# Notes - Analysis III TT23, Integration by substitution

### Flashcards

Can you state the “integration by substitution” theorem?

Suppose that $f: [a, b] \to \mathbb R$ is continuous and that $\phi : [c, d] \to [a, b]$ is continuous on $[c, d]$, has $\phi(c) = a$ and $\phi(d) = b$, and maps $(c, d)$ to $(a, b)$. Suppose moreover that $\phi$ is differentiable on $(c, d)$ and that its derivative $\phi’$ is integrable on this interval. Then

\[\int^b_a f(x)\text dx = \int^d_c f(\phi(t)) \phi(t)'\text{d}t\]### Proofs

Prove the “integration by substitution theorem”, i.e

Suppose that $f: [a, b] \to \mathbb R$ is continuous and that $\phi : [c, d] \to [a, b]$ is continuous on $[c, d]$, has $\phi(c) = a$ and $\phi(d) = b$, and maps $(c, d)$ to $(a, b)$. Suppose moreover that $\phi$ is differentiable on $(c, d)$ and that its derivative $\phi^‘$ is integrable on this interval. Then

\[\int^b _ a f(x)\text dx = \int^d _ c f(\phi(t)) \phi(t)'\text{d}t\]

Suppose that $f: [a, b] \to \mathbb R$ is continuous and that $\phi : [c, d] \to [a, b]$ is continuous on $[c, d]$, has $\phi(c) = a$ and $\phi(d) = b$, and maps $(c, d)$ to $(a, b)$. Suppose moreover that $\phi$ is differentiable on $(c, d)$ and that its derivative $\phi^‘$ is integrable on this interval. Then

Todo (analysis iii, page 24)