Double integral xy(x+y) between curves

Region. The curves y = x² and y = x meet where x² = x, i.e. at x = 0 and x = 1 (points (0,0) and (1,1)). Between them, x² ≤ y ≤ x for 0 ≤ x ≤ 1.

I = ∫₀¹ ∫{x²}^{x} xy(x + y) dy dx = ∫₀¹ ∫{x²}^{x} (x²y + xy²) dy dx

Inner integral (x constant)

∫ (x²y + xy²) dy = x² y²/2 + x y³/3

at y = x :   x⁴/2 + x⁴/3
at y = x² :  x⁶/2 + x⁷/3

inner = (x⁴/2 − x⁶/2) + (x⁴/3 − x⁷/3)
      = (5/6)x⁴ − x⁶/2 − x⁷/3

Outer integral

I = ∫₀¹ [ (5/6)x⁴ − x⁶/2 − x⁷/3 ] dx
  = (5/6)(1/5) − (1/2)(1/7) − (1/3)(1/8)
  = 1/6 − 1/14 − 1/24
  = 9/168 = 3/56

Using common denominator 168: 1/6 = 28/168, 1/14 = 12/168, 1/24 = 7/168, so I = (28 − 12 − 7)/168 = 9/168 = 3/56.

Answer: I = 3/56.