First of all I would like to start off by asking why do they have different change of variable formulas for definite integrals than indefinite...why cant we just integrate using U substitution as we normally do in indefinite integral and then sub the original U value back and use that integrand for definite integral?
I was at one point understanding integration but not when they started coming up with different formulas for definite integrals in U-substitution I got lost and resulted to just forcibly memorizing the formulas...
I dont 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.
I know that because if you dont want to sub the original value of U in and want to instead stick to U as your function you must use those new upper and lower bound but if you sub in the original value for U then you can use your old upper and lower bound values.
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 thats expressed as U...
Why do they make such a big deal out of it and complicate it when all they have to do is same U sub as indefinite integral and then plug original value of U in and go from there...are these math people just making excuses to come up with more work or is there more logic behind it?
$\endgroup$ 63 Answers
$\begingroup$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*}
$\endgroup$ 3 $\begingroup$For the original definite integral, the bounds are for the variable $x$. When you change variables from $x$ to $u$, you typically change the bounds to be in terms of the new variable. If you want, you can substitute back and should get the same answer.
$\frac{u^3}3=\frac{\cos^3 2x}{3}$
Now using the original bounds for $x$:
$\frac{\cos^32(\frac{\pi}4)}3-\frac{\cos^30}3=0-\frac13=-\frac13$
Evaluating it immediately in terms of u saves work.
$\endgroup$ $\begingroup$You say "all they have to do is same U sub as indefinite integral and then plug original value of U in and go from there"
But that's more work than not plugging in the expression for which $u$ was substituted. It's less complicated that way. To undo the substitution would add an extra complication that is not needed. Generally it's preferable not to add pointless complications.
$\endgroup$
