Question

ДАЮ 70 БАЛЛОВ. ПОМОГИТЕ ПОЖАЛУЙСТА!!!

Answer 1

Ответ:

2

1.

$f(x) = \int\limits(7 - 2x)dx = 7x - \frac{2 {x}^{2} }{2} +C = \\ = 7x - {x}^{2} + C$

2.

$f(x) = \int\limits(3 + 5x)dx = 3x + \frac{5 {x}^{2} }{2} + C$

3.

$f(x) = \int\limits(kx + b)dx = \frac{k {x}^{2} }{2} + bx + C$

4.

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

5.

$f(x) =\int\limits(4 - {x}^{3} )dx = 4x - \frac{ {x}^{4} }{4} + C$

6.

$f(x) = \int\limits( {x}^{2} + 4x - 7)dx = \\ = \frac{ {x}^{3} }{3} + \frac{4 {x}^{2} }{2} - 7x + C= \frac{ {x}^{3} }{3} + 2 {x}^{2} - 7x + C$

7.

$f(x) = \int\limits(a {x}^{2} + bx + c)dx = \\ = \frac{a {x}^{3} }{3} + \frac{b {x}^{2} }{2} + cx + C$

8.

$f(x) = \int\limits \frac{dx}{ {(3 + 2x)}^{4} } = \frac{1}{2} \int\limits \frac{d(2x)}{ {(3 + 2x)}^{4} } = \\ = \frac{1}{2} \int\limits {(3 + 2x)}^{ - 4}d (3 + 2x) = \\ = \frac{1}{2} \times \frac{ {(3 + 2x)}^{ - 3} }{( - 3)} + = \\ = - \frac{1}{6 {(3 + 2x)}^{3} } + C$

2

1.

$f(x) = \int\limits \sin(2x + 3) dx = \\ = \frac{1}{2} \int\limits \sin(2x + 3) d(2x + 3) = \\ = - \frac{1}{2} \cos(2x + 3) + C$

2.

$f(x) = \int\limits \cos(3x + 4) dx = \\ = \frac{1}{3} \cos(3x + 4) d(3x + 4) = \\ = \frac{1}{3} \sin(3x + 4) + C$

3.

$f(x) = \int\limits \cos( \frac{x}{2} - 1) dx= \\ = 2\int\limits \cos( \frac{x}{2} - 1) d( \frac{x}{2} - 1) = \\ = 2 \sin( \frac{x}{2} - 1) + C$

4.

$f(x) = 4 \int\limits \sin( \frac{x}{4} + 5 ) d( \frac{x}{4} + 5) = \\ = - 4 \cos( \frac{x}{4} ) + C$

5.

$f(x)\int\limits {e}^{ \frac{x + 1}{2} } dx = \int\limits {e}^{ \frac{x}{2} + \frac{1}{2} } dx = \\ =2 \int\limits {e}^{ \frac{x + 1}{2} } d( \frac{x + 1}{2} ) = \\ = 2 {e}^{ \frac{x + 1}{2} } + C$

6.

$f(x) = \frac{1}{3} \int\limits {e}^{3x - 5} d(3x - 5) = \frac{1}{3} {e}^{3x - 5} + C$

7.

$f(x)\int\limits \frac{1}{2x} dx = \frac{1}{2} \int\limits \frac{dx}{x} = \\ = \frac{1}{2} ln(x) + C$

8.

$f(x) = \frac{1}{3} \int\limits \frac{d(3x - 1)}{3x - 1} = \frac{1}{3} ln(3x - 1) + C$