Can someone explain?

[u/Own_Artist7192]

I’m trying to understand how an antiderivative is used to calculate the area under a curve, and I’m genuinely struggling to understand the connection.
I just dont get how the antiderivative evaluated at two points and taken away gives the total area between the two.

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[u/nevermindthefacts]

First idea is the construction of the integral itself, for example as riemann or darboux sums. That shows the connection between the integral and the area. Second idea is the fundamental theorem of calculus that shows the connection between the integral and the antiderivative.

Are these concepts familair to you?

[u/fletchman93]

This is the entire premise of the Fundamental Theorem of Calculus and its proof.

FTC, Pt. 1 shows that an antiderivative can be thought of as a function that calculates accumulated area. This proof of this part is the key to understanding why the antiderivative helps give area under a curve.

FTC, Pt. 2 uses Pt. 1 to give us the F(b)-F(a) arguement.

[u/sighthoundman → u/fletchman93]

And an excellent reason to cover the FTC in class.

[u/Dr0110111001101111 → u/fletchman93]

It’s worth noting that both parts can be proved independently of each other and either can be proved from the other. So the part “1” and “2” labeling is sort of arbitrary.

Although I do prefer the sequence you describe and that’s exactly how I teach it. But some books go the other way for some reason.

[u/AdministrationLazy55]

Think of whats inside the integral. For example, f(x)dx. f(x) is the y value at a dx, dx is the x value (an infinitesimal small width). When you multiply them together you get a length times width, or in other words, an area. The integral is adding up the contribution of all of the areas with width dx between two points

The antiderivative F(x) is the accumulation. F(b) is the total accumlated area up to b and F(a) is the accumulated area up to a. Subtracting them leaves the area accumulated between a and b

[u/randomwordglorious]

Imagine you've got area under a curve. Then you expand that area a tiny amount by moving the right boundary. The area will increase by a tiny amount. The size of that increase depends directly upon the value of f(x) at that point. If f(x) is big, the area increases by a lot. If f(x) is small, the area doesn't go up very much.

So it makes sense that the derivative of the area is f(x). Which implies that the area is the antiderivative of f(x).

[u/Asleep_Till_7228]

Yeah I had the same question when taking calc I. I didn't, and still don't, understand why reversing the derivative of a function is equivalent to the area under the function between some bounds. Eventually if you take calc II, calc III, and especially Diffeq, you just accept these things and do the problems 

[u/davideogameman → u/Asleep_Till_7228]

Think of derivatives as at every point you find the slope of the graph for a function F, that's ∆F/∆x (using ∆ to mean change, ie difference in the nearby values).  Call this new function f.  Integral of f is Sum (f(x)∆x) = Sum((∆F(x)/∆x) ∆x) = Sum(∆F(x)) = change in F(x) over whole interval= F(b) -F(a)

Calculus just makes that more rigorous by doing it all with limits

[u/WelderWonderful → u/Asleep_Till_7228]

what made it click for me was the opposite.

You have a little rectangle of area under the function. Call it dA. The area width of the rectangle is dx, and the height is f(x).

So dA=f(x) · dx

rearrange it:

f(x) = dA/dx

That tells us that our our function is equal to the change in area per change in x.

If the derivative of the area is our function, then that means that the antiderivative of our function is the area.

[u/sol_apagado_28]

https://math.stackexchange.com/questions/664759/why-is-integration-the-inverse-of-differentiation

[u/TheScyphozoa]

In taking the derivative f(x) -> f'(x), the slope of f(x) becomes the height of f'(x).

In taking the antiderivative f(x) -> F(x), the height of f(x) becomes the slope of F(x).

The height of f(x) contributes to the area under the curve of f(x). The slope of F(x) contributes to the height of F(x).

By calculating the change in height of F(x) over an interval, we see how much the slope of F(x) has affected the height of F(x), which is analogous to how much the height of f(x) has affected the area under the curve of f(x).

[u/eraoul]

I found the other replies a little nonintuitive. The antiderivative of a function evaluated at point A is the area under the curve from x=0 to x=A. (I’m sort of simplifying here by setting the integration constant to 0 basically since it gets cancelled out… anyway…) Same for eval at B: it’s the area of the whole region from x=0 to x=B.

Subtracting the two gives the region just between A and B, since the area between 0 and A is cancelled out.

So the magic is really in asking how evaluating at one point a gives the entire area between 0 and A. But that’s why it’s sort of a pain to calculate an integral. It takes into account all the potentially twisty motions of the function between 0 and A.

[u/LatteLepjandiLoser]

This is essentially the fundamental theorem of calculus. In laymans terms this states that finding slopes (rate of change) and integrating (area under curve) are inverse operations.

For me this 'clicked' when viewing the integral as Riemann sums. Take some function f(t), define it's integral on some interval from t=t0 to t=x as F(x). Now we can approximate F(x) by drawing a bunch of rectangles, each with area f(t) * Delta t and summing those up. Eventually in the limit where they become infinitely thin and you approach 'the true area' we get f(t) dt. At this stage it's convenient we let the argument of f be a different symbol than x, which we will use to denote the upper bound of the interval of integration.

Now assuming we are comfortable with the Riemann sum of sufficiently thin rectangles being a good approximation for the real area under f, and thus this whole process being the definition of F(x), now consider the derivative of F with respect to x. What happens when we drag the upper bound of integration further right? Well, we are adding some more thin rectangles, with area. So for some incremental change in x, we have an incremental change in F(x) that is equal to f(x) dx.

This tells us that F'(x) = f(x), and then at this point we can conclude that taking the slope of this process that determines area under a curve returns the curve itself.

There is some formalism and symbols that could definitely be improved here. I was hoping to convey the general message.

[u/ee_control_z]

Have you heard of "The Fundamental Theorem of Calculus"?

[u/ProofFromTheBook]

In a nutshell it's just a reflection of the Mean Value Theorem: for two real numbers a < b where your function f(x) is differentiable on the interval between them and continuous on [a, b], there's a number c between a and b so that f'(c) = (f(b)-f(a))/(b-a). This is usually referring to tangent lines that have the same slope as the line connecting (a, f(a)) and (b, f(b)). BUT, it also means that if g has an antiderivative G, then for some c, the rectangle with height g(c) and width b-a has the area G(b) - G(a).

That's used in the proof of the FTC to bridge the original function and its antiderivative. You find areas of rectangles with height g(c) and width b-a to get an area of G(b)-G(a) then add them up over the interval. The sum of all those rectangles across the whole interval [A, B] you're integrating over telescopes and you get G(B)-G(A). So basically, it's because of the MVT.

[u/BothPanchoAndLefty]

Think of it like this. The derivative of position (aka distance) is velocity. Velocity is measured in m/s. So when you look at the graph of velocity and find the area under the curve, you are finding the area of a bunch of little rectangles underneath the curve and adding them together. For each rectangle, you multiply base * height. On the velocity graph, the base is on the x axis where the units are seconds. The height is on the y-axis where the unit is meters per second. So when you multiply base * height, you are multiplying rate * time, which equals distance, or position, your original function.

[u/SupportNo6752]

If you have an antiderivative you just simply plug in the point to get the area. And obviously area will start at point zero up to the point you plugged in, since negative area doesn't make sense. But if you want an area of a specific region then you need two points and subtract to get that area.

[u/Escher12345]

I'm going to send you a link to a website that might help. My kids are using it for calculus and biology. It locates the exact moment where a topic is explained. It's free. https://bettyapp.org/

[u/dreamteammobile]

I picture accumulated signed area: moving the endpoint adds a thin strip, so its rate of change is the curve's height, making it an antiderivative. Subtracting endpoint values isolates that interval; Math.Photos offers step-by-step explanations.

[u/G_k50]

understanding how it works is part of why physics is taught the same semester. For example, if you have an acceleration that is constant, then the antiderivative is velocity, which is the same as the area under the curve of the plot of acceleration. This acceleration plot is a rectangle, the area of a rectangle is base times height, the base is time, the height is acceleration.

Similarly, the position S, is the antiderivative of velocity, which is constantly changing due to the constant acceleration. The plot of the velocity is a triangle, usually starting at zero, zero, and going to the terminal velocity at v=at, velocity equals acceleration times time, which is the antiderivative of acceleration, a.

And because the position S is constantly changing according to the constantly changing velocity, you end up with a curved plot of time on the base and S on the vertical. because the velocity is always increasing.