The value of is
are you sure you have the limits right? usually they put the smaller number at the bottom. should the -2 be 2?
int{-2:0}[(x + 2)^2]dx
we proceed by substitution
let u = x + 2
=> du = dx
so our integral becomes
int{u^2}du
= (1/3)u^3 + C
= (1/3)(x + 2)^3 + C
evaluating between -2 and 0 we get
int{-2:0}[(x + 2)^2]dx = (1/3)(0 + 2)^3 - (1/3)(-2 + 2)^3
................................= 8/3
i'm sure the integral is correct. switching around the limits to the order you used them just changes the answer to a negative value
int{0:-2}[(x + 2)^2]dx
we proceed by substitution
let u = x + 2
=> du = dx
so our integral becomes
int{u^2}du
= (1/3)u^3 + C
= (1/3)(x + 2)^3 + C
evaluating between 0 and -2 we get
int{0:-2}[(x + 2)^2]dx = (1/3)(-2 + 2)^3 - (1/3)(0 + 2)^3
................................= -8/3