How to use integration by substitution

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?

• 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 $[\alpha,\beta].$ – Allawonder Nov 23 '19 at 7:13

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.)
• Then what are the necessary and sufficient conditions on $g(t)$ while Substitution – user728159 Nov 23 '19 at 6:29
• 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. – LKM Klein Nov 23 '19 at 6:34
• Since if $g$ is not differentiable, it is nonsense to talk about multiplying by $g'$. – LKM Klein Nov 23 '19 at 6:34
• 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? – LKM Klein Nov 23 '19 at 6:39