user friendly proof of fundamental theorem of calculus

Silly question. Can someone show me a nice easy to follow proof on the fundamental theorem of calculus. More specifically,

$\displaystyle\int_{a}^{b}f(x)dx = F(b) - F(a)$

I know that by just googling fundamental theorem of calculus, one can get all sorts of answers, but for some odd reason I have a hard time following the arguments.

Solution 1:

To understand this you will also need to understand the first FTC, and I will show you this. (As you said there may be more than one proof, I will base mine off one similar given in the stewart calculus books).

You may notice during the proof of the first already the connectedness to the derivative which I assume you are familiar with.

$\mathbf{FTC 1:}$ If $f$ is continuous on $[a,b]$ then the function g defined by

$g(x)= \int_{a}^{x} f(t)dt$ for $a \le x \le b$ is continuos on $[a,b]$ and differentiable on $(a,b)$ , and $g'(x)=f(x)$.

$\mathbf{Proof:}$

Suppose we have $x$ and $x+h$ in $(a,b)$, then we have

$$g(x+h)-g(x)= \int_{a}^{x+h}f(t)dt-\int_{a}^{x}f(t)dt$$ $$=\int_{a}^{x}f(t)dt+\int_{x}^{x+h}f(t)dt-\int_{a}^{x}f(t)dt$$

$$=\int_{x}^{x+h}f(t)dt$$

Thus, we have for $h \neq 0$

$$\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_{x}^{x+h}f(t)dt$$

If we assume $h \gt 0$ , since f is continuous on $[x,x+h]$ the "Extreme Value Theorem" says that there exists some numbers, u and v in the closed interval $[x,x+h]$ such that $f(u)=m$ and and $f(v)=M$ where $m$ and $M$ represent the inf and sup of the interval respectively.

Thus, by further properties of Riemann integral

$$mh \le \int_{x}^{x+h}f(t)dt \le Mh$$

$\rightarrow$

$$f(u)h \le \int_{x}^{x+h}f(t)dt \le f(v)h$$

$\rightarrow$

$$f(u) \le \frac{1}{h} \int_{x}^{x+h}f(t)dt \le f(v)$$

ie,

$$f(u) \le \frac{g(x+h)-g(x)}{h} \le f(v)$$

( If $h \lt 0$ the same argument can be repeated with small changes)

so now we invoke limits,

Let $h \to 0$ , then you can see that $u \to x$ and $v \to x$ , ( it is clear from the interval they are in).

So both sides limits are the same, thus the limit exists and we have,

$$g'(x)=\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}=f(x)$$

thus , we finally have $$\frac{d}{dx}\int_{a}^{x}f(t)dt=f(x)$$

$\mathbf{FTC2:}$ If f is continuous on $[a,b]$, then $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where F is any antiderivative of f, that is $F'=f$.

$\mathbf{Proof:}$

Let $g(x)= \int_{a}^{x}f(t)dt$, then from FTC1 we have that $g'(x)=f(x)$

That is, g is an anti derivative of f. Now, if there is another antideravative of f, say F , on the same closed interval [a,b] then we know the only difference they can have is a constant,

$F(x)=g(x)+C$.

Consider $g(a)=\int_{a}^{a}f(t)dt=0$

Now consider,

$$F(b)-F(a)=[g(b)+C]-[g(a)+C]$$

$$=g(b)-g(a)=g(b)$$ $$= \int_{a}^{b} f(t)dt$$

Solution 2:

The key fact is that, if $f$ is continuous, the function $G(x)=\int_a^xf(t)\,dt$ is an antiderivative for $f$. For this, $$ \frac1h\,\left(\int_a^{x+h}f(t)\,dt-\int_a^xf(t)\,dt\right) =\frac1h\,\int_x^{x+h}f(t)\,dt\xrightarrow[h\to0]{}f(x). $$ The justification of the limit basically plays on the fact that $f$ is continuous. A formal proof requires dealing with the formal definition of continuity. Namely, given $\varepsilon>0$ by definition of continuity there exists $\delta>0$ such that $|f(x)-f(t)|<\varepsilon$ whenever $|x-y|<\delta$. If we choose $h<\delta$, then $|f(t)-f(x)|<\varepsilon$ for all $t\in [x,x+h]$. Then \begin{align} f(x)&=\frac1h\int_x^{x+h}f(x)\,dt\leq\frac1h\int_x^{x+h}(\varepsilon +f(t))\,dt=\varepsilon + \frac1h\int_x^{x+h}f(t)\,dt\\[0.3cm] &\leq2\varepsilon + \frac1h\int_x^{x+h}f(t)\,dt=2\varepsilon+f(x). \end{align} Thus $$ f(x)-\varepsilon\leq \frac1h\int_x^{x+h}f(t)\,dt\leq f(x)+\varepsilon, $$ showing the convergence.

Now $G(a)=0$, and $G(b)-G(a)=G(b)=\int_a^bf(t)\,dt$. If $F$ is any other antiderivative of $f$, we have $F'=f=G'$, so $(G-F)'=G'-F'=0$, i .e. $G(x)-F(x)=c$ for some constant. That is, $F(x)=G(x)-c$. Then $$ F(b)-F(a)=(G(b)-c)-(G(a)-c)=G(b)-G(a)=\int_a^bf(t)\,dt. $$