I feel similar about the trace of a matrix being equal to the sum of eigenvalues.
Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.
the discrete version is much clearer to me. Suppose you have a function f(n) defined at integer positions n. Its "derivative" is just the difference of consecutive values
f'(n) = f(n+1) - f(n)
f(b) - f(a) = \sum_a^b f'(n)The fact that the derivative of this accumulator function is equal to the original function, this is the fundamental theorem of calculus, and I violently agree with you that this part is shockingly, unexpectedly beautiful
Where it gets interesting is that if I were to naively construct this accumulator in a non-computational, infinitesimal manner, I would probably start using the idea that the infinitesimal accumulation is related to the instantaneous slope and using that somehow with the previous accumulation. But this is going the opposite direction of derivatives that the fundamental theorem uses.
I understand there are a few other types of integrals that slice things different ways that might be more intuitive. I vaguely recall some (non fundamental theorem) proof that made Riemann integrals click for a while.
To see why \int_a^b f(x) dx = F(b) - F(a) with F'(x) = f(x),
we replace f with f' (and hence F with f) and get
\int_a^b f'(x) dx = f(b) - f(a).
Re-arranging terms, we get
f(b) = f(a) + \int_a^b f'(x) dx.
The last line just says: The value of function f at point b is is the value at point a plus the sum of all the infinitely many changes the function goes through on its path from a to b.
+5 +10 +0 -25 = -15
+5 +12 -2 -25 = -15
They have different graphs if you consider the values above sampling points, this is what parent is asking.
let x_i be in (a,b) with any i drawn from [0,N] and x_0=a and x_N=b
then
int_a^b d/dx F = (F(x_1) - F(x_0))+ (F(x_i+1) - F(x_i))+ (F(x_n) - F(x_n-1)) = F(x_N) - F(x_0)
the derivative produces a ratio of differences and the integral is a weighted sum of those same differences
every middle term is added and subtracted. and only the endpoints remain with opposite sign
So to get the area under the curve between a and b, you calculate the area under the curve from 0 to b (antiderivative at b) and subtract the area under the curve from 0 to a (antiderivative at a).
At least that's my sleep deprived take.
There are many types of examples, and many different reasons why I don't find a particular connection or connection type surprising. So I can concentrate on memorising them, and building intuition.
For the Fundamental Theorem of Calculus:
- int f' = f: the sum of the change is the thing itself. E.g. pour water in the bathtub, if you sum the rate you pour, that's the total water in the bathtub
- int f' = f(t2) - f(t1) : same but water differences between 2 times.
- (int f)' = f: the rate of the sum is the function itself. If you go and integrate your function f, the integrate function's change rate at x is f(x)
- and so on.
> the area under an entire curve being related to the derivative at only two points
This is a very wrong sentence. The area under f on [a,b] is not related to the derivative of f at a and b. The area under f on [0,x] is a real function F(x) by definition, and there is nothing surprising that the area of f on [a,b] is F(b)-F(a). Simple interval arithmetic. Granted F, the integral function, is related to f: F' = f.
tl;dr : in the "fundamental theorem of calculus" there are 2 main observations:
- the operators 'sum of' and 'change rate' are each other's inverse and commute:
(int f)' = int (f') = f
F' = f <==> int f = F
- from interval arithmetics:
S(a,b) = S(0,b) - S(0,a)