the real fundamental theorem of calculus

Every time I teach calculus there's one lecture that I love and the students meet with disdain.  Well, maybe not disdain  exactly, but indifference.  Yesterday was that lecture and this year was no exception--the low murmur of inattention was all the feedback I needed.

Why all the hate?  The Mean Value Theorem can't help it.  Oh, and by the way, it's probably the most fundamental result in differential calculus.  It makes most of the rest of calculus work, a point that was driven home to me the first time I taught advanced calculus and had to pull it out in every other proof.  One of these days I'm going to write my calculus for kids book, Where's the Mean Value Theorem? , a Where's Waldo? -style exploration of the subject in which the reader is invited to spot the MVT wherever it's lurking.

The MVT is the Tesla of calculus--so thoroughly embedded you don't even realize it.  In case you've forgotten, here's a picture. The statement of the MVT  is that for a differentiable function $$f$$ on a closed interval $$[a,b]$$ there is a $$c$$ in the interior where $$f'(c)(b-a) = f(b)-f(a)$$.  In other words, somewhere in the interior, a tangent line has the same slope as the line joining the endpoints of the curve.  In other other words, somewhere in the interior the instantaneous rate of change of the function equals the average rate of change over the whole interval.

the graph of a function with some lines on it.

That might not seem like a big deal, really, but it is.  Here's the first baby application.  Suppose you take a drive and you travel 120 miles in 2 hours.  How many times did your speedometer read 60 mph?  The MVT says it happens at least once:  since the average speed over the whole 2 hours is 60 mph there is at least one spot where the instantaneous velocity (i.e., what your speedometer reads) equals the total average.  Now really it must happen at least twice since you probably go above 60 at some point and then drop back down to 0.  Or if you have your cruise control set to 60 then it happens infinitely often, but you get the point.

Here's another math-y application:  suppose a function $$f$$ satisfies $$f'(x) = 0$$ on some interval.  Then $$f$$ is constant on that interval.  We know that constant functions have zero derivative; this is the converse.  It's easy to prove, too: take $$x_1 < x_2$$ in the interval.  By MVT, there is a $$c$$ between them with $$f'(c)(x_2-x_1) = f(x_2)-f(x_1)$$.  But the left hand side of that equation is $$0$$ by assumption and so $$f(x_1)=f(x_2)$$; that is, $$f$$ is constant.  Corollary:  if $$f'(x) = g'(x)$$ on an interval, then $$f(x) = g(x) + C$$ for some constant $$C$$.  This is proved by letting $$F(x) = f(x) - g(x)$$ and applying the previous result.  The big consequence of this is that any two antiderivatives of a function on an interval differ by a constant, thereby unleashing a tidal wave of $$1$$-point deductions on calculus tests everywhere when students omit the $$+ C$$.

Want to prove a function only has one real root?  Mean Value Theorem (well, really Rolle's Theorem, the special case where the function has the same value at the endpoints).  L'Hospital's Rule in its full generality?  Mean Value Theorem.  The Fundamental Theorem of Calculus?  Yep, you use the Mean Value Theorem.  Well, there are a couple of proofs.  The first uses the MVT for integrals (which is a consequence of the Intermediate Value Theorem, and asserts that a continuous function takes on its average value on an interval).  The second is more direct and uses the MVT directly.  You can read it here.

There's also a several variables version which implies that the mixed partials of a function $$f:{\mathbb R}^n \to {\mathbb R}$$ are equal: $\frac{\partial^2 f}{\partial x_i\partial x_j} = \frac{\partial^2 f}{\partial x_j\partial x_i}.$  So, while I know it will never catch on (and it probably shouldn't) I will continue to call the MVT the Real Fundamental Theorem of Calculus.