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To solve the following integral, one can use u-substitution: $$\int_2^3 \frac{9}{\sqrt[4]{x-2}} \,dx,$$ With $u = \sqrt[4]{x-2}$, our bounds become 0 and 1 respectively. Thus, we end up with the following: $$36\int_0^1{u^2} \,du$$ In the first case, the lower bound is a vertical asymptote so we would have to use limits to find the answer. However, in the second case there's no longer an asymptotal bound - would you still have to write the limits since the original function would've needed limits, or can you just solve this by plugging in the substituted bounds? I know the final answer will be the same either way, but I want to know if it can be considered correct to exclude the limits in the 2nd integral from a technical perspective since the function has been changed. Many thanks in advance!

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Without too much simplification, the substitution you cite yields

$$\int_2^3\frac9{(x-2)^{1/4}}\,\mathrm dx=36\int_0^1\frac{u^3}u\,\mathrm du$$

which you certainly welcome to treat as an improper integral,

$$36\left(\frac13-\lim_{u\to0^+}\frac{u^3}3\right)$$

but since $u=0$ is a removable discontinuity and the limand reduces to $u^2$, you may as skip this treatment altogether.

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    $\begingroup$ I'm confused about your last expression. It seems like you want to still have $\int$ present, but the $1-$ seems like you've already taken the integral, but the $\frac{u^3}u$ is from before you've taken the integral... $\endgroup$ – Teepeemm Feb 20 at 2:05
  • $\begingroup$ Yes, it was wrong as written. I have edited it. $\endgroup$ – Martin Argerami Feb 20 at 3:51
  • $\begingroup$ Thanks @MartinArgerami ! $\endgroup$ – user170231 Feb 20 at 14:55
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There are certain conditions that must be met for a substitution to be "legal". In most circumstances these conditions are naturally met and so they are not emphasized; you have here one situation in which they are not.

For instance, one set of conditions is given in Anton, Anderson, and Bivens (Calculus, Early Transcendentals, 11th Edition), Theorem 5.9.1:

Theorem 5.9.1. If $g'(x)$ is continuous on $[a,b]$, and $f$ is continuous on an interval containing the values of $g(x)$ for $a\leq x\leq b$, then $$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$$

Here you have $g(x) = \sqrt[4]{x-2}$, so $g'(x) = \frac{1}{4}(x-2)^{-3/4}$... which is not continuous on the interval $[2,3]$ that you are working on.

In fact, as you note, the initial integral is improper, which means you aren't really evaluating that integral: you are evaluting a limit, $$\lim_{h\to 2^+}\int_h^3 \frac{9}{\sqrt[4]{x-2}}\,dx.$$ The integral in the limit does satisfy the conditions of the theorem above, so you can make the substitution to get $$\lim_{h\to 2^+}\int_{\sqrt[4]{h-2}}^1 \frac{u^3}{u}\,du = \lim_{a\to 0^+}\int_a^1 u^2\,du,$$ and proceed from there.

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  • $\begingroup$ Thank you very much Arturo, your answer was very in-depth and provided a lot of reinforcement for what you were saying. I wish I could select multiple answers to be correct, I thought your inclusion of the cited textbook snippet was really nicely done too. I chose the other answer only because I found it easier to digest/understand at first glance, but after I properly read your comment I understood it more fully. Thank you for your quick and effective contribution :-) $\endgroup$ – Lord Kanelsnegle Feb 19 at 17:32
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Suppose you must. Then we have $$9\lim_{a\to2^+}\int_a^3\frac1{\sqrt[4]{x-2}}\,dx.$$

Let $u=x-2$. Then we have $$9\lim_{a\to2^+}\int_{a-2}^1 u^{-1/4}\, du.$$

By the power rule, we have $$9\lim_{a\to2^+} \left.\frac43u^{3/4}\right]_{a-2}^1=9\left[\frac43(1-\lim_{a\to2^+}(a-2)^{3/4})\right]=9\left[\frac43(1-0)\right]=12.$$

Notice that it does not make a difference whether you use $\lim_{a\to2^+}(a-2)$ or $0$.

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    $\begingroup$ Thanks for the quick and simple response! I found the answer I selected to be more aligned with the question, but I wanted to comment because I appreciate the time you took to write this. Thank you once again. $\endgroup$ – Lord Kanelsnegle Feb 19 at 17:29

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