Question

Интегралы. Помогите !!

Answer 1

Ответ:

a

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

б

$\int\limits^{ 1 } _ { \frac{1}{2} }( \frac{2}{ {x}^{3} } + 8)dx = \int\limits^{ 1} _ { \frac{1}{2} }(2 {x}^{ - 3} + 8)dx = \\ = ( \frac{2 {x}^{ - 2} }{( - 2)} + 8x) | ^{1} _ { \frac{1}{2} } = ( - \frac{1}{ {x}^{2} } + 8x) | ^{1} _ { \frac{1}{2} } = \\ = - 1 + 8 - ( - 4 + 4) = 7$