any advice ? im bad at math and the highest math for me is geometry in my sophomore year and im graduating highschool early. and im doing cal 1 and cal 2 in college next month

@@josem-wx7uz your Math skills are undeveloped for Calc.

@@plozar Yeah i started taking college algebra and i take trig next

YOUR PAST NEVER EQUALS YOUR FUTURE.....GOOD OR BAD. Keep breaking down the walls.

🙏

Note taking takes a lot of self discipline and skill

I'm 72, and i flunked College calculus 'cause i ran a Surf shop at the same time and found surfing more fun than math...

Southern Cal??

I’m 76 and found this explanation great, except what C was

Could you tell me what GCSE’s you did for this (if you live in the uk) and a levels for computer programmer cause I will not be doing calculus no way

You don’t really need calculus to become a software developer but you do need good logic plus the ability to copy and paste. Not forgetting AI code generators 🤔

totally agree

Mathematics is a language😊

Fr

It is said that working math problems may keep our brains in great condition in senior years.😊

A very brilliant math student once told me that if you want to learn something..teach it...😅

whoever's skipping, i recommend watching it entirely

Quite helpful 😢

Thank you! I'm beyond exasperated with the ridiculous blather! Maybe I'll hang on a bit.

Thank you!!! I understand why the intro was there, but that intro wasn't for me.

are u with us 😂😂 or in the Pluto planent😂

Lol, sounds like you don’t need help.

The fact I can’t even tell if you’re speaking English or not says a lot 😭 lagrangians what now?

God bless you as well 😁💗

I agree that note-taking is hard in math. Maybe some handouts with mini-quizzes built into them.

Specifically looking to solve the x squared(^2) function, see 10:06 into video

Don’t know if I am understanding your question correctly… If you integrate the function from 1 to 3 you would simply subtract 1 from 3 in doing the mathematical calculation after integrating. Here, the area of the function being integrated is from 0 to 2, so you would subtract 0 from 2 by that same process. If it were from 1/4 to 1/2, you would employ the very same process! Does that answer help any? For better and greater clarity, perhaps, let’s use a simpler more straightforward geometrical figure, a rectangle, and suppose you integrated the shaded area of the function y = 2. “Its integral would be 2x.” And let us suppose that the boundaries of the function were the x-axis from x = 0 to x = 4, the y-axis from y = 0 to y = 2, the line x = 4, from y = 0 to y = 2, and the line y = 2 from x = 0 to x = 4. Now, we already know, even before doing any calculus at all, that area has to be a rectangle with a length of 4 units and a width of 2 units, and that the product of those dimensions must give us an area of 8 units^ 2. Now if you integrated that area, which should give the integral of 2 from 2 to 3 to get 2x from 2 to 3 by subtracting 2 from 3 you should get 1/4 the area, which would be 2 units ^2. Whereas on integrating the same function from 0 to 4, which is the entire area of the rectangle, by subtracting 0 from 4 in doing the math you should get 8 units^2, which, again, is the entire area. That’s how it’s done… Hope that clarifies things somewhat rather than confuses you even more. No sé si entiendo bien tu pregunta... Si integras la función de 1 a 3, simplemente restarás 1 de 3 al realizar el cálculo matemático mientras integras. Aquí, el área de la función que se está integrando es de 0 a 2, por lo que restarías 0 de 2 mediante el mismo proceso. ¡Si fuera de 1/4 a 1/2, emplearías exactamente el mismo proceso! ¿Esa respuesta ayuda en algo? Para mayor claridad, tal vez usemos una figura geométrica más simple y directa, un rectángulo, y supongamos que integra el área sombreada de la función y = 2. “Su integral sería 2x”. Y supongamos que los límites de la función fueran el eje x de x = 0 a x = 4, el eje y de y = 0 a y = 2, la recta x = 4, de y = 0 a y = 2, y la recta y = 2 desde x = 0 hasta x = 4. Ahora, ya sabemos, incluso antes de hacer cualquier cálculo, que esa área tiene que ser un rectángulo con una longitud de 4 unidades y un ancho de 2 unidades. , y que el producto de esas dimensiones debe darnos un área de 8 unidades ^ 2. Ahora, si integraste esa área, lo que debería dar la integral de 2 de 2 a 3 para obtener 2x de 2 a 3 restando 2 de 3, debería obtener 1/4 del área, que serían 2 unidades ^2. Mientras que al integrar la misma función de 0 a 4, que es el área completa del rectángulo, al restar 0 de 4 al hacer los cálculos deberías obtener 8 unidades ^ 2, que, nuevamente, es el área completa. Así es como se hace... Espero que esto aclare un poco las cosas en lugar de confundirte aún más.

@ William Rudman Which “units” are you referring to? If you mean x^2, he simply chose 2 arbitrarily to substitute for n. In other words, he, or you, could’ve chose any number (power), he could’ve chosen 1, or 4, for instance. X^n is the general formula or form that allows us to choose any number in the place of n, the n simply allows us to choose the number (power) we’ll replace it with to find the area under the particular curve our choice gives us. If the formula was x^5, say, that would be fixed and final; we couldn’t substitute any number for 5, we could only use 5; n is an unknown, it’s variable and therefore allows us to do our own choosing as to what number it’ll be. 5, on the other hand, or any other number, for that matter, is fixed and final and a done deal and allows us to do nothing other than use it. Therefore we could never use it to find areas under curves raised to different powers other than 5. It would be useless for anything else but itself! But general formulas using unknown values such as n allows us to arbitrarily choose what we want, we could make n 2, 4, whatever! In other words, if we say f(x) = 2x + 1, just for example of the importance of variability, we could choose an infinite number of values for x. But if we limited x to only equal 4 what useful value would such a formula be to us when we wanted to find f(x) at all values other than 4? Well that’s what’s going on here with n! It allows us to pick a value in its place, is all… n simply says, go on why don’t ya, have fun, create your own graph by plugging in a number substituted for me, is all! But if you mean the expression x^3/3 - x^3/3 from 0 to 2, he simply followed the rule for integrating functions: x^n goes to x^(n + 1)/n + 1, which gives, when we follow the rule, and when we choose n to equal, or said another way, to be 2, for instance, and as done here, x^2 becomes x^3/3. Then we evaluate that expression from 0 to 2 since the area under the curve that we’re trying to find is bounded by the curve f(x) = x^2 and x = 0, and x = 2. Then he takes that integrated expression, x^3/3 and evaluate it from 0 to 2 like so: x^3/3 - x^3/3 becomes 2^3/3 - 0^3/3 = 8/3 - 0/3 = 8/3 - 0 = 8/3 ≈ 2.67. Or perhaps in asking how does he end up with the units squared, the answer you’re looking for is because the size of an area is found by multiplying the lengths of two sides of something. For instance, a board that measures 3ft. X 4ft. = 12 ft.^2, or 12 square feet… the units square because ft. X ft. = ft.^2, just as b • b = b^2. For the same reason, volume is given in cubic units because in that case the lengths of 3 sides are multiplied together to give units^3, that is, ft. • ft. • ft. = ft^3! Now… if you didn’t ask for all that, then, my apologies.

It’s used in computer science.

@@jeffthevomitguy1178 OK it’s used in computer science. so with the computer I would think that a cactus calculation could be shown in an animation real time event. For example, launch a SpaceX rocket into space take it through Mexicue until second stage separation start up. Show all of the calculus calculations going on and change the values and part of the screen while the rocket goes into orbit or stage

X^3/3 becomes 2^3/3 when 2 is substituted for x to become 2^3/3 which becomes 2•2•2/3 which becomes 8/3. 2 times 2 times 2 = 8. You don’t add the 2’s, you multiply them!

@ Robert Fanfalone Geometrical argument and analysis in trying to find the areas of odd shapes that have no equations but are built through geometrical analysis and scientific investigation and observation that began with the Greeks, especially with Archimedes, who lived around 287-212 B.C. A brief research of the topic of “The Quadrature of Parabola” of Archimedes will give you some insight and mental grasp into how the formula came about centuries ago with Archimedes’ discovery and development of the formula x^3/3 + C through his work by way of geometrical analysis of the quadrature problem. And it is said that had Archimedes known the things learn in the later centuries he would’ve discovered the calculus… way back then! Geometry, algebra, geometrical analysis, trigonometry, to name a few, and of course scientific investigation all serve as the origin and foundation not only calculus, but also functions as the one seen here. This stuff goes way back for centuries. There’s a scientific video series by Cal State titled “The Mechanical Universe and Beyond” which debuted around 1985 or so. And it’s done in a most wonderful narrative form as the narrator goes through the history of the sciences and mathematics from the earliest recorded time in history up to the mid to late 1980’s! It has 56 episodes of roughly 30 minutes each! Check it out here on KZfaq! It gives you the history of all mathematics, all the sciences and brief biographies of all the great mathematicians and scientists, especially, it’s narrates the history of the efforts of proclaimed inventors of the calculus, Newton and Liebnitz, and the little spat and beef over the race between them to be the first to invent the calculus, interesting, it narrates the discovery of “The Law of Falling Bodies” which maps out the path of a parabola and discovered by Galileo, as well as his life trials and other discoveries, a genius he was, as well as the others, but he’s one of my favorites, if not my favorite… sometimes it’s Newton, though. And it gives you both the origin and the process of developing mathematical functions, and the likes. And to top things off, it teaches calculus and physics as you go through the entire series! Incomparably valuable resource… don’t miss it! And you’ll never regret it!

​@@ndailorw5079 Great info Robert! Thank you for sharing the details.