# change of variables for definite integrals

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?

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Just a note: the stated $du$ appears to be wrong, although they seem to use the correct version in the calculation. – Dylan Moreland Dec 8 '11 at 21:10
To answer your last question, you can certainly find an indefinite integral and then evaluate at the bounds. But doing things this way can save calculation, in my experience. – Dylan Moreland Dec 8 '11 at 21:12
has anyone noticed, they forgot about the 2 from the main question when jumping to u and du – Raynos Dec 8 '11 at 22:21
A simple answer is that if you don't change the limits, you don't have equality. If we let $u=x-2$, then whereas$$\int_0^2(x-2)\,dx \text{ is equal to }\int_{-2}^0 u\,du,$$ $$\int_0^2(x-2)\,dx\text{ is not equal to }\int_0^2u\,du.$$Also, if this were part of a long, complex computation, involving several changes of variable and/or integration by parts, are you really confident in your ability to "undo" all the substitutions and get back to the original? What if you forget? Better to keep track of those changes in the limits. – Arturo Magidin Dec 9 '11 at 6:58
$du=-2\sin 2x\, dx\neq -\sin 2x\, dx$. – user236182 Oct 19 '15 at 21:52

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.

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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.

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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:

1. Undo the original substitution by setting $u = g(x)$, and then plug in $x = b$ and $x = a$.

2. 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*}

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I'm rather not a fan of notation such as $\int_{x=a}^{x=b} f(u)\, du$, which some of my colleagues use to avoid an explicit change of bounds. – Mark McClure Jan 6 '15 at 16:04
@MarkMcClure: I don't advocate the notation for normal use either, but it's convenient for clarifying this particular question. :) – Andrew D. Hwang Jan 6 '15 at 16:12
@user84518 - Point taken! – Mark McClure Jan 6 '15 at 16:15