Задача 69576 ...

6355a2d83db1a168cec87f02

25.02.2023 17:37:10

вычислить определенный интеграл

∫ (x^2+7x+12) cosxdx

5f3ea7e3faf909182968ddd9

26.02.2023 13:20:51

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Интегрирование по частям:

[m]u=x^2+7x+12[/m] ⇒ [m]du=(x^2+7x+12)`dx[/m] ⇒ [m]du=(2x+7)dx[/m]

[m]dv=cosxdx[/m] ⇒ [m]v= ∫ dv[/m] ⇒ [m]v= ∫ cosxdx=sinx[/m]


[m] ∫ (x^2+7x+12)cosxdx=(x^2+7x+12)\cdot sinx- ∫ (2x+7)sinxdx[/m]

Интегрирование по частям:
[m]u=2x+7x[/m] ⇒ [m]du=(2x+7)`dx[/m] ⇒ [m]du=2dx[/m]

[m]dv=sinxdx[/m] ⇒ [m]v= ∫ dv[/m] ⇒ [m]v= ∫ sinxdx=-cos[/m]


[m] ∫ (x^2+7x+12)cosxdx=(x^2+7x+12)\cdot sinx- ∫ (2x+7)sinxdx=(x^2+7x+12)\cdot sinx- ((2x+7)\cdot (-cosx)- ∫(-cosx)2dx)= [/m]

[m]=(x^2+7x+12)\cdot sinx+(2x+7)\cdot cosx-2∫cosxdx=(x^2+7x+12)\cdot sinx+(2x+7)\cdot cosx-2sinx +C [/m]

[m]=(x^2+7x+10)\cdot sinx+(2x+7)\cdot cosx +C [/m]

По формуле Ньютона Лейбница:

[m] ∫^{0}_{-4} (x^2+7x+12)cosxdx=((x^2+7x+10)\cdot sinx+(2x+7)\cdot cosx )|^{0}_{-4}=[/m]


[m]=10sin0+7cos0-(((-4)^2+7\cdot (-4)+10)\cdot sin(-4)+(2\cdot (-4)+7)\cdot cos(-4))=7-(2sin4-cos4)=7+cos4-2sin4[/m]

626fab9a354ab65fb2f43abb

25.02.2023 18:08:13

[m]\int_{-4}^0 (x^2 + 7x + 12)cos(x)\ dx = \int_{-4}^0 x^2cos(x)\ dx + 7\int_{-4}^0 x \cdot cos(x)\ dx + 12\int_{-4}^0 cos(x)\ dx [/m]

1 интеграл нужно решать по частям:
[m]\int_{-4}^0 x^2 \cdot cos(x)\ dx = |u=x^2; dv = cos(x)\ dx; du = 2x\ dx; v = sin(x)| =[/m]
[m]= x^2 \cdot sin(x)|_{-4}^0 - \int_{-4}^0 2x \cdot sin(x)\ dx =[/m]
Снова по частям:
[m]= |u = x; dv = sin(x)\ dx; du = dx; v = -cos(x)| =[/m]
[m]= 0 - (-4)^2 \cdot sin(-4) - 2x(-cos(x))|_{-4}^0 + \int_{-4}^0 (-cos(x))\ dx =[/m]
[m]= 16sin(4) + 0 - 2(-4)cos(-4) - (sin(0) - sin(-4)) = [/m]
[m] = 16sin(4) + 8cos(4) - sin(4) = 15sin(4) + 8cos(4)[/m]

2 интеграл тоже решаем по частям:
[m]\int_{-4}^0 x \cdot cos(x)\ dx = |u = x; dv = cos(x)\ dx; du = dx; v = sin(x)| =[/m]
[m]= x \cdot sin(x)|_{-4}^0 - \int_{-4}^0 sin(x)\ dx =[/m]
[m]= 0 - (-4) \cdot sin(-4) - (-cos(x))|_{-4}^0 =[/m]
[m]= -4sin(4) + cos(0) - cos(-4) = -4sin(4) + 1 - cos(4)[/m]

3 интеграл - табличный:
[m]\int_{-4}^0 cos(x)\ dx = sin(x)|_{-4}^0 = sin(0) - sin(-4) = sin(4)[/m]
Итого получаем:
[m]\int_{-4}^0 x^2cos(x)\ dx + 7\int_{-4}^0 x \cdot cos(x)\ dx + 12\int_{-4}^0 cos(x)\ dx = [/m]
[m]= 15sin(4) + 8cos(4) + 7(-4sin(4) + 1 - cos(4)) + 12sin(4) =[/m]
[m]= 27sin(4) + 8cos(4) - 28sin(4) - 7cos(4) + 7 =[/m]
[m]= 7 + cos(4) - sin(4)[/m]