Question

Помогите решить дам 30 баллов

Answer 1

Ответ:

1.

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

A(-2;1)

$1 = - \frac{1}{2} \times {( - 2)}^{3} + \frac{3}{2} \times {( - 2)}^{2} - 14 + C \\ 1 = 4 + 6 - 14 + C\\ C = 1 + 4 = 5$

$F(x) = - \frac{ {x}^{4} }{2} + \frac{3 {x}^{2} }{2} + 7x + 5$

2.

$F(x) = \int\limits( {x}^{4} - 2)dx = \frac{ {x}^{5} }{5} - 2x + C$

M(0;2)

$2 = 0 - 0 + C \\ C = 2$

$F(x) = \frac{ {x}^{5} }{5} - 2x + 2$

3.

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

B(-1;1)

$- 1 = \frac{1}{3} + \frac{3}{2} - 4 + c \\ C = 3 - \frac{1}{3} - \frac{3}{2} = \frac{18 - 2 - 9}{6} = \frac{7}{6}$

$F(x) = \frac{ {x}^{3} }{3} + \frac{3 {x}^{2} }{2} - 4x + \frac{7}{6}$

a)

$\int\limits(5 {x}^{4} - 4x + 3)dx = \frac{5 {x}^{5} }{5} - \frac{4 {x}^{2} }{2} + 3x + C = \\ = {x}^{5} - 2 {x}^{2} + 3x + C$

б)

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

в)

$\int\limits \frac{x \sqrt[5]{ {x}^{3} } }{ \sqrt{x} } dx = \int\limits {x}^{1 + \frac{3}{5} - \frac{1}{2} } dx = \\ = \int\limits {x}^{ \frac{10 + 6 - 5}{10} } dx = \int\limits {x}^{ \frac{11}{10} } dx = \\ = \frac{ {x}^{ \frac{21}{10} } }{ \frac{21}{10} } + C= \frac{10}{21} {x}^{2} \sqrt[10]{x} + C$

г)

$\int\limits \frac{ \sqrt{x} \sqrt[5]{x} }{ \sqrt[3]{x} } dx = \int\limits {x}^{ \frac{1}{2} + \frac{1}{5} - \frac{1}{3} } dx = \\ = \int\limits {x}^{ \frac{15 + 6 - 10}{30} } dx = \int\limits {x}^{ \frac{11}{30} } dx = \\ = \frac{ {x}^{ \frac{43}{30} } }{ \frac{41}{30} } + C= \frac{30}{41} x \sqrt[30]{ {x}^{11} } + C$