Analysis III TT23, Integration by parts
Flashcards
Can you state the “integration by parts” theorem?
Suppose $f, g : [a, b] \to \mathbb R$ are continuous functions that are differentiable on $(a, b)$. Suppose further that the derivatives $f’$ and $g’$ are integrable on $(a, b)$. Then $fg’$ and $f’g$ are integrable on $(a, b)$, and \(\int^b _ a f g' = f(b)g(b) - f(a)g(a) - \int _ a^b f'g\)
Proofs
Prove the “integration by parts” theorem:
Suppose $f, g : [a, b] \to \mathbb R$ are continuous functions that are differentiable on $(a, b)$. Suppose further that the derivatives $f’$ and $g’$ are integrable on $(a, b)$. Then $fg’$ and $f’g$ are integrable on $(a, b)$, and
\(\int^b _ a f g' = f(b)g(b) - f(a)g(a) - \int _ a^b f'g\)
Suppose $f, g : [a, b] \to \mathbb R$ are continuous functions that are differentiable on $(a, b)$. Suppose further that the derivatives $f’$ and $g’$ are integrable on $(a, b)$. Then $fg’$ and $f’g$ are integrable on $(a, b)$, and \(\int^b _ a f g' = f(b)g(b) - f(a)g(a) - \int _ a^b f'g\)
Todo.