Question

Вычислить площади фигур ограниченных графиками функции !!!!Очень срочно ​

Answer 1

Будем искать сумму площадей:

$S = S_a + S_b$

$S_a = S_1 - S_2 - S_3$

$S_b = S_4 - S_5$

$S_1 = \int\limits^{ - 3 } _ { - 6}( \frac{4}{3} x + 13)dx = ( \frac{4 {x}^{2} }{3 \times 2} + 13x) | ^{ - 3} _ { - 6} = \\ = ( \frac{2 {x}^{2} }{3} + 13x) | ^{ - 3} _ { - 6} = \\ = \frac{2}{3} \times 9 - 39 - ( \frac{2}{3} \times 36 - 78) = \\ = 6 - 39 - 24 + 78 = 21$

$S_2= \int\limits^{ - 4 } _ { - 6} ( \frac{ {x}^{2} }{4} - 4)dx |^{ 2 } _ {1} = ( \frac{ {x}^{3} }{12} - 4x) | ^{ - 4 } _ { - 6} = \\ = - \frac{ {4}^{3} }{12} + 16 - ( - \frac{ {6}^{3} }{12} + 24) = \\ = - \frac{16}{3} + 16 + 18 - 24 = 10 - \frac{16}{3} = \frac{14}{3}$

$S_3 = \int\limits^{ - 3 } _ { - 4}(2x + 8)dx = ( {x}^{2} + 8x) | ^{ - 3} _ { - 4} = \\ = 9 - 24 - (16 - 32) = \\ = - 15 + 16 = 1$

$S_a = 21 - \frac{14}{3} - 1 = 20 - \frac{14}{3} = \frac{46}{3}$

_________________________

$S_4 = \int\limits^{ - 2 } _ { - 3} {x}^{2} dx = \frac{ {x}^{3} }{3} | ^{ - 2 } _ { - 3} = \\ = - \frac{8}{3} + 9 = \frac{19}{3}$

$S_5 = \int\limits^{ - 2 } _ { - 3}(2x + 8)dx = ( {x}^{2} + 8x) | ^{ - 2 } _ { - 3} = \\ = 4 - 16 - (9 - 24) = - 12 + 15 = 3$

$S_b = \frac{19}{3} - 3 = \frac{10}{3}$

_____________

$S= S_a + S_b = \frac{46}{3} + \frac{10}{3} = \frac{56}{3} = 18 \frac{2}{3}$