3 = - 6$

$f(x) = {x}^{3} - {x}^{2} + 3x - 6$

2.

$\int\limits^{ 2 } _ {1}(2x - \frac{1}{ {x}^{2} })dx = \int\limits^{ 2 } _ {1}(2x - {x}^{ - 2} )dx = \\ = (\frac{2 {x}^{2} }{2} - \frac{ {x}^{ - 1} }{( - 1)})| ^{2} _ {1} = ( {x}^{2} + \frac{1}{x}) | ^{2} _ {1} = \\ = {2}^{2} + \frac{1}{2} - 1 - 1 = 4.5 - 2 = 2.5$

$\int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{dx}{ \sin {}^{2} (x) } = - ctgx | ^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } = \\ = - ctg( \frac{\pi}{3} ) + ctg( \frac{\pi}{4} ) = \\ = - \frac{ \sqrt{3} }{3} + 1 = 1 - \frac{ \sqrt{3} }{3}$