Anyone have a good proof for the second part of FTC?

Question

Does anyone have a good proof for the second part of the fundamental theorem of calculus? I haven't been able to find any good videos on it so far so I'd like someone to write it down and I can throw some questions on it if that's okay?

Prove: if $F'=f$ is Riemann integrable on $[a,b]$ then $$ \int_{a}^b f(x)\, dx = F(b) - F(a) .$$

Answer 1

If you are asking for the proof of $\int f(x)\,dx=F(x)\implies F'(x)=f(x)$

$$F(b)-F(a)=\lim_{n\to\infty}\frac1{n}\sum_{r=na}^{nb}f\left(\frac{r}{n}\right)$$

Letting $b=a+h$ with $h=\frac{1}{n}$

$$\lim_{h\to0}F(a+h)-F(a)=\lim_{n\to\infty}\frac{1}{n}\sum_{r=na}^{na+1}f\left(\frac{r}{n}\right)$$

$$\lim_{h\to0}F(a+h)-F(a)=\lim_{n\to\infty}\frac1{n}f(a)$$

$$\lim_{h\to0}\frac{F(a+h)-F(a)}{h}=f(a)$$

$$F'(a)=f(a)$$

Answer 2

My favorite proof doesn't use any Riemann sums, so I like it more. It is as follows:

We start by proving part one.

Let $G(x)=\int_a^{x}f(t)\,dt$. It then follows that $$G(x+h)-G(x)=\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x}f(t)\,dt=\int_{x}^{x+h}f(t)\,dt.$$ Or, equivalently, $$\frac{1}{h}(G(x+h)-G(x))=\frac{1}{h}\int_{x}^{x+h}f(t)\,dt.$$ Now, consider the interval $I=[x,x+h]$. The function $f(t)$ has a maximum on this interval, so we will call it $M$. It also has a minimum on this interval, we'll call it $N$. It is a known fact that $Nh\leq\int_{x}^{x+h}f(t)\,dt\leq Mh$. Dividing through by $h$ gives $N\leq\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\leq M$. Clearly, $\lim_{h\to 0}M=\lim_{h\to 0}N=f(x)$. Thus, by the squeeze theorem, $$G'(x)=\lim_{h\to 0}\frac{1}{h}(G(x+h)-G(x))=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)\,dt=f(x).$$

Now for part two.

Since $G(x)$ is an antiderivative of $f(x)$, any other antiderivative, $F(x)$, of $f(x)$ satisfies the relation $F(x)+C=G(x)$. Since $G(b)-G(a)=\int_{a}^{b}f(t)\,dt$, it follows that $\int_{a}^{b}f(t)\,dt=(F(b)+C)-(F(a)+C)=F(b)-F(a)$. The $t$ can now be changed to an $x$ because it is a dummy variable. Thus, $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$.

Answer 3

If you will forgive me for linking to my own site, I wrote a blog post for my students about understanding the fundamental ideas of one variable calculus.

The proof the the second fundamental theorem of calculus takes place before what I called definition 4 (defining integrals as areas) and theorem 5 (the second fundamental theorem).

Answer 4

Intuitively, this theorem says that "the total change is the sum of all the little changes", and this can be made into a rigorous proof.

Here is some intuition. Chop the interval $[a,b]$ up into tiny pieces: $a = x_0 < x_1 < \cdots < x_N = b$. Let $\Delta F_i$ be the change in $F$ as its input changes from $x_i$ to $x_{i+1}$. Of course, $F(b) - F(a) = \sum_i \Delta F_i$. ("The total change is the sum of all the little changes.") But $\Delta F_i \approx F'(\xi_i) \Delta x_i$, where $\xi_i$ is any point in $[x_i,x_{i+1}]$. Hence \begin{align*} F(b) - F(a) &= \sum_i \Delta F_i \\ & \approx \sum_i F'(\xi_i) \Delta x_i \\ & \approx \int_a^b F'(x) \, dx. \end{align*}

This intuitive argument can be made into a rigorous proof by using the Mean Value Theorem to write $\Delta F_i = F'(\xi_i) \Delta x_i$ with exact equality, for some $\xi_i \in (x_i, x_{i+1})$.

Assume $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F' = f$ is integrable. Let $\epsilon > 0$. There exists $\delta > 0$ such that if $a = x_0 < x_1 < \ldots < x_N = b$ is a partition of $[a,b]$ with mesh less than $\delta$, and $\xi_i \in [x_i,x_{i+1}]$ for $i = 0, \ldots, N - 1$, then the Riemann sum $\sum_{i=0}^{N-1} f(\xi_i) \Delta x_i$ is within $\epsilon$ of $\int_a^b f(x) \, dx$.

So, let $a = x_0 < x_1 < \ldots < x_N = b$ be a partition of $[a,b]$ with mesh less than $\delta$, and select $\xi_i \in (x_i,x_{i+1})$ so that $\Delta F_i = F'(\xi_i) \Delta x_i$. (This is possible by the Mean Value Theorem.) Then we have that \begin{align} F(b) - F(a) &= \sum_i \Delta F_i \\ &= \sum_i F'(\xi_i) \Delta x_i \end{align} is within $\epsilon$ of $\int_a^b F'(x) \, dx$.

Because $\epsilon > 0$ was arbitary, we see that $F(b) - F(a) = \int_a^b F'(x) \, dx$.

Answer 5

What about using the first part of the Fundamental Theorem of Calculus? It seems you have no problem with that part. The first part of FTC says that the function $G(x)=\int_a^x f(t) dt$ is the unique antiderivative of $f$ (so $G'(x)=f(x)$ for any $x$) satisfying $G(a)=0$.

Remark also that $\int_a^b f(t) dt=G(b)=G(b)-G(a)$.

Now, you now that if $F$ is another antiderivative of $f$ on $[a,b]$, then $F$ and $G$ differ from a constant. So there exists a constant $C$ such that $F(x)=G(x)+C$ for any $x$.

So $F(b)-F(a)=G(b)-G(a)=\int_a^b f(t) dt$ for any antiderivative of $f$ on $[a,b]$.