Вариант 1 , СРОЧНО Алгебра. Created by nactya201613. algebra-ru

Ответ:1.= > F(x) является первообразной для f(x).2.3.- общий видВ точке А:4.

Вариант 1 , СРОЧНО Алгебра...

nactya201613 @nactya201613

November 2021 1 19 Report

Вариант 1 , СРОЧНО
Алгебра

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Miroslava227

Ответ:

$F'(x) = 3 {x}^{2} + 0 = 3 {x}^{2} = f(x)$

= > F(x) является первообразной для f(x).

$F(x) = \int\limits{x}^{5} dx = \frac{ {x}^{6} }{6} + C\\$

$F(x) = \int\limits3 {x}^{2} dx = \frac{3 {x}^{3} }{3} + C = {x}^{3} + C\\$

- общий вид

В точке А:

$25 = {3}^{3} + C \\ C= 25 - 27 = - 2$

$F(x) = {x}^{3} - 2 \\$

$F(x) = \int\limits(2 \cos(2x) + 3 \sin(3x) )dx = \\ =\int\limits \cos(2x) d(2x) +\int\limits \sin(3x) d(3x) = \\ = \sin(2x) - \cos(3x) + C$

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