Calculus, explained at a very BASIC level… 

I still have anxiety connected to grammer, more accurately writing, composition... heck language in general.

I am an English language teacher working in an Italian high school. You are not alone. I understand your anxieties. I can assure you that gradually, your anxieties about grammar and language learning will fade away. Grammar difficulties are overcome with time, patience, and practice. The same goes for writing compositions. The important thing is to believe in yourself and be patient. Take care and all the best.@@GG-ud1ib

I learned it in 1997 through a professor with two PhDs who taught us “his way”. I quickly forgot setting the moment after in part his final exam and never heard it again. I’d love to learn it…again.

So true.

That is the operation called integration. I can't explain why it works, but Newton and Liebnitz figured it out. When you integrate a function you can use that integral to get the area under the curve.

@@danolson1467 But there must be a reason why the number is one and not some other number.

Yeah, I’m not sure if he did that on purpose because he’s a teacher……or if it was a Freudian slip.

If you do a search for the beaning of STEM, you get Engineering not education.

It is. I prefer STEAM.

Engineering.

I hardly remember studying calculus. I got some help after school but I was unsuccessful. That was 36 years ago. Maybe if I had taken it later I would have been allright.

You are looking for a deeper explanation of the Power Rule. For the given function, (x^2) the new formula (called the anti-derivative) will always be (x^3/3). It doesn't change. (Won't go into why that works, here. Maybe try Khan academy ) Your region of interest, in this case from x=2 to x=4, are called the limits of integration, and define where you evaluate (plug into) your formula for the anti-derivative: To calculate the area between your new limits of integration, say between 2 and 4, you will plug in the value of the upper limit 4^3/3 minus the value of the lower limit (2^3/3). The formula for the anti-derivative only depends on the form of the original function. It does not depend on the region you are integrating over. You adjust the region you are integrating over by plugging the desired limits into the formula for the anti-derivative. You get the formula for the anti-derivative by applying a simple rule, in this case the Power Rule, to your original function, that is all he is saying here.

@@choister522 thank you. I just got notification of you post. Sorry for the delay of my responce.

Was hoping for a success story man.. you can do it!

Yes, I am not against foreigners at all, but a hard subject like this is impossible when you can hardly understand the words spoken , because of the accents/ weird sounding pronunciation! !!!

@@tesmith47 that's entirely the problem. If he had been teaching from the textbook I might have been able to piece it together. But between the heavy African accent and not teaching from the textbook, it was pretty much impossible

I totally agree. Understanding the concept of limits is essential.

This is not the sort of problem that requires differentiation; this is the other half of calculus called integral.

He just gave you a formula for the curve he drew. The shape and the formula are arbitrary. Funny tho, he started out saying you didn’t have a formula then he made one up.

@@the_best__of_me345 can’t learn calculus by following rules. Can’t learn anything by following rules.

It is the simplest version of a parabola. He didn't just make it up.

For calculating the properties of random strange shapes vector calculus might be a good option. You can construct parameterisations that make pretty weird shapes

Then you approximate the shape as the sum of known, well behaved shapes. That way you can apply the simple rules to the well behaved functions! Sometimes, you have to add an infinite number of well behaved shapes to get it to match just right! We call this an infinite series expansion. (You'll learn these tricks later on (great question, by the way) such as Taylor series or Fourier series expansion. (This may sound intimidating, but Fourier series, for example, is really just a bunch of sine waves) The idea is just the same: add simple functions together to get where you're going! 🙂

I couldn't believe he didn't cruch all of calculus in a 20:00 minute video either

To understand that, you actually have to take the course - it is called the Power Rule, and is one of the most widely used rules in intro calculus. For this video, he just wanted to show you that if you follow a simple rule, you can transform a formula (x^2) into another one (X^3/3) that will quickly give you the area under the curve. And therefore, that calculus doesn't have to be intimidating. 🙂 Can you guess the form of the rule? You actually said it - Increase the exponent by 1, (and divide by n+1). This works for all simple polynomials. Most intro calc students memorize it like this, and it is very easy to apply. However, understanding why the rule works... is a bit more involved... ... but it's a really cool proof!

@@choister522 Thanks. I always loved math but my last course was Trig in college, just short of pre calculus, but I always needed to know why something was done, didn't accept rules without explanation. Thanks again.

Not arbitrary. What it does is to find the area under a curve or graph for which you have a formula - the curve can be a straight line, a parabola, a circle, a sinusoid etc. Once you have the formula for the curve, you can look up the table of integrals to get the "rule", for example if the curve is y = x^n, then the integral is x^(n+1)/(n+1) . After this, to find the area under the graph above the x-axis from say x=3 to x=5, just substitute x=5 and x=3 in the integral fromula and find their difference.

dx is an infinitesimally smal division of the x axis such that it is almost 0. Area in general is the y dimension multiplied by the x dimension. Since the y is changing as per the equation for the curve, to find the area from say, x=3 to x=5 manually, you would first find the first y point by substituting x=3 in the equation and multiply that with a dx of your choice, say dx=0.001 (1 milli-unit of x, whether meter, seconds or whatever) to find the area of the first infinitesimal (very small) rectangle. Then you will increment x to 3+dx = 3+0.001 and find the area of the next small rectangle and so on till after adding up 2000 such rectangles, you get an approximate area under the curve. The integral formula and the integration within bounds (x=3 to 5 here) makes the manual process very accurate by mathematically making use of dx that is very very small (refrerred to as dx tends to 0, or dx is very close to 0 so that it is not the 2000 divisions that we used in the manual integration we did above, but infinite umber of small rectangles are used to find the area).