найдем точки пересечения
x^2 - 4x + 3 = 8
x^2 - 4x -5=0
х= -1 х = 5
x^2 - 12x + 35 = 8
x^2 - 12x + 27=0
х = 3 х= 9
x^2 - 4x + 3 =x^2 - 12x + 35
8х = 32
х = 4
1) интеграл от 4 до 5 (8-(x^2 - 4x + 3 ))= 8х -x^3 /3 +2x^2 -3x = 25 -125/3 +50 - 32 +64/3 -32 =11 61/3 = 31 1/3
2) интеграл от3 до 4 (8-(x^2 - 12x + 35)) = 8х - x ^3 /3 +6x^2 -35x = -27*4 -64/3 +96 +27*3 +9 -54 = 24 -21 1/3 =2 2/3
31 1/3 +3 2/3 = 35
найдем точки пересечения
x^2 - 4x + 3 = 8
x^2 - 4x -5=0
х= -1 х = 5
x^2 - 12x + 35 = 8
x^2 - 12x + 27=0
х = 3 х= 9
x^2 - 4x + 3 =x^2 - 12x + 35
8х = 32
х = 4
1) интеграл от 4 до 5 (8-(x^2 - 4x + 3 ))= 8х -x^3 /3 +2x^2 -3x = 25 -125/3 +50 - 32 +64/3 -32 =11 61/3 = 31 1/3
2) интеграл от3 до 4 (8-(x^2 - 12x + 35)) = 8х - x ^3 /3 +6x^2 -35x = -27*4 -64/3 +96 +27*3 +9 -54 = 24 -21 1/3 =2 2/3
31 1/3 +3 2/3 = 35
1) 2cosx-1 < 0
cosx < 1/2
arccos(1/2) + 2πn < x < 2π - arccos(1/2) + 2πn, n ∈ Z
π/3 + 2πn < x < 2π - π/3 + 2πn, n ∈ Z
π/3 + 2πn < x < 5π/3 + 2πn, n ∈ Z
2) sin2x - √2/2 < 0
sin2x < √2/2
- π - arcsin(√2/2) + 2πk < 2x < arcsin(√2/2) + 2πk, k ∈ Z
- π - π/4 + 2πk < 2x < π/4 + 2πk, k ∈ Z
- 5π/4 + 2πk < 2x < π/4 + 2πk, k ∈ Z
- 5π/8 + πk < x < π/8 + πk, k ∈ Z
3) tgx<1
- π/2 + πn < x < arctg(1) + πn, n ∈ Z
- π/2 + πn < x < π/4 + πn, n ∈ Z