∫[g(a)→g(b)] f(t) dt = ∫[a→b] f(g(u)) g'(u) du となる理由

f(t) = F'(t) とすると、
  ∫[g(a)→g(b)] f (t) dt
= ∫[g(a)→g(b)] F'(t) dt
= F(g(b)) - F(g(a))
= ∫[a→b] (d/du F(g(u))) du
= ∫[a→b] F'(g(u)) g'(u) du
= ∫[a→b] f (g(u)) g'(u) du