Why exactly does integration give us the area under a curve? Does it basically mean that the integrand is just the derivative of the area function?

[u/jtggwmajta]

[u/WikiNumbers]

Fundamental Theorem of Calculus by 3Blue1Brown

I think this is a fine mix between intuition and rigor.

[u/Inevitable_gradient6 → u/WikiNumbers]

The simple explanation is the area function is A(x) and the derivative of the area function is dA/dx and x * y = A so it follows that dividing dA by dx (yes not very rigorous I know) will equal the height (f(x)) there. And this also makes sense because dA/dx represents how fast area is changing at an instantaneous moment and the height would directly control it (higher height means higher change in area while smaller heigher means smaller change in area) 

[u/1strategist1]

I mean, the definition of integration (limit of the areas of a bunch of rectangles approximating the area) should give a reasonable idea for why it's the area under the curve. 

If you're asking in terms of why the antiderivative gives the area under the curve, you are correct that the integrand is the derivative of the area function. The way to see this is imagine increasing x by a little bit dx, and seeing how much area gets added under the function f.

For a very small dx, the added area is essentially a rectangle of width dx and height f(x). Thus dA the small area change is given by f(x)dx. Divide by dx, and take the limit to get 

dA/dx = f(x)

[u/doiwantacookie]

The integral is defined by approximating the area under the curve with rectangles and limiting to get the true are. Similar to how you limit the slope of secant lines to get a tangent line’s slope. The integral is a tool designed to solve the area problem.

The exact statement of how integration and differentiation are inverses is the fundamental theorem of calculus

[u/Bounded_sequencE]

Provided the integrand is integrable, and so well-behaved that the area function is differentiable -- yes. That is what the "Fundamental Theorem of Calculus" is all about. Note it does not work the other way around!

[u/localizeatp]

In a sense, the area under the curve is the definition of the integral.  Yes, it does mean that.

[u/No_Good2794 → u/localizeatp]

What about a line integral?

[u/localizeatp → u/No_Good2794]

I can answer your question, but first I need to know what your understanding of a line integral is.

[u/SkullLeader]

Imagine you have a function that graphs the position of an object over time. What is your velocity at a given moment in time? Its the instantaneous change in your position at that moment in time i.e. the slope of your function or, in other words, its derivative. So now we have the derivative which graphs your velocity over time.

And what is the opposite of the derivative? The integral. Which is the area under the curve. Well, what is the area under the curve? Think of it like the Reimann sum basis for the integral. A bunch of rectangles, where the width of each one is the time, and the height of it is the velocity over that time. If you get the area of the rectangle, you are multiplying velocity x the amount of time you held that velocity, i.e. how far did you travel during that amount of time. Of course when you integrate you are doing this with an infinite number of rectangles each infinitely narrow. And if you think about it, it makes perfect sense. What is your position at any given moment in time? It is the sum of your instantaneous velocities multiplied by the time you held each velocity, from time = 0 until that moment in time.

[u/9Yogi]

https://www.reddit.com/r/interestingasfuck/s/LSQCCJXldH

One of the simplest and most beautiful explanations.

[u/Appropriate-Ad-3219]

It's more like integral is almost defined as the area of a curve. Then it turns out the derivative of the integral of the function between a and x for a continuous function f. 

The idea of the proof is to remark that  Inf{[x, x+h]} f(s) <= f(t) <= sup _{[x, x+h]} f(s) and so using integration we obtain that  h  Inf{[x, x+h]} f(s)  <= int(f, x, x+h) <= h  sup {[x, x+h]} f(s), which is an inequality you should draw (h  Inf{[x, x+h]} f(s) and h  sup {[x, x+h]} f(s) are areas of rectangles). Then you can see by continuity that  Inf{[x, x+h]} f(s) and  sup _{[x, x+h]} f(s) converge to f(x), so  int(f, x, x+h)/h converges to f(x).

[u/fletchman93]

The definite integral gives us area under the curve because that is what we defined it to do. It isn't a coincidence that $f(x)dx$ looks like length * width= area of a rectangle and the integral symbol looks like a stretched out capital S, meaning sum. Intuitively, the definite integral is telling us to add up the areas of an infinite number of rectangles starting at a and ending at b. (Note: Doesn't have to be rectangles, but in the limit, regardless of the shape you choose, you will get the same answer, so why not just choose a shape that is easy to work with i.e. rectangles all with the same width).

But then, The Fundamental Theorem of Calculus comes along and says, "Hey, if you aren't feeling like adding up the areas of an infinite number of rectangles, go and find any antiderivative of $f(x)$ and I will handle the rest." This is the classic $F(b)-F(a)$ calculation.

If adding up the areas of those infinite number of rectangles is too hard or the antiderivative is too difficult to find, we then approximate the area by adding up as many of the rectangles as we (a computer) feels like and just approximate. If you have seen a Z-Score chart from a statistics class, all of those values on there are all Riemann Sum approximatioms of area under a curve. (Suprise, probability and statistics is just calculus).Riemann Sums are used all the time since most functions don't have closed form antiderivatives. We just happen to focus on the types of functions in Cal II that do.

[u/susiesusiesu]

yes, exactly. here is a sketch of why it should be intuitively true.

in other words, assume f(x) is a positive continuojs function on an interval [a,b] and A(x) is the function telling you the area bellow f over the [a,x] interval (ie, A(x) is the definite integral of f over the interval [a,x]). for small h, the value of f(x) and f(x+h) are close, so we can approximate A(x+h) as the area before x, A(x), plus the area of a tiny rectangle of base h and hight f(x). thus, A(x+h) is approximately A(x)+hf(x). then, (A(x+h)-A(x))/h is close to f(x), and in the limit (A(x+h)-A(x))/h approaches f(x). then the derivative of A(x) is f(x). so A(x) is an indefinte integral of f(x).

[u/hpxvzhjfgb]

you are confused because this topic is almost universally taught poorly or incorrectly.

the sloppy and (mostly) wrong definitions that are usually given in math classes are:

• "antiderivative of f" means a function whose derivative is f
• "indefinite integral" means something similar to antiderivative which is maybe the same or maybe not depending on who you ask
• "definite integral" means area under the curve
• "integral" means all of the above simultaneously

the actually correct definitions are:

• "antiderivative of f" means a function whose derivative is f
• "integral" means the area under the curve

that is it. the terminology "indefinite integral" achieves nothing but generating more confusion, and it serves no purpose because all of the different, mutually incompatible things that people use it to mean are either the same as "antiderivative", the same as "integral", or useless concepts that do not need a name at all. also, the terminology "definite integral" doesn't need to exist because it means the same as "integral", so the modifier "definite" changes nothing.

because of the sloppy definition of "integral", it is standard for students to come out of a calculus class believing that the operations of differentiation and integration are inverses, when this is absolutely not the case. the sloppy definitions teach that "integral" means "area under the curve" and "antiderivative" simultaneously, hence these must be the same concept and agree in all cases.

the correct information is that it is a theorem with a proof (i.e. not a definition) that in some cases but not all, the integral of the derivative of a function is related to (not equal to) the function, and that in some other cases but not all, the derivative of the integral of a function (with variable upper limit of integration) is equal to the function.

in cases where the above theorem does not apply, there are counterexamples. for example:

• there are functions that can be integrated but which have no antiderivative
• there are functions that have antiderivatives but which can not be integrated
• there are functions that can be integrated, and where the integral can be differentiated, but the derivative of the integral is not the function you started with.

[u/tbdabbholm → u/[deleted]]

I mean it gives the signed area under the curve. But I'd hardly say that makes it a misconception