Question

Решить двойные интегралы

Answer 1

Ответ:

1

$\int\limits^{ 2 } _ {1}\int\limits^{ 2 } _ {1}(x + y)dxdy = \int\limits^{ 2 } _ {1}dx\int\limits^{ 2 } _ {1}(x + y)dy = \\ = \int\limits^{ 2 } _ {1}dx \times (xy + \frac{ {y}^{2} }{2} )|^{ 2 } _ {1} = \int\limits^{ 2 } _ {1}dx \times (2x + \frac{4}{2} - x - \frac{1}{2} ) = \\ = \int\limits^{ 2 } _ {1}(x + \frac{3}{2} )dx = ( \frac{ {x}^{2} }{2} + \frac{3x}{2} )|^{ 2 } _ {1} = \frac{ {x}^{2} + 3x}{2} |^{ 2 } _ {1} = \\ = \frac{1}{2} (4 + 6 - 1 - 3) = \frac{6}{2} = 3$

2

$\int\limits^{ 1 } _ {0}\int\limits^{ 3 } _ {1}(x + y)dxdy = \int\limits^{ 1} _ {0}dx\int\limits^{ 3 } _ {1}(x + y)dy = \\ = \int\limits^{ 1 } _ {0}dx \times (xy + \frac{ {y}^{2} }{2} )|^{ 3 } _ {1} = \int\limits^{ 1} _ {0}dx \times (3x + \frac{9}{2} - x - \frac{1}{2} ) = \\ = \int\limits^{ 1 } _ {0}(2x + 4 )dx = ( {x}^{2} + 4x)|^{ 1 } _ {0} = (1 + 4 - 0) = 5$

3

$\int\limits^{ 2 } _ {1}dx\int\limits^{ 1 } _ {0}(5x - y)dy = \int\limits^{ 2 } _ {1}dx \times (5xy - \frac{ {y}^{2} }{2})|^{ 1} _ {0} = \\ = \int\limits^{ 2 } _ {1}dx \times (5x - \frac{1}{2} - 0) = \int\limits^{ 2 } _ {1}(5x - \frac{1}{2} )dx = \\ = ( \frac{5 {x}^{2} }{2} - \frac{x}{2} )|^{ 2 } _ {1} = \frac{1}{2} (10 - 1 - \frac{5}{2} + \frac{1}{2} ) = \\ = \frac{1}{2} (9 - 2) = \frac{7}{2} = 3.5$