>>258
z = x-y
でyからzに置換すれば
∫_{y = 0 to x} f(x-y) dy
= -∫_{z = x to 0} f(z) dz
= ∫_{z = 0 to x} f(z) dz

∫_{x = 0 to 1} ( ∫_{z = 0 to x} f(z) dz) dx

は、xz平面に積分領域を書いてみれば
(0,0),(1,0),(1,1)を頂点とする三角形になり
これを見ながら積分順序の入れ替えを行えば
∫_{x = 0 to 1} ( ∫_{z = 0 to x} f(z) dz) dx
= ∫_{z = 0 to 1} ( ∫_{x = z to 1} f(z) dx) dz
= ∫_{z = 0 to 1} (1-z)f(z) dz