TH1: `-2 < a < 0 => f(x) > 0  forall  x in (-2;a)`

`S_D=int_(-2)^a |f(x)|dx=int_(-2)^a  f(x)dx`

`=(1/3x^3-x^2)|_(-2)^a`

`=(1/3a^3-a^2)-(-20/3)`

`=1/3a^3-a^2+20/3`

`=>` `1/3a^3-a^2+20/3=22/3`

`=>` `a ~~ 3,2` (Loại)

TH2: `0 <= a < 2`

`S_D=int_(-2)^a |f(x)|dx`

`=int_(-2)^0 |f(x)|dx+int_(0)^a |f(x)|dx`

`=int_(-2)^0 f(x)dx-int_(0)^a f(x)dx` (Do `f(x) <= 0  forall x in [0;a]`)

`=20/3-(1/3x^3-x^2)|_0^a`

`=20/3-(1/3a^3-a^2)`

`=>` `20/3-(1/3a^3-a^2)=22/3`

`=>` `[(a=1  (tm)),(a=1+sqrt3 > 2  (l)),(a=1-sqrt3 < 0  (l)):}`

TH3: `a >= 2`

`S_D=int_(-2)^a |f(x)|dx`

`=int_(-2)^1 |f(x)|dx+int_1^a |f(x)|dx`

`=22/3+int_1^a |f(x)|dx`

Do `int_1^a |f(x)|dx > 0 => S_D > 22/3`

Vậy `a=1` thỏa mãn bài toán