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In high school I learned that if I want to solve $$ \int_0^R f(x) dx $$ we can do $x = \psi (\theta)$ and $dx = \psi'(\theta) d \theta$ and the integral becomes $$ \int_0^R f(x) dx = \int_{a_1}^{a_2} f(\psi(\theta)) \psi'(\theta) d \theta, $$ say. Depending on $f$ and a suitable choice of $\psi$, this second integral can be easier to solve. Here $\psi(a_1)= 0$ and $\psi(a_2) = R$.

My question. Does the end point matter as long as it the values of $\psi$ are $0$ and $R$? In other words, can I pick any other two points $b_1$ $b_2$ such that $\psi(b_1) = 0$ and $\psi(b_2) = R$ and we have $$ \int_0^R f(x) dx = \int_{b_1}^{b_2} f(\psi(\theta)) \psi'(\theta) d \theta? $$

SJY
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1 Answers1

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The change-of-variables theorem for Riemann integration specifies that

$$\tag{*}\int_{\psi(b_1)}^{\psi(b_2)}f(x) \, dx = \int_{b_1}^{b_2} f(\psi(\theta)) \,\psi'(\theta) \,d \theta,$$

under suitable conditions on $f$ and $\psi$. Strong conditions for which this holds are that $f$ is continuous and $\psi$ is continuously differentiable on $[b_1,b_2]$ with $\psi([b_1,b_2])$ contained in the domain of $f$. Furthermore, the conditions can be weakened somewhat. For example, we can drop the requirement of continuity for $f$ and replace it with Riemann integrability as long as $\psi$ is restricted appropriately, as discussed here.

As far as your question is concerned, the limits of integration on the left side of (*) will be $0$ and $R$ for the pair of functions $f$ and $\psi$, when $\psi(b_1) = 0$ and $\psi(b_2) = R$. However, the behavior of $\psi$ over the entire interval $b_1$ and $b_2$ must meet the requirements (strong or weak) for which the theorem holds. Basically this means that the product of the composition and the derivative $(f \circ \psi)\psi'$ must be integrable over $[b_1,b_2]$. If for another interval $[a_1,a_2]$ we have $\psi(a_1) = 0 $ and $\psi(a_2) = R$, then as long as the inegrability requirement is met, we will have

$$\tag{*}\int_{0}^{R}f(x) \, dx = \int_{a_1}^{a_2} f(\psi(\theta)) \,\psi'(\theta) \,d \theta,$$

RRL
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  • If the transformation is not bijection, we get different results. For instance, let $u=x^2$ then $4/3=\int_{-1}^1 2x^2dx=\int_{1}^1\sqrt{u}du=0$ – Bilal Jafar Karaki Sep 20 '21 at 03:02
  • @BilalJafarKaraki: Presumably $R \neq 0$ in the question so we don't have a case where $\psi(a_1) = \psi(a_2)$ here. – RRL Sep 20 '21 at 04:33
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    More importantly your example translates in the notation of the OP into $f: x \mapsto \sqrt{x}$ and $\psi: \theta \mapsto \theta^2$ with $a_1 = -1 $ and $a_2 = 1$. The integral in question is $ \int_{a_1}^{a_2} f(\psi(\theta)) ,\psi'(\theta) ,d \theta = \int_{-1}^ 1 \sqrt{\theta^2} 2 \theta , d \theta$. But to be correct in interpreting the square root as a function we must have $\sqrt{\theta^2} = |\theta|$ and $ \int_{-1}^ 1 \sqrt{\theta^2} 2 \theta , d \theta = \int_{-1}^1 2|\theta|\theta , d\theta = 0 = \int_1^1 \sqrt{x} , dx$ so there is no contradiction. – RRL Sep 20 '21 at 04:37
  • Even Wolfram alpha gets this right: https://www.wolframalpha.com/input/?i=integral+of+sqrt%28x%5E2%29+2x++from+-1+to+1. You have to be careful about correctly interpreting the statement of the change-of-variables theorem and the "suitable conditions" for it to hold. – RRL Sep 20 '21 at 04:39