From change of variables formula to $\int_{\varphi(a)}^{\varphi (b)}f(v)\,\mathrm d v= \int_a^b f (\varphi (u)) \varphi ' (u) \, \mathrm d u$

Question

Let $a,b \in \mathbb R$ with $a < b$. Let $U := (a, b)$. Let $f:U \to \mathbb R$ be integrable. Let $\varphi:U \to \mathbb R$ be injective and continuously differentiable.

• By change of variables formula, we have $$ \int_{\varphi (U)} f(v) \, \mathrm d v = \int_a^b f (\varphi (u)) |\varphi ' (u)| \, \mathrm d u. \tag{1} $$

• In this note, we have $$ \int_{\varphi (a)}^{\varphi (b)} f(v) \, \mathrm d v = \int_a^b f (\varphi (u)) \varphi ' (u) \, \mathrm d u. \tag{2} $$

Could you explain how to go from (1) to (2)?

Answer 1

If $\varphi$ is increasing, there is no difference between those two expressions.

Now, suppose that $\varphi$ is decreasing and let $[c,d]=\varphi\bigl([a,b]\bigr)$. Then \begin{align} \int_{\varphi(U)}f(v)\,dv&=\int_c^df(v)\,dv\\ &=-\int_d^cf(v)\,dv.\\ &=-\int_{\varphi(a)}^{\varphi(b)}f(v)\,dv \end{align} whereas $$ \int_a^bf(\varphi(u))|\varphi'(u)|\,du=-\int_a^bf(\varphi(u))\varphi'(u)\,du. $$ And $\int_{\varphi(a)}^{\varphi(b)}f(v)\,dv=\int_a^bf(\varphi(u))\varphi'(u)\,du$.