I don't get why for $u$ substitution they sub the upper and lower bounds
into $u$ from the original function to find the new upper and lower
bounds with the function $u$.
[...]
My question is what or how does plugging your old lower and upper
bound values into $u$ give you the new values of your new function that's
expressed as $u$...
The blunt answer is "That's what the change of variables theorem says."
But here's a more conceptual and notational explanation. The notation $\int_a^b f(x)\, dx$ connotes "integrating from $x = a$ to $x = b$". To emphasize this, let's write
$$
\int_{x=a}^{x=b} f(x)\, dx.
$$
Now suppose you want to evaluate
$$
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx.
$$
If you make the substitution $u = g(x)$, then $du = g'(x)\, dx$ ("by the chain rule"). If $F$ denotes an antiderivative of $f$, the preceding becomes
$$
\int_{x=a}^{x=b} f(u)\, du = F(u) \Big|_{x=a}^{x=b}.
$$
Now, it should be notationally clear that setting $u = a$ and $u = b$ does not (in general) "give the right answer": Those are not the limits specified in the original integral.
To proceed, you have two choices:
Undo the original substitution by setting $u = g(x)$, and then plug in $x = b$ and $x = a$.
Find the "new limits of integration", $u = g(a)$ and $u = g(b)$, by plugging the "old" limits $x = a$ and $x = b$ into the substitution $u = g(x)$.
The second makes notational sense because "when $x = a$, we have $u = g(a)$" and "when $x = b$, we have $u = g(b)$". It should be procedurally clear the two methods are mathematically equivalent. Computationally, the second is usually less work (as both prior respondents note); the first amounts to writing something down, then erasing it.
In symbols, either approach gives
\begin{align*}
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx
&= \int_{x=a}^{x=b} f(u)\, du && \text{Substitute $u = g(x)$;} \\
&= F(u) \Big|_{x=a}^{x=b} && \text{Antidifferentiate;} \\
&= F(u) \Big|_{u=g(a)}^{u=g(b)} && \text{Change limits;} \\
&= F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) && \text{Plug in;} \\
&= \int_{u=g(a)}^{u=g(b)} f(u)\, du. && \text{Reinterpret the fundamental theorem.}
\end{align*}
So much for the explanation; what about Real Life? The notation
$$
\int_{x=a}^{x=b} f(x)\, dx
$$
is redundant, and in practice, out of laziness^H^H^H elegance, we fall back on $\int_a^b f(x)\, dx$ (sometimes to the confusion of calculus students). But when you're learning substitution the first time, it may help to write in the variable corresponding to the numerical limits, as in:
\begin{align*}
\int_{x=0}^{x=\pi/4} -2\cos^{2}(2x) \sin(2x)\, dx
&= \int_{x=0}^{x=\pi/4} u^{2}\, du && u = \cos(2x),\quad du = -2\sin(2x)\, dx; \\
&= \int_{u=1}^{u=0} u^{2}\, du && \text{When $x = 0$, $u = 1$; when $x = \pi/4$, $u = 0$.}
\end{align*}
Let $a=x_0<x_1<\dots<x_{n-1}<x_n=b$ be a decomposition $P$ of $[a,b]$ then then the lower Darboux sum with respect to $P$ is
$$L(f,P)=\sum_{k=1}^n \inf_{x\in [x_{k-1},x_k]}f(x)\cdot (x_k-x_{k-1})=\sum_{k=1}^n 0\cdot (x_k-x_{k-1})=0.$$
On the other hand for upper Darboux sum with respect to $P$ we have two cases:
$$U(f,P)=\sum_{k=1}^n \sup_{x\in [x_{k-1},x_k]}f(x)\cdot (x_k-x_{k-1})=
\begin{cases}
1\cdot (x_i-x_{i-1})\quad\text{if $x_{i-1}<c<x_{i}$,}\\
1\cdot (x_{i+1}-x_{i-1}) \text{ if $c=x_{i}$.}
\end{cases}$$
Can you take it from here and find $U(f)$ and $L(f)$?
Best Answer
The upper and lower bounds of definite integrals are presented as: $$ (b-a)\inf_{x\in [a,b]}f(x)\le\int_a^b f(x)dx\le(b-a)\sup_{x\in [a,b]}f(x). $$ So according to your example let's find the $\inf_{x\in [0,1]} f(x)$ and $\sup_{x\in [0,1]} f(x)$ for $f(x)=\frac{1}{\sqrt{1+x^4}}$. Let us calculate $\frac{df}{dx}$ $$ f'(x)=\frac{df}{dx}=\frac{-2x^3}{(1+x^4)\sqrt{1+x^4}}. $$ By equating $f'(x)=0$, we only get $x=0$ as critical point. By checking the bounds of the interval $f(0)=1$ and $f(1)=\frac{1}{\sqrt{2}}$, we conclude that $\inf_{x\in [0,1]} f(x)=\frac{1}{\sqrt{2}}$ and $\sup_{x\in [0,1]} f(x)=1$. Therefore $$ \frac{1}{\sqrt{2}}\le\int_a^b f(x)dx\le1. $$