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Second Fundamental Theorem Of Calculus (The Evaluation Theorem):

For the function $f$ continuous everywhere in the $[A, B]$, and the function $F$ whose derivative with respect to $x$ equals to $f$:

$$\int_A^B f(x)dx = F(B) - F(A)$$


Let $P = \{x_i\}$ to be a regular partition of the closed interval from $A$ to $B$, where $i = 1, 2, 3, ..., n$. Then lets construct the Riemann Series for the function $f$:

$$\sum_{i=1}^n f(x_i)\Delta x$$

By the definition, limit of the series will be equal to the integral of $f$ within the same closed interval:

$$\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x = \int_A^B f(x)dx$$

Where

$$\Delta x = \frac{B - A}{n}$$ $$x_i = A + i*\Delta x$$

and so that

$$x_0 = A$$ $$x_n = A + n*\frac{B - A}{n} = A + B - A = B.$$

Since we also know that

$$n \to \infty \Rightarrow \Delta x \to 0 $$ $$f(x_i) = F'(x_i)$$ $$f(x_i) = \lim_{\Delta x \to 0} \frac{F(x_i) - F(x_i - \Delta x)}{\Delta x},$$

we can write that

$$\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x = \lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B - A}{\Delta x}} f(x_i)\Delta x.$$

And following that we can notate the series without need of a second limit via applying the limit of $\Delta x$ to each element of the sum (Also look to the clarificiation at the bottom of the question) as

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B - A}{\Delta x}} \frac{F(x_i) - F(x_i - \Delta x)}{\Delta x}\Delta x = \lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B - A}{\Delta x}} F(x_i) - F(x_i - \Delta x),$$

which is that

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B - A}{\Delta x}} F(x_i) - F(x_i - \Delta x) = \lim_{n \to \infty}\sum_{i=1}^n F(x_i) - F(x_{i - 1}).$$

And so what we obtain is equivalent with

$$(F_1 - F_0) + (F_2 - F_1) + (F_3 - F_2) + ... + (F_n - F_{n-1})$$

which is nothing but

$$F_n - F_0 = F(x_n) - f(x_0) = F(B) - F(A).$$

Following that:

$$\int_A^B f(x)dx = F(B) - F(A)$$

Q.E.D.

Clarification for the part "without need of a second limit via applying the limit of $\Delta x$ to each element of sum":

$$\lim_{\Delta x \to 0} f(x_i)\Delta x = \lim_{\Delta x \to 0} (f(x_i)) * \lim_{\Delta x \to 0} (\Delta x) = \lim_{\Delta x \to 0} (\frac{F(x_i) - F(x_i - \Delta x)}{\Delta x}) * \lim_{\Delta x \to 0} (\Delta x) = \lim_{\Delta x \to 0} (\frac{F(x_i) - F(x_i - \Delta x)}{\Delta x} * \Delta x). $$

So we can continue with

$$\lim_{\Delta x \to 0}\sum_{i=1}^{\frac{B - A}{\Delta x}} \frac{F(x_i) - F(x_i - \Delta x)}{\Delta x}\Delta x. $$

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    $\begingroup$ In spirit, yes. To get it airtight, you need to invoke the MVT somewhere. $\endgroup$ – Randall Jan 18 '19 at 16:24
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    $\begingroup$ $F_n = F(x_n)$ and $F_0 = F(x_0)$. Also $x_n = B$ and $x_0 = A$. So notation error i think? $\endgroup$ – İbrahim İpek Jan 18 '19 at 17:05
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    $\begingroup$ I mean technically you need the mean value theorem to be able to say $f(x_i)\Delta x=F(x_i)-F(x_i-\Delta x)$, right? $\endgroup$ – Cardioid_Ass_22 Jan 18 '19 at 17:30
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    $\begingroup$ @İbrahimİpek: I am well aware that sigma notation represents a sum, thank you. The point I made, though, is that you have a limit (as $\Delta x\to 0$) of a limit (as $t\to 0$), which you are saying can be done as a single limit. I am not saying you cannot, but you have to justify it; in general, nested limits are not equivalent to single limits, and hand waving by itself does not justify it. $\endgroup$ – Arturo Magidin Jan 18 '19 at 20:47
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    $\begingroup$ Note that your “clarification” just compounds the abuse of notation, because the $\Delta x$ in $\lim_{\Delta x\to 0}f(x_i)$ in the second expression is not the same $\Delta x$ in $\lim_{\Delta x\to 0}(F(x_i)-F(x_i-\Delta x))/\Delta x)$ that appears in the third expression. Your $f(x_i)$ is independent of $\Delta x$ in the second expression, but the term in the third expression is not independent of $\Delta x$, so they cannot represent the same limit variable. It’s just handwaving to pretend that they are the same. $\endgroup$ – Arturo Magidin Jan 18 '19 at 20:52
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Your proof would be correct, if you can first prove something like the following:

Theorem? For any increasing integer sequence $b(n)\to \infty$ and double-index sequence $A(k,n)$ such that [insert nice assumptions here...], $$ \lim_{n\to\infty} \sum_{k=1}^{b(n)}\frac{A(k,n)}n=\lim_{n\to\infty} \sum_{k=1}^{b(n)} \frac{\lim_{m\to\infty}A(k,m)}n $$

For you, $A(k,m)$ would be something like $\frac{f(x_k + 1/m) - f(x_k)}{1/m}$. As an indication that this isn't easy, note that it isn't true without additional assumptions:

$$ \frac12 = \lim_{n\to\infty} \sum_{k=1}^n \frac{k/n}n \overset{???}= \lim_{n\to\infty} \sum_{k=1}^n \frac{\lim_{m\to\infty} k/m}n = \lim_{n\to\infty} \sum_{k=0}^n \frac{0}n = 0\dots $$

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I once read a similar approach in one of the textbooks meant for high school students of age 16-17 years. As noted in comments the proof is not really a proof but more of a hand-waving.

Since the function $f$ is continuous it is best to notice that in this case the first and second theorems of calculus are essentially the same. But even then one can't avoid mean value theorem. Here is how you can proceed.

First Fundamental Theorem of Calculus for Continuous Functions: Let the function $f:[a, b] \to\mathbb{R} $ be continuous on $[a, b] $. Then the function $G:[a, b] \to\mathbb {R} $ defined by $$G(x) =\int_{a} ^{x} f(t) \, dt$$ is differentiable on $[a, b] $ with $G'(x) =f(x) \, \forall x\in[a, b] $.

Now we prove the second fundamental theorem as mentioned in your question using the first theorem mentioned above. Consider the function $h(x) =F(x) - G(x) $ then $h$ is continuous and differentiable on $[a, b] $ with $h'(x) =f(x)-f(x)=0$. By mean value theorem $h$ is constant on $[a, b] $ and let's say $h(x) =k$ for all all $x\in[a, b] $ so that $F(x) =G(x) +k$ for all $x\in[a, b] $.

By definition of $G$ we have $G(a) =0$ and $$\int_{a} ^{b} f(x) \, dx=G(b) =G(b) - G(a) =G(b) +k-(G(a) +k) =F(b) - F(a) $$

It is only when the function $f$ is discontinuous the two fundamental theorems of calculus are different and have independent proofs.

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