Why don't we define the definite integral as the second FTOC?

Let's first confine ourselves to a continuous function $f:\mathbb{R}\to\mathbb{R}$ over an interval $[a,b]$. When we define $$\int_a^bf(x) dx$$ we usually consider the Darboux integral or the Riemann Integral. And, further using these definitions, we prove the properties of definite integrals. And then, there is the second FTOC, which states that if $F$ is anti-derivative of $f$, then $$\int_a^bf(x)dx=F(b)-F(a)$$ I was wondering, for a definition, why we don't use this. I don't really know, but if this theorem works for all functions, whether continuous or not, which have a definite integral, then using this theorem as the definition would really make the proofs of properties of integrals easier.

So, my question: Why don't we use this as a definition?

• Not every integrable function has an antiderivative, that's one reason. – Tim B. Dec 30 '16 at 8:24
• BTW most introductory calculus textbooks in India do define definite integrals as difference between values of anti-derivarive. But this is a totally non rigorous which misses the true aspect of integration which is related to summation and also this approach makes the Fundamental Theorem of Calculus a trivial result of no significance. This in my opinion is intellectual dishonesty on part of textbook authors. – Paramanand Singh Dec 31 '16 at 6:16
• See this answer math.stackexchange.com/a/1768053/72031 to a related question. – Paramanand Singh Dec 31 '16 at 6:19

First of all because FTOC has requirements on the function, that is it does not work for all functions.

Second there are formulations of FTOC (for example on wikipedia) which involves measurement theory which means that the formulation basically means that you have to have the integral concept in place. It would be a circular reasoning.

For example were it fails is for the step function (ie $u(x)=1$ for positive $x$ and zero otherwise). We have:

$$U(x)=\int_{-\infty}^x u(\xi)d\xi = \begin{cases}x& \text{if } x>0\\0 & \text{otherwise}\end{cases}$$

Here we have that $U'(x) = u(x)$ whenever $x\ne 0$, which is the closest we can get to an antiderivate, but it is not quite that. In this case we could ignore the deviation at $x=0$, but what can be ignored is expressed in terms of integrals. The second FTOC says that if $U(x)$ is continuous and define $\chi(x)$ as:

$$\chi(x) = \begin{cases} 0 & \text{if }U'(x) = u(x) \\ 1 & \text{otherwise}\end{cases}$$

then if $\int_{-\infty}^{+\infty}\chi(x)dx = 0$ then the deviation is small enough, but then that means we're doing circular definition.

Another reason is that using antiderivates doesn't expand that well. You seen it already in above example. It does work with nice functions of one variable, but already then it has it's problems for other functions. When extending the concept to multi-varible functions the definition does not expand that nicely. The standard definition however extends nicely to multi-variable functions.