Question

Помогите выполнить работу,по теме "Интегралы" с 1 по 19 пункт

Answer 1

Ответ:

$1)\int\limits \: xdx = \frac{ {x}^{2} }{2} + c$

$2)\int\limits {x}^{3} dx = \frac{ {x}^{4} } {4} + c$

$3)\int\limits {x}^{6} dx = \frac{ {x}^{7} }{7} + c$

$4)\int\limits3dx = 3x + c$

$5)\int\limits5xdx = \frac{5 {x}^{2} }{2} + c$

$6)\int\limits \frac{1}{3} {t}^{3}dt = \frac{ {t}^{4} }{12} + c$

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

$8)\int\limits(5x - {x}^{2} )dx = \frac{5 {x}^{2} }{2} - \frac{ {x}^{3} }{3} + c$

$9)\int\limits3(x - 3)dx =\int\limits(3x - 9) dx = \\ = \frac{3 {x}^{2} }{2} - 9x + c$

$10)\int\limits(4 {x}^{3} + 8x - 2)dx = \frac{4 {x}^{4} }{4} + \frac{8 {x}^{2} }{2} - 2x + c = \\ = {x}^{4} + 4 {x}^{2} - 2x + c$

$11)\int\limits {x}^{2} (1 + 4x)dx =\int\limits( {x}^{2} + 4 {x}^{3} )dx = \\ = \frac{ {x}^{3} }{3} + \frac{4 {x}^{4} }{4} + c = \frac{ {x}^{3} }{3} + {x}^{4} + c$

$12)\int\limits {(x - 2)}^{2} dx = \int\limits {(x - 2)}^{2} d(x - 2) = \\ = \frac{ {(x - 2)}^{3} }{3} + c$

$13)\int\limits4(3x - 2)^{2} dx = \int\limits4(9 {x}^{2} - 12x + 4)dx = \\ = \int\limits(36 {x}^{2} - 48x + 16)dx = \frac{36 {x}^{3} }{3} - \frac{48 {x}^{2} }{2} + 16x + c = \\ = 12 {x}^{3} - 24 {x}^{2} + 16x + c$

$14)\int\limits \: x {(5 - x)}^{2} dx = \int\limits \: x(25 - 10x + {x}^{2} )dx = \\ = \int\limits(25x - 10 {x}^{2} + {x}^{3} )dx = \\ = \frac{25 {x}^{2} }{2} - \frac{10 {x}^{3} } { 3} + \frac{ {x}^{4} }{4} + c$

$15)\int\limits2 {x}^{ \frac{1}{2} }dx = 2 \times \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } + c = \\ = \frac{4}{3} x \sqrt{x} + c$

$16)\int\limits {x}^ {\frac{2}{3} } dx = \frac{ {x}^{ \frac{5}{3} } }{ \frac{5}{3} } + c = \\ = \frac{3}{5} x \sqrt[3]{ {x}^{2} } + c$

$17)\int\limits {x}^{ - 2} dx = \frac{ {x}^{ - 1} }{ - 1} + c = - \frac{1}{x} + c$

$18)\int\limits {x}^{ - \frac{1}{2} } dx = \frac{ {x}^{ \frac{1}{2} } }{ \frac{1}{2} } +c = 2 \sqrt{x} + c$

$19)\int\limits {x}^{ - \frac{2}{3} } dx = \frac{ {x}^{ \frac{1}{3} } }{ \frac{1}{3} } +c = 3 \sqrt[3]{x} + c$