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368468
@368468
November 2021
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Найдите площадь заштрихованной фигуры
Ответ:52/3
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11qqaa
Из графика видно точки пересечения линий (-1; 3) (1; 7) (3; 3)
S1 ищем на отрезке [-1; 1]
S2 ищем на отрезке [1; 3]
S = S1+S2 = [-1; 1] ∫ (2x+5 -(x^2-2x)) dx + [1; 3] ∫ (x^2 -6x +12 -(x^2-2x)) dx =
= [-1; 1] ∫ ( -x^2 +4x +5 ) dx + [1; 3] ∫ ( -4x +12 ) dx =
= ( -x^3/3 +2x^2 +5x) [-1; 1] + ( -2x^2 +12x ) [1; 3] =
= ( -1^3/3 +2*1^2 +5*1) - ( -(-1)^3/3 +2*(-1)^2 +5*(-1)) +
+ ( -2*3^2 +12*3 ) - ( -2*1^2 +12*1 ) = 20/3 + 8/3 + 18 - 10 =
= 28/3 + 8 = 52/3
ответ 52/3
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Answers & Comments
S1 ищем на отрезке [-1; 1]
S2 ищем на отрезке [1; 3]
S = S1+S2 = [-1; 1] ∫ (2x+5 -(x^2-2x)) dx + [1; 3] ∫ (x^2 -6x +12 -(x^2-2x)) dx =
= [-1; 1] ∫ ( -x^2 +4x +5 ) dx + [1; 3] ∫ ( -4x +12 ) dx =
= ( -x^3/3 +2x^2 +5x) [-1; 1] + ( -2x^2 +12x ) [1; 3] =
= ( -1^3/3 +2*1^2 +5*1) - ( -(-1)^3/3 +2*(-1)^2 +5*(-1)) +
+ ( -2*3^2 +12*3 ) - ( -2*1^2 +12*1 ) = 20/3 + 8/3 + 18 - 10 =
= 28/3 + 8 = 52/3
ответ 52/3