ddx∫xaf(t)dt

I would love to to understand what exactly is the point of FTC. I’m not interested in mechanically churning out solutions to problems. It doesn’t state anything that isn’t already known. Prior to reading about FTC, the integral is defined as the anti-derivative. So, it’s basically an operator. “Take the anti-derivative by figuring out whose derivative this is!” Simple. So, what is so “fundamental” about redundantly restating the very definition of the integral? (The derivative of the anti-derivative is the function). This to me is like saying −(−1)=+1. Not exactly earth shattering.

Am I missing something with regard to the indefinite vs. definite integral?

If we look at a simple example,

ddx∫x1t2dt=⋯=x2Can we discuss what exactly this is representing?

- Why would you even write this? Why would you take the rate of change

of an area under the curve? Why would you want to take the

derivative of an integral? Or, is this just done to prove something

else? When would you even come across this situation in Math? Taking the rate of change of the area under a curve and/or total displacement? (derivative of the definite integral)- Also, what is the significance of using t as a variable?
- Why would you integrate from a constant to a function in the first

place? (take area under the curve or compute total displacement)I don’t understand what exactly things FTC even allows anyone to do. Without FTC, I can already evaluate definite integrals. Without FTC, I can already take derivatives. So, with FTC, I can take an integral then take a derivative? So, what’s even the point of FTC? I really don’t see anything “fundamental” whatsoever about this redundant self-evident “theorem”. This is like taking the inverse of an inverse. Right back to f(x), but that’s simply a “neat trick” vs. a “Fundamental Theorem of Algebra”.

**Answer**

I am guessing that you have been taught that an integral **is** an antiderivative, and in these terms your complaint is completely justified: this makes the FTC a triviality.

However the “proper” definition of an integral is quite different from this and is based upon *Riemann sums*. Too long to explain here but there will be many references online.

Something else you might like to think about however. The way you have been taught makes it obvious that an integral is the opposite of a derivative. But then, if the integral is the opposite of a derivative, this makes it extremely non-obvious that the integral can be used to calculate areas!

**Comment**: to keep the real experts happy, replace “the proper definition” by “one of the proper definitions” in my second sentence.

**Attribution***Source : Link , Question Author : JackOfAll , Answer Author : David*