Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

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Recommends Mr Pie · 2018-02 · Mathematics Stack Exchange · on “How to view 'dx' in an Integral?”

The following was originally my comment, but I want to put it in an answer because a conclusion has been reached, which is contrary to the answer most above. Go here $\longrightarrow$ [LINK] for a meaning of $\text{d}x$ in a general integral. In a Riemann Integral , the definition of $\text{d}x$ is a little bit different as it is derived from the Riemann Sum . For that, I suggest going here $\longrightarrow$ [LINK] . There have been definitions like treat $\text{d}x$ as a full stop but this is only to help notating an integral in symbols. This is not the definition of $\text{d}x$. I also found another link to a YouTube video that is part of a series called Essence of Calculus . In this video

Recommends Mr Pie · 2018-02 · Mathematics Stack Exchange · on “What is an integral?”

Sorry about that, then. Perhaps this might help $\longrightarrow$ [LINK] This video mainly talks about why integrals have $\text{d}x$ at the end, but it does brush over on what an integral is. However, for a more advanced and in-depth explanation on integrals and the essence of calculus (mainly the fundamental theorem of calculus), then go here $\longrightarrow$ [this video]

Recommends user532874 · 2019-10 · Mathematics Stack Exchange · on “3Blue1Brown's cloudy argument on the Fundamental Theorem of Calculus”

After watching 3blue1brown's Integration and the fundamental theorem of calculus [this video] , I got confused at around 11:00 [this video] , where Grant says the derivative of the area function is exactly the function itself. The reason for this is since when $dT$ approaches zero, the rectangular approximation $v(T)dT$ resembles $dS$ more and more, it makes sense then that at $dT=0$ , the derivative of $s(T)$ is exactly $v(T)$ . However, $dT=0$ invalidates the equality because a zero denominator is undefined. I think this contradiction possibly highlights a consequence of treating differentials as variables when they're actually not. Is my reasoning correct?

Recommends Ben Grossmann · 2020-02 · Mathematics Stack Exchange · on “Is it normal if proofs dont help me understand a concept better?”

Here is a popular video [this video] that explains the FTOC very well. Note that the segment from 9:40 - 11:20 is really a (big picture, somewhat simplified) explanation of the proof. The video is made better, of course, with the addition of visuals, a physical application, and an example.

Recommends Adam Rubinson · 2020-12 · Mathematics Stack Exchange · on “Integrals and the fundamental theorem of calculus”

"rigorous algebraic proof" and "straightforward explanations" is sort of a dichotomy. The following youtube video by 3blue1brown I found gave a decent overview of FTOC: [this video]

Cites Byte11 · 2018-05 · Mathematics Stack Exchange · on “What happens when I convert a Taylor series into an integral?”

I believe that the summation is an approximation for the integral. Basically, the summation only counts up in whole number increments. The integral counts all values between these numbers all the way up to infinity. This means that the sum will approximate the integral. Here's a visual representation: [LINK] The sum is the summation of the areas of all the green rectangle. The integral is the area under the curve. That's why the two values are similar but not the same. Image source: Integration and the Fundamental Theorem of Calculus by 3blue1brown [this video]

Cites widge · 2020-08 · Mathematics Stack Exchange · on “When we evaluate an indefinite integral of one variable, what area does this yield?”

For example, take the function $ v(t) = t(8 - t) $ from Grant Sanderson's video on integration [this video] . Then its antiderivative is $ x(t) = -\frac{1}{3}t^3 + 4t^2 + C $ . If I evaluate this antiderivative at $ t = 2 $ , so $ x(2) = \frac{40}{3} + C $ , does it make sense to talk of this quantity as an area? Is there an implicit lower bound $ c $ , and if so, what is that bound? If we are given that $ \int_a^b f'(x) \mathrm{d}x = f(b) - f(a) $ , how can we be assured that $ c \le a \le b $ for some lower bound $ c $ that is the lower bound of the areas $ f(b) $ and $ f(a) $ ? I apologize if this question is nonsensical. I've made it through Calculus III in a somewhat rote fashion, and I stil

Cites Álvaro Rodrigo · 2022-02 · Mathematics Stack Exchange · on “A Little Trouble With the Fundamental Theorem of Calculus.”

I'm currently studying calculus. I understand pretty well with the math and with the main concepts. But there is something I don't fully understand, "The fundamental Theorem of Calculus". Not in a comprehensive way, but in a conceptual way. I saw this two videos: [this video] [LINK] And I completely understand the math and the proof. But it leaves me with a bad taste on my mouth, they seem completely unrelated, but they explain the same thing. How it is posible that an area under the velocity curve gives (in the case of a car moving) its position at that current time. And how it is possible just to know how much space someone has covered just by computing this: $$ \int^T_0{v(t) \; \mathrm d t} =

Cites PrincessEev · 2024-01 · Mathematics Stack Exchange · on “How Can Infinite Sums of Zero Sum to 1?”

So again, infinite sums of 0 appear to be equal to a non-zero result. Not quite. Your focus was on $\frac{b-a}{n}$ , but $n$ appears two times in that expression: the second time is in the upper bound of the summation: $$ \frac{b-a}{\color{blue}{n}} \sum_{i=1}^{\color{blue}{n}} f(x_i) $$ So while the factor out front shrinks, you're summing up even more values of the function. For well-behaved functions, these two phenomena "destructively interfere" in a sense, usually reaching a finite value and often nonzero. For others, it can blow up altogether. (To see why, note that the integral can often be thought of as the area under a curve.) In general, the logical flaw here is the claim $$ \lim_{