Question

Вычислите определенный интеграл:​

Answer 1

а

$\int\limits^{ 0} _ { - 2}( {x}^{2} + 2x)dx = ( \frac{ {x}^{3} }{3} + \frac{2 {x}^{2} }{2} )|^{ 0 } _ { - 2} = \\ = ( \frac{ {x}^{3} }{3} + {x}^{2} )|^{ 0 } _ { - 2} = \\ = 0 - ( - \frac{8}{3} + 4) = - \frac{12 - 8}{3} = - \frac{4}{3}$

б

$\int\limits^{ 2 } _ {1}(4 \sqrt{x} + 2x)dx = \int\limits^{ 2 } _ {1}(4 {x}^{ \frac{1}{2} } + \frac{ {x}^{2} }{2} ) dz= \\ = ( \frac{4 {x}^{ \frac{3}{2} } }{ \frac{3}{2} } + {x}^{2} ) |^{ 2 } _ {1} = ( \frac{8}{3}x \sqrt{x} + {x}^{2} )|^{ 2 } _ {1} = \\ = \frac{8}{3} \times 2 \times \sqrt{2} + 4 - \frac{8}{3} - 1 = \\ = \frac{16 \sqrt{3} - 8}{3} + 3$

с

$\int\limits^{ 0} _ { - 1}( {x}^{3} + {x}^{2}) dx = ( \frac{ {x}^{4} }{4} + \frac{ {x}^{3} }{3} )| ^{ 0 } _ { - 1} = \\ = 0 - ( \frac{1}{4} - \frac{1}{3} ) = - \frac{3 - 4}{12} = \frac{1}{12}$

д

$\int\limits^{ 1} _ {0}( {x}^{ \frac{1}{3} } + x)dx = ( \frac{ {x}^{ \frac{4}{3} } }{ \frac{4}{3} } + \frac{ {x}^{2} }{2} )|^{ 1} _ {0} = \\ = ( \frac{3}{4}x \sqrt[3]{x} + \frac{ {x}^{2} }{2}) |^{ 1 } _ {0} = \\ = \frac{3}{4} + \frac{1}{2} - 0 = \frac{5}{4} =1.25$