Trying to explore definite integration and its source of definition.

Question

Suppose if $\phi(x)$ and its derivative is $f(x)$

Let's try to find following $$\int f(x)$$ $$\int f(x)=\phi(x)+c$$

So in indefinite integration we say that integrand would be the derivative of integral.

But as we raise limits to integration like following

$$\int_{a}^{b}f(x)$$

We say that it is actually equal to net area under $f(x)$ from $a$ to $b$ provided $f(x)$ is continuous in $(a,b)$. By "net area", I mean $\text {area}$ above $X \text { axis }-\text{area}$ below $X \text{ axis }$.

So by this definition we can say,

$$\int_{a}^{b}f(x)=\lim_{h\to 0}h\left(f(a)+f(a+h)+f(a+2h)\cdots\cdots f(a+(n-1)h)\right)\tag{1}$$

In above equation $n\rightarrow \infty$

But we also magically say that $$\int_{a}^{b}f(x)=\phi(b)-\phi(a)$$, how we can say this because by definition this looks totally different. Any proof? I didn't find the explanation for $$\phi(b)-\phi(a)$$.

I tried to prove it. My attempt is as follows:-

$$f(a)=\lim_{h\to 0}\dfrac{\phi(a+h)-\phi(a)}{h}$$ $$f(a+h)=\lim_{h\to 0}\dfrac{\phi(a+2h)-\phi(a+h)}{h}$$ $$f(a+2h)=\lim_{h\to 0}\dfrac{\phi(a+3h)-\phi(a+2h)}{h}$$

So by this we can write $f(a+(n-1)h)=\lim_{h\to 0}\dfrac{\phi(a+nh)-\phi(a+(n-1)h)}{h}$

Putting these values in equation $1$

$$\int_{a}^{b}f(x)=\lim_{h\to 0}h\cdot\dfrac{\left(\phi(a+h)-\phi(a)+\phi(a+2h)-\phi(a+h)+\phi(a+3h)-\phi(a+2h)\cdot\cdots+\phi(a+nh)-\phi(a+(n-1)h)\right)}{h}$$ $$\int_{a}^{b}f(x)=\lim_{h\to 0}\phi(a+nh)-\phi(a)$$

As we know $a+nh=b$

$$\int_{a}^{b}f(x)=\phi(b)-\phi(a)$$

So we proved everything but still we were able to prove this when we had the definition of $\int_{a}^{b}f(x)$ in place. How on earth would somebody be able to construct this definition when we just know $\int f(x)=\phi(x)+c$?

Answer 1

Let's limit ourselves to continuous functions over intervals.

Your “proof” is essentially correct, but it begs the question. More precisely, how do you know that every continuous function $f$ over the interval $[a,b]$ has an antiderivative?

If you can produce one, then there is no problem, but can you show a function $\varphi$ such that its derivative is $$ f(x)=\sqrt[3]{\arctan\bigl(\sqrt{x^2+9}-\log(13^x+1)\bigr)} $$ or, more simply, $f(x)=e^{-x^2}$?

The fundamental theorem of calculus does exactly that: it shows that every continuous function over an interval $I$ has an antiderivative.

Let $f$ be a continuous function over the interval $[a,b]$. A partition $P$ of $[a,b]$ is a finite sequence $P=\{x_0=a,x_1,\dots,x_{n-1},x_n=b\}$, with $x_0<x_1<\dots<x_{n-1}<x_n$; the lower sum relative to $P$ is $$ L(P)=\sum_{k=1}^n m_k(x_{k}-x_{k-1}) $$ where $m_k$ is the minimum of $f$ over the interval $[x_{k-1},x_k]$. The upper sum is $$ U(P)=\sum_{k=1}^n M_k(x_{k}-x_{k-1}) $$ where $M_k$ is the maximum of $f$ over the interval $[x_{k-1},x_k]$. The big result is that

the infimum of the upper sums equals the supremum of the lower sums.

Then we can define $$ \int_a^b f(t)\,dt $$ to be this common infimum and supremum. Next, we can define $$ \varphi(x)=\int_a^x f(t)\,dt $$ for $x\in[a,b]$ and prove that $\varphi'(x)=f(x)$, for every $x\in(a,b)$ (the fundamental theorem of calculus). As a consequence of the fact that two antiderivatives of $f$ differ by a constant, we can now state that, if $\varphi$ is any antiderivative of $f$, then $$ \int_a^b f(t)\,dt=\varphi(b)-\varphi(a) $$ Without the FTC and the previous definition of the integral, you cannot know that an antiderivative exists to begin with.