15
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Are you familiar with the Riemann sum definition of the integral? $\endgroup$
    – user99914
    Commented May 2, 2015 at 2:24
  • 4
    $\begingroup$ There is no real easy proof of this, it's a deep fact. The best intuition is to think of the anti-derivative as keeping a running sum of the accumulation of the area, where the sum is taken to the limit. $\endgroup$ Commented May 2, 2015 at 2:29
  • 1
    $\begingroup$ What conditions do you want on $f$? $\endgroup$
    – Chappers
    Commented May 2, 2015 at 2:36

3 Answers 3

18
$\begingroup$

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 on 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 exist some numbers, u and v in the closed interval $[x,x+h]$ such that $f(u)=m$ 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 antiderivative of f. Now, if there is another antiderivative 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,

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

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

$\endgroup$
2
$\begingroup$

i will assume that $f$ is continuous, $F$ is differentiable in $(a,b)$ and $$F'(x)=f(x). $$ we will need the mean value theorem in the form $$\text{ for any } a \le c, d < \le b,\quad F(d) - F(c) = (d-c)F'(t) = (d-c)f(t) \text{ for some } c < t < d.$$

let $a \le x_0 \le x_1\le \cdots\le x_n = b$ any partition of $[a, b].$ then $$\begin{align}F(b) - F(a) &= \left(F(x_n) - F(x_{n-1}\right) + \cdots +\left(F(x_j) - F(x_{j-1}\right)+\cdots+\left(F(x_1) - F(x_{0}\right)\\ &=(x_n-x_{n-1})f(t_n)+\cdots+(x_j-x_{j-1})f(t_j)+\cdots+(x_1-x_0)f(t_1)\\ &=\int_a^bf(x)\, dx.\end{align} $$

$\endgroup$
1
  • $\begingroup$ I think you will need a bit more justification in the last line. What you have shown is that if $P$ is any partition of $[a, b]$ then $F(b) - F(a)$ can be regarded as a Riemann Sum of $f$ over $[a, b]$. But a Riemann Sum is not necessary equal to Riemann integral. In this case since $F(b) - F(a)$ is a Riemann for every partition $P$, it follows that it lies between every lower darboux sum and every upper darboux sum and since $f$ is integrable, $F(b) - F(a)$ is the unique limit of lower and upper darboux sum and therefore is the Riemann integral of $f$ over $[a, b]$. $\endgroup$
    – Paramanand Singh
    Commented May 3, 2015 at 9:34
1
$\begingroup$

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 $$ F(b)-F(a)=(G(b)-c)-(G(a)-c)=G(b)-G(a)=\int_a^bf(t)\,dt. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .