Well, the FTC is broken into two parts. In the first part, what we end up seeing is that differentiation and integration are inverse operations of each other. So if you know that the derivative of sin(x) is cos(x), you can then say the integral of cos(x) is sin(x) + c. (The constant appears due to the fact it would disappear in the differentiation).
The second part of the FTC allows us to understand what a definite integral actually refers to. Basically, given that f(x) = g'(x) for all x in some interval [a,b], the integral from a to b of f(x) dx is exactly g(b) - g(a). That is, intuitively, the definite integral of any function is really the area between the x axis and the function itself. Note, that the integral will return a negative answer for a function below the x axis, and if there is an equal area both above and below, will return a 0.
As well. I feel you. My supervisor for my graduate degree is my age (26). I am scared shitless of him for many reasons, but he's by far one of the most amazing people I've met.
The ideas are simple: differentiation and integration are inverses of each other -- just like adding and subtracting are inverses, or that multiplying and dividing are inverses. If you're able to use them in your practice problems, then you probably know it well enough for this kind of course...
First and foremost, the proof of the FTC is really NOT that important. If this is a major stumbling block to your understanding of FTC, I would ignore it
( ... )
Comments 5
The second part of the FTC allows us to understand what a definite integral actually refers to. Basically, given that f(x) = g'(x) for all x in some interval [a,b], the integral from a to b of f(x) dx is exactly g(b) - g(a). That is, intuitively, the definite integral of any function is really the area between the x axis and the function itself. Note, that the integral will return a negative answer for a function below the x axis, and if there is an equal area both above and below, will return a 0.
Reply
Reply
The ideas are simple: differentiation and integration are inverses of each other -- just like adding and subtracting are inverses, or that multiplying and dividing are inverses. If you're able to use them in your practice problems, then you probably know it well enough for this kind of course...
Reply
Reply
Reply
Leave a comment