evaluate ∫∫R xy if R is the region bounded by y = x^3 and y = x^2.

I am assuming you want to evaluate ∫∫R xy dA.
y = x^3 and y = x^2 intersect when:
x^3 = x^2 ==> x = 0 and x = 1.
Note that R is a typeI region because we can represent R as:
R = {(x, y)  0 <= x <= 1, x^2 <= y <= x^3}.
So, we have:
∫∫ R xy dA = ∫(from x=0 to 1) ∫(from y=x^2 to x^3) xy dy dx
= ∫(from x=0 to 1) [(1/2)xy^2] (eval from y=x^2 to x^3) dx, by doing the inner integral
= 1/2 ∫(from x=0 to 1) x[(x^3)^2  (x^2)^2] dx
= 1/2 ∫(from x=0 to 1) (x^7  x^5) dx, by expanding
= (1/2)[(1/8)x^8  (1/6)x^6] (eval from x=0 to 1), by evaluating this integral
= (1/2)[(1/8  1/6)  (0  0)]
= 1/48.
I hope this helps!
y = x^3 and y = x^2 intersect when:
x^3 = x^2 ==> x = 0 and x = 1.
Note that R is a typeI region because we can represent R as:
R = {(x, y)  0 <= x <= 1, x^2 <= y <= x^3}.
So, we have:
∫∫ R xy dA = ∫(from x=0 to 1) ∫(from y=x^2 to x^3) xy dy dx
= ∫(from x=0 to 1) [(1/2)xy^2] (eval from y=x^2 to x^3) dx, by doing the inner integral
= 1/2 ∫(from x=0 to 1) x[(x^3)^2  (x^2)^2] dx
= 1/2 ∫(from x=0 to 1) (x^7  x^5) dx, by expanding
= (1/2)[(1/8)x^8  (1/6)x^6] (eval from x=0 to 1), by evaluating this integral
= (1/2)[(1/8  1/6)  (0  0)]
= 1/48.
I hope this helps!