Why don't the bounds in this definite integral change?

Question

The question

This is probably a very basic question but I'm having a brain lapse and don't know why they didn't change the definite integral bounds from ($0 \rightarrow4$) to ($4 \rightarrow20$). I know that using $u$-sub, you would have $u = x^2 + 4$. So why are the bounds seemingly not affected?

Answer 1

First, note that a $u$-substitution was not (explicitly) used, though it was implicitly used and the details somewhat skipped over. However, they converted back into terms of $x$, instead of staying in terms of $u$.

In the first integral, you have that $u=x^2+4$ gives $\mathrm{d} u = 2x \, \mathrm{d} x$, so

$$\int_0^4 \frac{x}{x^2 + 4} \, \mathrm{d} x = \int_4^{20} \frac{1/2}{u} \, \mathrm{d} u$$

You can finish this in, evidently, your preferred method:

$$\int_4^{20} \frac{1/2}{u} \, \mathrm{d} u = \frac 1 2 \ln|u| \bigg|_{u=4}^{u=20} = \frac 12 \ln(20) - \frac 1 2 \ln(4)$$

You can also use the shown method of converting back to $x$, but you will get the same answer.

$$\int_4^{20} \frac{1/2}{u} \, \mathrm{d} u = \frac 12 \ln|u| \bigg|_{u=4}^{u=20} = \frac 1 2 \ln(x^2+1) \bigg|_{x=0}^{x=4}$$

This idea holds in general: if you do a $u$-substitution, you don't have to change the bounds, so long as you convert back into terms of $x$ the moment you've found an antiderivative, before you plug in the bounds of the integral.

In my opinion this is undesirable since $u$-substitutions are meant to simplify the form of the antiderivative, which means converting back to $x$ can result in something unpleasant to plug numbers into, while the expression in terms of $u$ may be more pleasant. (Of course there can be a slight trade-off of the bounds often being more unpleasant in the $u$ case, but I think the elegance of the antiderivative outweighs that.)

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Addendum 1: A third means to handle the integral is that

$$\int \frac{f'(x)}{f(x)} \, \mathrm{d} x = \ln |f(x)| + C$$

for suitable $f$, so one can circumvent implied or explicit $u$-substitution altogether by the identity above and realizing

$$\int_0^4 \frac{x}{x^2 + 4} \, \mathrm{d} x = \frac 1 2 \int_0^4 \frac{2x}{x^2 + 4} \, \mathrm{d} x$$

Granted, the stated identity is also just an implication of $u$-sub, so I'd hesitate to say this method truly avoids $u$-substitution.

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Addendum 2: Be mindful of which integral you're using the $u$-substitution on; I see this often with my Calculus I/II students where they identify the $u$-substitution but don't realize its resulting differential just does not mesh well with the integral unless further work is done. In the original integral of

$$\int_0^4 \frac{x+2}{x^2 + 4} \, \mathrm{d} x$$

$u=x^2+4$ does not work, since there is no pleasant way to handle the $+2$ in this scenario beyond noting that $x = \pm \sqrt{u-4}$, but this only makes things messier.

I felt like bringing this up since technically, one can argue the second integral (that becomes $\arctan(\cdots)$) also implicitly uses $u$-substitution in an analogous way, but the substitution would be very differently.

Answer 2

Because there is no $u$ substitution written down.

If substitution is used, then it was done in the head and reversed for the text.

$$\begin{align}\int_0^4 \dfrac{x~\mathrm d x}{x^2+4} &=\dfrac 12\int_4^{20}\dfrac{\mathrm d u}{u}\\&=\left.\dfrac 12\ln(u)\right\rvert_4^{20}\\&=\left.\dfrac 12\ln(x^2+4)\right\rvert_0^{4} \end{align}$$