Why is the derivative at an instantaneous point, but the integral is the area of the whole function?

Question

I'm reviewing Calc theory for an upcoming course and I can't really piece this part together.

So, the derivative of a function is another function that can tell you the instantaneous slope of two related variables.

But the integral of the same function is another function that will tell you the total area under the graph, or the sum of all instantaneous multiplicative relations of the two variables.

So the derivative can tell you how a specific set of variables are related, their slope, at that point in the function but it won't tell you how the slope of the function behaves before that point. Whereas the integral will tell you another relationship between the variables, the area, but it does it for the whole function, sort of like a summary of the function.

It seems like a discrepancy. Like either the result of the derivative at a certain point should be a summary of the instantaneous slope and the previous slopes, like the integral. Or the the integral should be just the area at that point like the derivative.

Maybe I'm overthinking it or getting to caught up in the graph representation of the relationship. But if anybody could shed some light on this it would be greatly appreciated. I've been using calc in Uni courses for 3 years so maybe I already know this and I'm just not seeing it.

"the integral should be just the area at that point" - The area below a curve at a single point is $0$. — fosho, Jan 14 '16 at 10:02
Not sure what you are exactly saying, but differentiation is a local property, while integration is not. It is nonsense to say "area at a point". If this is meaningful (so that it has some value instead of $0$), then the "area of local region" can be defined by no means. How will you define uncountable sum of nonzero numbers? — Rubertos, Jan 14 '16 at 10:06
Why is a horse not a crocodile? I'm familiar with both of them for over 75 years now, but the discrepancy has remained. — Christian Blatter, Jan 14 '16 at 10:26
Related questions: Why is intergration so much harder than differentiation?, Integration being the opposite of differentiation, Meaning of integration — Winther, Jan 14 '16 at 10:44
Your question is a bit like asking 'Isn't the fundamental theorem of calculus surprising?' Yes, but there's a proof. — , Jan 14 '16 at 10:50

Emilio Novati · Answer 1 · 2016-01-14T10:34:11.537

A simple answer is that the slope of a tangent to a curve (that is essentially its derivative from a geometrical point of view) is a local property while the area enclosed by the curve is a global property.

In a more analytical way we can note that the value of a function at a point $b$, say $f(b)$, is given by: $$ f(b)=f(a)+\int_a^b f'(x)dx $$ where the integral is essentially the sum of all the increments of the function from $a$ to $b$. So, if we know all the increments of the function ( i.e. its derivative) we can find the value of the function at a point, but we need also the value at some other point.

score 0 · Answer 2 · answered Jan 14 '16 at 10:07

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Note the bounds $a$ and $b$ in $\int_b^a f dx$, compared to $f'(a)$ (just one point is evaluated).

answered Jan 14 '16 at 10:07

Elle Najt

20,740

Why is the derivative at an instantaneous point, but the integral is the area of the whole function?

2 Answers2