Решите интегралы очень подробно 7 штук решение фотографиеем отправляем

Ответы

MomoiSatsuki16 18.02.2021 05:47

$\int\limits ^{2 } _ { - 1}2dx = 2x| ^{ 2} _ { - 1} = 2(2 - ( - 1)) = 2 \times 3 = 6 \\$

$\int\limits ^{ 2} _ { - 2 }(3 - x)dx = (3x - \frac{ {x}^{2} }{2}) | ^{ 2} _ { - 2} = \\ = 6 - \frac{ {2}^{2} }{2} - ( - 6 - \frac{ {2}^{2} }{2} ) = \\ = 6 - 2 + 6 + 2 = 12$

$\int\limits ^{3 } _ { 1}( {x}^{2} - 2x)dx = ( \frac{ {x}^{3} }{3} - \frac{2 {x}^{2} }{2}) | ^{3 } _ {1} = \\ = ( \frac{ {x}^{3} }{3} - {x}^{2} ) | ^{3 } _ {1} = \\ = \frac{ {3}^{3} }{3} - {3}^{2} - ( \frac{1}{3} - 1) = \\ = 9 - 9 - \frac{1}{3} + 1 = \frac{2}{3}$

$\int\limits ^{1 } _ { - 1}(2x - 3 {x}^{2} ) = ( \frac{2 {x}^{2} }{2} - \frac{3 {x}^{3} }{3} ) | ^{ 1} _ { - 1} = \\ =( {x}^{2} - {x}^{3} ) | ^{ 1} _ { - 1} = \\ = 1 - 1 - (1 + 1) = - 2$

$\int\limits ^{ 8} _ { 1} \sqrt[3]{x} dx =\int\limits ^{8 } _ { 1} {x}^{ \frac{1}{3} }dx = \frac{ {x}^{ \frac{4}{3} } }{ \frac{4}{3} } | ^{8 } _ {1} = \\ = \frac{3}{4}x \sqrt[3]{x} | ^{8 } _ {1} = \\ = \frac{3}{4} (8 \times \sqrt[3]{8} - 1) = \frac{3}{4} (16 - 1) = \\ = \frac{3}{4} \times 15 = \frac{45}{4}$

$\int\limits ^{2 } _ { 1} \frac{dx}{ {x}^{3} } = \int\limits ^{ 2} _ {1 } {x}^{ - 3}dx = \frac{ {x}^{ - 2} }{ - 2} | ^{2 } _ {1} = \\ = - \frac{1}{2 {x}^{2} } | ^{ 2} _ {1} = - \frac{1}{2} ( \frac{1}{ {2}^{2} } - 1) = \\ = - \frac{1}{2} \times ( \frac{1}{4} - 1) = \\ = - \frac{1}{2} \times ( - \frac{3}{4} ) = \frac{3}{8}$

$\int\limits ^{ \frac{\pi}{2} } _ { - \frac{\pi}{2} } \cos(x)dx = \sin(x) | ^{ \frac{\pi}{2} } _ { - \frac{\pi}{2} } = \\ = \sin( \frac{\pi}{2} ) - \sin( - \frac{\pi}{2} ) = \\ = 1 - ( - 1) = 2$