The Unapologetic Mathematician

Mathematics for the interested outsider

The Fundamental Theorem of Calculus II

And now we come to the second part of the FToC. This takes the first part and flips it around.

We again start with a continuous function f:\left[a,b\right]\rightarrow\mathbb{R}, but now we take any antiderivative F, so that f(x)=F'(x). Then the FToC asserts that

\displaystyle F(b)-F(a)=\int\limits_a^bf(x)dx

Before we differentiated a function we got by integrating to get back where we started. Now we’re integrating a function we get by differentiating, and again get back where we started. Integration and differentiation are two sides of the same coin.

Let’s consider a partition of \left[a,b\right] with points a=x_0,x_1,...,x_{n-1},x_n=b. Then we see that F(b)-F(a)=F(x_n)-F(x_0). We can add and subtract the value of F at each of the intermediate points to see that

\displaystyle F(b)-F(a)=F(x_n)-F(x_{n-1})+F(x_{n-1})-...-F(x_1)+F(x_1)-F(x_0)
\displaystyle F(b)-F(a)=\sum\limits_{i=1}^nF(x_i)-F(x_{i-1})

Now the Differential Mean Value Theorem tells us that there’s a point c_i\in\left[x_{i-1},x_i\right] so that F(x_i)-F(x_{i-1})=(x_i-x_{i-1})F'(c_i). And we assumed that F'(c_i)=f(c_i), so we have

\displaystyle F(b)-F(a)=\sum\limits_{i=1}^n(x_i-x_{i-1})f(c_i)

But this is a Riemann sum for the partition we chose, using the points c_i as the tags. Since every partition, no matter how fine, has such a Riemann sum, the integral must take this value, and the second part of the FToC holds.

February 14, 2008 - Posted by John Armstrong | Analysis, Calculus | | 9 Comments

9 Comments »

  1. Duh…. The mean value theorem again, dragged in, irrelevant and sticking out like sore thumb… When are you going to learn some good mathematical taste, professor?

    Comment by Michael Livshits | February 15, 2008

  2. I forgot again to prepend “yes, but,” my apologies.

    Comment by Michael Livshits | February 15, 2008

  3. You know, John, usually I disagree with your mathematical taste in the strongest possible terms, but I really want to applaud you here for proving the 2nd fundamental theorem of calculus directly from the mean value theorem rather than by applying the 1st fundamental theorem. I have no idea why it became traditional to derive the 2nd from the 1st — it doesn’t really save time, and in my experience students find it universally confusing (moreover, you get a stronger theorem — one doesn’t need to assume that f is continuous, merely that it is Riemann integrable and the derivative of
    a differentiable function…of course, this is overkill for freshman, but I’m just saying ). I’ve been fighting for this strategy for a long time…

    Comment by Andy P. | February 15, 2008

  4. I think my “grin” mark got knocked off after my “I’m just saying” comment…

    Comment by Andy P. | February 15, 2008

  5. Well, Andy, you’re going to hate tomorrow, when I go ahead and prove each part from the other.

    Here’s why the approach of yesterday’s approach followed by part 1 -> part 2 became the standard: because we’ve been dumbing down calculus for years. We barely mention any proper Riemann sums anymore, much less any rigorous way of taking their limit.

    Unfortunately, a calculus class is no longer about understanding the calculus, but about getting enough of the tools into the kids’ hands so they can go off into engineering classes and build.. a.. plane… and I’m flying back to DC for spring break?!?

    Comment by John Armstrong | February 15, 2008

  6. Yes, but understanding calculus is not the same as eating the limit-continuity-MVT homogenized garbage, served the math department, and asking for more.

    Comment by Michael Livshits | February 15, 2008

  7. I don’t understand what all this rigmagole is about, guys. FTCI says that G(x)=\int_a^xf(u)du is a primitive of f, i.e., G'=f. Since the difference between any 2 primitives is a constant, we are done.

    Comment by Michael Livshits | February 15, 2008

  8. [...] Flame War Wrap-Up Well, we’ve certainly had a lively time the last few days. Regular commentercomplainer Michael Livshits kicked it off by noting that I presented The [...]

    Pingback by FToC Flame War Wrap-Up « The Unapologetic Mathematician | February 18, 2008

  9. [...] as . Note that we aren’t saying which antiderivative we mean, and for the purposes of the FToC (part 2), we don’t need to be. It’s customary, though, to write the result generically by adding [...]

    Pingback by Indefinite Integration « The Unapologetic Mathematician | March 6, 2008

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