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In the method of integration by substitution while substituting $x=g(t)$ we must have g(t) a continuously differentiable function on $[\alpha, \beta]$ an so we transform the integral $$\int_a^bf(x)dx=\int_{\alpha}^{\beta}f[g(t)]g'(t)dt$$ but in a book i have seen that while evaluating the integral $\int_2^3(x-2)^2dx$ the substitution made was $x=2+\sqrt{t}$ with $x$ varying in $[2, 3]$ and $t$ varying in $[0, 1]$ but in this interval $\sqrt{t}$ is not even differentiable then why this substitution was made?

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  • $\begingroup$ You can always push your tools to the limit. (Pun intended.) An important condition on $g$ that you forgot is that it must be monotonic in the interval $[\alpha,\beta].$ $\endgroup$ – Allawonder Nov 23 '19 at 7:13
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When $t$ varies over $[0,1]$, think of it as improper integral.

Like let $\epsilon \in (0,1]$, substitution rule can be applied easily on $\int_\epsilon^1 t \Big (\dfrac{1}{2\sqrt(t)} \Big )dt$. As this has a limit, and taking limit as $t$ tends to $0$, gives the results. (Since if $f$ is integrable, its improper integral gives same value.)

Rmk: Better integrate directly, this gives answer in less steps and more accurate.

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  • $\begingroup$ Then what are the necessary and sufficient conditions on $g(t)$ while Substitution $\endgroup$ – user728159 Nov 23 '19 at 6:29
  • $\begingroup$ Integration by subsitution is proven for $g$ which has integrable derivatives (when $f$ continuous), and used as a way to replace the integrator with $'dx'$, hence this is necessary and sufficient. $\endgroup$ – LKM Klein Nov 23 '19 at 6:34
  • $\begingroup$ Since if $g$ is not differentiable, it is nonsense to talk about multiplying by $g'$. $\endgroup$ – LKM Klein Nov 23 '19 at 6:34
  • $\begingroup$ But in ur answer it was not differentiable $\endgroup$ – user728159 Nov 23 '19 at 6:37
  • $\begingroup$ then this is a question regarding improper integral, but not integration by a substitution (which is a theorem itself). Related question is, if $g$ is differentiable except on few points, and $f*g'$ integrable, then can integration by substitution be done? $\endgroup$ – LKM Klein Nov 23 '19 at 6:39

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