It looks difficult but is it ?

Question

Let ƒ be continuous on [0,1] and let Ω be the triangular region in the xy- plane with

vertices (0,0) , (1,0) , and (0,1).

Show that ∫∫_Ωƒ(x+y) dA = ∫ ₀¹uƒ(u )d u

Answer 1

Let x +y =u and x-y =v

Under this transformation the triangular region Ω is mapped into the triangular D region bounded by the lines

u = v , v= -u and u =1 in the uv-plane.

x = [ u+v ] /2 and y = [ u-v] /2

Then the Jacobian ∂( x, y ) / ∂ ( u,v ) =- 1 / 2

Hence ∫_u=0^u=1 ∫^v=u_v=-u ƒ( u) | -1 /2 | d v d u =

∫_u=0^u=1 [ v ƒ( u) ( 1 /2 ) ] |^v=u_v=-u d u

∫_u=0^u=1 2uƒ( u) ( 1 /2 ) d u

∫_u=0^u=1 u ƒ( u) d u

Answer 2

The key is letting u = x+y, then integrating over 0 <= x <= u, 0 <= u <= 1 of f(x+y) = f(z)

Then, I[0,1][0,u]f(u)dxdu = I[0,1]xf(u) E[0, u] du = I[0,1](u - 0) f(u) du = I[0,1] u f(u) du