Same here though I don't live in Europe anymore. I soooo much would love this American math teacher give some lectures on physics and astronomy, I finally might get something 😊
English is not my first language, i have lived in europe my whole life, but you mr, you explained this 100x better than what my teacher ever has done even though i am a racist IRL.
Dear John, some teachers know calculus very well, but they do not know how to teach calculus to other people. You know how to teach calculus. Keep up the good work!!
I think I would have excelled in Calculus when I was in college esp with videos like this. I had to learn it on my own with a prof who had a thick Egyptian accent; it was torture but I survived!
they don't consider how the language problem factors in. not just thick accents that you have to focus on to understand instead of the subject, but also the problems in expressing themselves fluently = clearly
Love your calculus videos. I got all of the math classes up to calculus. Analytical geometry was my last class graphing equations on the x-y axis. Never got to the next step which would be calculus. I understand your explanations with ease. That's saying a lot since it's been almost 50 years ago. I have always loved math and believe everything exist because of it. Keep up the good work.
Thanks for the help! You made it simple to understand! Some teachers only want to impress you about how smart they are without really explaining anything in in a simple way!
I don't get why you ignore the dx unless it's the only thing in the integral. For example: integral(xdx) is (x^2)/2, but integral(dx) is just x. How come the solution to first one isn't ((x^2)/2) * x, rewritten as (x^3)/2?
@FirstnameLastnamefn6ik Many many months later, but… I deleted my initial reply to you because I misunderstood your question. You MUST MUST understand and always remember the power rules for differentiation and interpretation, and the rule for constants inside the integral sign! The integral of dx = x because it’s really understood to be the integral of 1dx. Which further goes to equal and mean 1 times the integral dx. The constant comes outside of the integral sign, by the rule for constants inside the integral sign, and is moved to its left side to give 1 times the integral of dx, and the result equals x. Remember, the integral asks,” what function gives 1 when differentiated?” The answer is, a variable raised to the first power. For example, the derivative of y = x is really the derivative of y = x^1. So by the rules of differentiation, the derivative of y = x^1 is y’ = (x^1)’ = dy/dx = (x^1)’ = dy/dx = x^(1-1) = dy/dx = x^0 = dy/dx = 1 = dy = 1dx = dy = dx! So now, when we integrate both sides of that final equation we get, (the integral) of dy = (the integral) of dx, which gives, y = x! With all that in mind, notice the power rule for integration…? The rule says, x^n >>> [(x^(n + 1)]/(n + 1)! Well… by that rule, 1 = x^0 >>> [x^(0 + 1)]/(0 + 1) = x^1/1 = x, so the integral of 1dx = the integral of dx = x! That’s where 1 comes from! So, bearing in mind the power rule for integration and the rule for constants inside the integral sign, and to answer your question more directly, the integral of the expression written on the board becomes, x^3 - 2x^2 + 5x + C.
What are the “little dx” things for? How come they’re there but seem to have nothing to do with the problem as they just disappear? And how did the 5 become 5x?
@4XLibelle “the “little dx” things” simply indicate that the function being integrated, called “the integrand,” is the derivative of the function that results from the integration, is all. And that being the case, that is, and by the power rules for integration and differentiation being inverse processes one to the other, then, and therefore, the processes of differentiation and integration are inverses one to the other. That is, one process is used to find the original of the other process’ result. So, by integration, which is the inverse of differentiation, the integral of 5 becomes 5x, because the derivative of 5x = 5, that’s where 5 comes from; and, by differentiation, which is the inverse of integration, the derivative of 5x becomes 5, because the integral of 5 = 5x, that’s where 5x comes from… they’re inverse processes one to the other! But bear in mind that the answer to an “indefinite integral,” such as the one here, is a “function, in some form f(x) + C. Whereas the answer to a “definite integral” is a “number,” some numerical value. A “definite integral” is so called because it’s “definite,” it’s.. “defined.” That is, it’s arbitrary such that when we’re told to find the “definite integral” of a function, say, the function f(x) = x^2, from x = 1 to x = 4 (1 and 4 are the lower and upper limits, or boundaries, of the integration and are called “the limits of integration”), what we’re being told to do is to find the “numerical” value, or sum, of the area beneath the curve of that particular function bounded by those points 1 and 4, the x-axis itself at the bottom of that area between and up to those points 1 and 4, inclusively, and the curve above the area beneath it between and up to those same points 1 and 4 for that one particular function f(x) = x^2 only, and not for any of the other functions that are family members with the integral of that particular function but differ some constant C and therefore have the form f(x) = x^3/3 + C because they too have the same derivative f’(x) = x^2 because the derivative of their constants is also 0. All other solutions, in other words, of the form f(x) = x^3/3 + C are absolutely excluded, arbitrarily, from consideration as a solution to a definite integral, or an answer to that one particular integrated function simply because it’s already “defined,” “definite;” we only want and therefore need, that is, the integral whose function is f(x) = x^3/3 as the one and only and final answer and solution, that is, we only want the numerical value, or integral of the function f(x) = x^2 and none of the other family members of the form f(x) = x^3/3 + C, period! In other words, we don’t want to know the numerical value of the area under the curve of the function f(x) = x^2 + 1, or the function f(x) = (x^2 - 7 from x = a to x = b, or the numerical values of the areas under the curves of any of the other functions of the family of functions having the form of f(x) = x^2 + C from x = 1 to x = 4 for this particular, “definite,” “defined,” exampled function. We only want the numerical value of the areas beneath the curve of the function f(x) = x^2 from 1 to 4, the “definite,” or “defined,” or, perhaps, arbitrary, or already chosen function, f(x) = x^2. So, the “constant of integration” C, therefore, is there and must always be included along with the integral expression as an answer and solution for an indefinite because it allows and accounts for, and therefore accommodates the fact that an “indefinite integral” is “indefinite” and so can have a family of “functions” as its solutions, and not just one particular function as it’s solution. The derivative of the function y = 2x, for example, is 2. But 2x as the only solution for the indefinite integral of 2dx = 2x would be an incomplete answer since we’re looking for all “functions” that are solutions of the integrated function since, for instance, the derivative of the function y = 2x - 3, and a whole entire host and family of other functions in the form of 2x + C also = 2. So, the integral is “indefinite,” and is, F(x) = f(x) + C, F(x) simply being and meaning the antiderivative of the function being integrated. So the “constant of integration” “C” must be included with the expression for the integral of the function being integrated in acknowledgement and accommodation of that fact. That is, that there are also other “functions” that serve as solutions to the function being integrated since it’s “indefinite,” and as solutions differ from the integral of the integrated function, 2x, only by a constant C, since the derivative of a constant is 0 and therefore all have the exact same derivative …if that’s not confusing…? But more specifically, and technically, integration is summation, it’s adding. That’s why the elongated S, which stands for and means summation. And as such, “the little dx things” also represent the infinitesimally small width of infinitesimally small rectangles beneath the curve of a function f(x) whose function values f(x) serve as the lengths of those rectangles. And the product of those two terms “f(x),” the length, and “dx,” the width, f(x)dx, the product of the areas of those rectangles, inherently and intuitively gives the areas of each of those infinitesimally small rectangles whose areas are then “summed” together, “added” together, to give by approximation the total area beneath the curve of the function f(x) as the approximations approach the total amount or area beneath the curve as the “limit” of the exact area beneath the curve! That method is a geometrical analysis long discovered and established centuries ago by Archimedes, and the crew of other geniuses following him up through the centuries to this day. So when we see the integral of 2dx, that is, the elongated S of 2dx, say, what we’re really looking at is a statement which expresses “the sum of the products of each of the areas of all of the infinitesimally small rectangles beneath the curve y = 2 whose lengths, or heights are 2, which is simply f(x), or the function values of the integrated function, and whose widths are dx, and so whose areas are the products 2•dx, or, 2dx. Circles, triangles, and rectangles, and the likes, have area formulas; but areas under quadratic and cubic functions, and the like, don’t have formulas. So the definite integral of f(x)dx serves as a formula for such functions since they represent and indeed are the length “f(x)” and the width “dx” of the rectangles of an infinitesimal number of infinitesimally small rectangles beneath a curve. And that works! Because remember, a rectangle does have a formula for its area, given by A = lw, that is, its length times its width! And the sum of the areas of all the rectangles under a curve given by the products of each of those infinitesimally small rectangles under that curve gives us the total area beneath that curve! Flippin simplistic genius, isn’t it! The very area itself beneath a curve becomes and indeed is the “limit” of the number of infinitesimally small rectangles that be be inscribed beneath it to give its exact area! The concept of the “limit” is the foundation of differential and integral calculus! In differential calculus, when we differentiate a function we reach a “limit” in approximating the slopes of secant lines of a function closer and closer down to a specific, instantaneous point on a function in finding the slope, or the “derivative” of that function at that specific point on the function, and in integral calculus, when we integrate a function we also reach a “limit” in approximating the area under a curve by inscribing an infinite number of infinitesimally small rectangles beneath it in finding the “ integral” of the function being integrated, or, that is, in finding the area beneath that curve. I know… long-winded…?
at 6.55 we have the result of ∫ x^2 dx is (x^3)/3 Where is the constant? In the final answer we have a constant C, but this integral should have a constant in the answer as well.
@harpleblues he that. ∫ x^2 dx is (x^3)/3 If he thinks he is talking about intergrals that's wrong. ∫ x^2 dx = (x^3)/3. + C If he is talking about antiderivatives he is also wrong because the antiderivatives of x^2 are (X^3)/3. + C With the antiderivative we always mean all functions with the correct derivative, that's why we use the C . (X^3)/3 is just one of them. There are two kind of integrals: indefinite and definite. He is teaching indefinite integrals by the way. And the indefinite integral of a function is the set of all antiderivatives of that function, so that is why it includes the C
@harpleblues An antiderivative is the same thing as both an indefinite integral, and a definite integral. Saying “take the antiderivative of a function” is simply another way of saying, “take the integral of that function,” be that integral a definite integral or an indefinite integral! Integration (integrals) and differentiation (derivatives) are inverse processes of each other. That’s “ Part 1, or, The First Part of The Fundamental Theorem of Calculus”! The one undoes what the other does, mutually. If we took the derivative of the integrated function g(x) = x^3/3 + C, we’d get g’(x) = f(x) = x^2, bearing in mind that C is a constant, and that the derivative of a constant is 0. But we also get f(x) = x^2 when we take the derivative of g(x) = x^3/3, which doesn’t have the constant C in its expression. So that both expressions are solutions of the indefinite integral, except that one differs in value from the other by a constant of C. That being the case, the constant of integration C in the expression for the most general form of an indefinite integral [(x^(n + 1)]/(n + 1) ] + C allows for, and “must” allow for all solutions whose derivatives are the exact same. The very fact that the derivative of C is 0 makes both x^n and x^n + C both be solutions of an indefinite integral. So the constant of integration C in the expression for the most general form of the antiderivative allows for and “must” allow for that fact when taking the indefinite integral of a function. Therefore the constant of integration C “cannot” be omitted from the expression for the most general form of the indefinite integral. or the indefinite integral. And since definite integrals are “defined” by their “limits of integration,” that is, since the definite integral of some function g(x) is “defined” from a to b, or, in other words or terms, “defined” from 0 to 1, say, for example, the constant of integration C is not needed at the end of the integration because there’s only one solution for definite integrals simply because they’re already “defined.” It’s also there to find the particular solution as opposed to a general solution, of the antiderivative when we’re given information in terms of “conditions.” For example, we have the function f(x) = 3x^2. And we’re given an initial condition such that its indefinite integral g(x) = -1 when x = 1. So we first find the most general solution by taking the integral of f(x) = 3x^2, which equals g(x) = x^3 + C. Next, to find the particular solution of that general indefinite integral we substitute our given initial condition into the equation for that general indefinite integral f(x) = x^3 + C, which gives us, g(1) = (1)^3 + C = -1, which gives 1 + C = -1, which gives C = -2. So we’ve found that our C = -2. Then, Finally, we substitute -2 for C in the general solution given by g(x) = x^3 + C to get g(x) = x^3 -2. Which is the particular solution having that initial condition, as opposed to a general solution. Which all tells us that the particular solution we sought is f(x) = x^3 -2, that the derivative, f’(x) = 3x^2, is the slope of that curve, and finally, that the curve passes through the point (1, -1), which was our given initial condition given at the beginning of the problem. That’s the other use for and importance of the constant of integration C.
@@ndailorw5079 At the end you write "we know that x^2 is the anti-derivative of 2x". Why? If I start with f(x)=x^2 + 8. the derivative is f’(x) =2x. So that means that x^2 + 8 is also "the" derivative of 2x? You are wrong in assuming that there is just one antiderivative ("the" anti-derivative" as you say). You can't write about calculating "the" anti-derivative as the inverse process of finding the derivative. That is because if you know the derivative you are not able to tell what your original function was.
@@henkhu100 And I absolutely agree, you’re absolutely correct. It’s incorrect to speak of antiderivatives absent of speaking of the constant of integration C that belongs to and is also crucial to that discussion… unless we’re discussing the definite integral, of course. So… I stand corrected. I tried to mention as much about the constant of integration C around the middle section of my above post. But it’s true that I did a horrible job at it near the end of that post where I failed to make further mention of the crucial importance of adding the constant of integration C when discussing or doing the indefinite integral. But, and at any rate, yeah, you’re absolutely correct! I stated above towards the middle section of my above post that since x^2 has 2x as its derivative, the general form of the expression for the antiderivative “must” reflect, allow and make accommodations for that fact by adding the constant of integration C to the expression [(x^n + 1)/n + 1] + C since an infinite family of x^2 (+/-) C’s also have 2x as their derivative since the derivative of any constant C is 0. But I got sloppy and careless near the end of that post and incorrectly failed to continue and carry along in that portion of it any and all discussion of the constant C, since, as you correctly point out, the derivative of x^2 + C, for all C’s, has the exact same derivative as x^2, that derivative being 2x! So that for that very reason the general form of the antiderivative, again, must be expressed by adding C to the expression like so [(x^n+ 1)/n + 1] + C to allow for that truth. But I’ll revisit my above comment and edit where needed and or delete where needed or delete altogether. Thanks. Don’t want to confuse and mislead.
To a lot of students who say Why do I need to take all this math algebra 1 and 2 Trigonometry And Geometry ?? You need to know ALL of that to do Calculus you will be using it all With calculus you can do and solve things that you as young math students don’t fully understand yet Calculus is truly mind blowing in sooo many ways Isaac Newton came up with this Thought of this .Calculus is not easy not by a long shot it wasn’t for me …Newton was a friken supper genius I can’t comprehend that kind of intelligence I have since forgotten a lot of what All the math I had once known In order to succeed in calculus you have to be an outstanding Algebra and Trigonometry student You simply can’t to calculus if you are not proficient in Algebra and Trigonometry Calculus is the key to unlocking the mathematical universe
I agree. I tried teaching my friends who aren’t taking calculus some things and they were completely lost so then I had to go back and show them OTHER rules and mathematical concepts and that just repeated.
also the reason you're made to learn things in school you'll never use is some people in the class will use it, but they must be exposed to it before they can do that...
I don’t get it so when using the power rule why in some instances you subtract one from the exponent but in other cases you add one to the exponent someone explain 😢
@huriznggraphics8200 The two processes of differentiation and integration are inverses of each other. For example, if we have a function y = x^2, say, then, its derivative, upon differentiating that function is, dy/dx = 2x, by the power rule for differentiation (x^n >> nx^(n - 1). But for integration, the power rule says do the inverse of what’s done with the power rule for differentiation, x^n >> x^(n + 1)/(n + 1). That is, when applying the power rule for differentiation and integration, and since the two processes are inverses of one another, where in differentiation we subtract 1 from the exponent then multiply the base of that exponent by the result we get from subtracting 1 from the exponent, in integration we do the inverse and add 1 to the exponent and divide the base by the result we get from adding 1 to the exponent. And “notice” that in doing so the operations of subtraction and addition, and the operations of multiplication and division are themselves inverse operations, which makes the whole thing and process work! So what good is all that?” If we needed to find the function that was differentiated and gave y = 2x as it’s result (remember that the function was y = x^2), we’d simply do the inverse of differentiation and integrate the new function y = 2x which we got by differentiating the function y = x^2. So we simply integrate y = 2x and we get (2x^2)/2 = x^2, by the power rule for integration, the inverse of differentiation, and get right back to the very original function y = x^2 which we started with at the very outset. That’s why you’re subtracting or adding 1, subtracting for differentiation, and adding for integration. Again, the two processes are inverses of each other. So, likewise, if I integrated a function I could differentiate the result and get back to the original function that was integrated. A practical example, the equation for the law of falling bodies, wind is neglected, is the distance function s(t) = 1/2gt^2. The derivative of that function, the distance function, is the velocity function v(t) = gt and is called the first derivative of the distance function. The derivative of the velocity function is the acceleration function a(t) = g and is called the first derivative of the velocity function and the second derivative of the distance function. Now, suppose you only had the acceleration function but wanted to know how far a body falls in 8 seconds, but you only have the acceleration function a(t) = g. Well, you’d simply integrate the acceleration function a(t) = g to get v(t) = gt. But that’s just the velocity function, and it only shows how fast a body falls at some exact, specific instant of time, but you want to know how far a body falls in a given amount of time. No problem! You’d simply integrate once more for a second time by integrating the velocity function v(t) = gt to get the distance function for falling bodies, s(t) = 1/2gt^2, then plug in your numbers for time and the acceleration due to gravity and find how far a body falls within a given amount of time. And you’re done! Now in that entire process of getting back to the distance function from the acceleration function we were using the power rule for integration. For a(t) = g, to integrate back to the velocity function, g is a constant. But what function has a constant as a derivative? g times some variable. The only variable in this whole entire process is t! So the integral of a(t) = g is v(t) = gt, our velocity function! But we want the distance function. Ok. What function has as its derivative v(t) = gt? By the power rule for integration the integral of v(t) = gt is, s(t) =1/2gt^2! And that’s our distance function m. So that now the distance a body will fall in a given amount of time can be found! But notice at each integration the power rule for integration first added the variable t to the acceleration function to get us to the velocity function, and that same power rule for integration increased the exponent of the velocity function and divided it by 2 to gives us the distance function, which is the function we wanted! So we differentiated twice to get the acceleration function, then went backwards by two integrations to get back to the distance function, thanks to the power rule for both differentiation and integration! So, if you were given the velocity function and you wanted to know a body’s acceleration at an exact time, you’d simply differentiate the velocity function because acceleration is the derivative of velocity. But if you were given that same velocity function and were told to find the distance a body falls over a given amount of time you’d integrate the velocity function because distance is the integral of velocity, or, velocity is the derivative of distance! So, for acceleration, you’d integrate once to get to velocity, and twice, by next integrating velocity, to get distance. For distance, you’d differentiate once to get to velocity, and twice, buy differentiating velocity, to get to acceleration because distance is an integral of velocity. And for velocity, you’d simply integrate once to get to distance since distance is the integral of velocity, and differentiate once to get to acceleration since acceleration is the derivative of velocity. So, by the power rules for differentiation and integration, that’s what all the subtracting and adding the 1, and the multiplying and dividing by the results of that subtracting or adding 1 are all about. Hope this helps ..and doesn’t confuse…
First attribute a math teacher must have is Good English, good pronunciation, You are very deficient in that aspect, It is very distracting that you keep producing so much saliva in your mouth and you keep choking on it and swallowing it with difficulty. Also you say "prom" instead of "problem"
English is not my first language, i have lived in europe my whole life, but you mr, you explained this 100x better than what my teacher ever has done
Same here though I don't live in Europe anymore.
I soooo much would love this American math teacher give some lectures on physics and astronomy, I finally might get something 😊
English is not my first language, i have lived in europe my whole life, but you mr, you explained this 100x better than what my teacher ever has done even though i am a racist IRL.
Dear John, some teachers know calculus very well, but they do not know how to teach calculus to other people. You know how to teach calculus. Keep up the good work!!
I totally fell in love with maths after listening to your lesson. You are such a brilliant teacher. I really appreciate what you are doing.
I think I would have excelled in Calculus when I was in college esp with videos like this. I had to learn it on my own with a prof who had a thick Egyptian accent; it was torture but I survived!
they don't consider how the language problem factors in. not just thick accents that you have to focus on to understand instead of the subject, but also the problems in expressing themselves fluently = clearly
Love your calculus videos. I got all of the math classes up to calculus. Analytical geometry was my last class graphing equations on the x-y axis. Never got to the next step which would be calculus. I understand your explanations with ease. That's saying a lot since it's been almost 50 years ago. I have always loved math and believe everything exist because of it. Keep up the good work.
I totally agree! His videos are always so transparent and it helps me understand everything!!!!!
Thanks for the help! You made it simple to understand! Some teachers only want to impress you about how smart they are without really explaining anything in in a simple way!
Sounds like they need to relearn what teaching is
my school really neglects the steps leading to the answer they just give you the quick route, i enjoy these type of videos quite alot
I'm 62 and I am enjoying revisiting "The Calculus"
great video! much better than the eplanation i got at uni... thanks!
Wish you were my math teacher back then
Even as a eighth grader, watching through this video helped me understand it more than any other.
Dude’s really a good teacher
Like fr he 100x better than the shits in my school
Excellent video!!!
I don't get why you ignore the dx unless it's the only thing in the integral. For example: integral(xdx) is (x^2)/2, but integral(dx) is just x. How come the solution to first one isn't ((x^2)/2) * x, rewritten as (x^3)/2?
@FirstnameLastnamefn6ik
Many many months later, but… I deleted my initial reply to you because I misunderstood your question. You MUST MUST understand and always remember the power rules for differentiation and interpretation, and the rule for constants inside the integral sign! The integral of dx = x because it’s really understood to be the integral of 1dx. Which further goes to equal and mean 1 times the integral dx. The constant comes outside of the integral sign, by the rule for constants inside the integral sign, and is moved to its left side to give 1 times the integral of dx, and the result equals x. Remember, the integral asks,” what function gives 1 when differentiated?” The answer is, a variable raised to the first power. For example, the derivative of y = x is really the derivative of y = x^1. So by the rules of differentiation, the derivative of y = x^1 is y’ = (x^1)’ = dy/dx = (x^1)’ = dy/dx = x^(1-1) = dy/dx = x^0 = dy/dx = 1 = dy = 1dx = dy = dx! So now, when we integrate both sides of that final equation we get, (the integral) of dy = (the integral) of dx, which gives, y = x! With all that in mind, notice the power rule for integration…? The rule says, x^n >>> [(x^(n + 1)]/(n + 1)! Well… by that rule, 1 = x^0 >>> [x^(0 + 1)]/(0 + 1) = x^1/1 = x, so the integral of 1dx = the integral of dx = x! That’s where 1 comes from!
So, bearing in mind the power rule for integration and the rule for constants inside the integral sign, and to answer your question more directly, the integral of the expression written on the board becomes, x^3 - 2x^2 + 5x + C.
Greetings. The integral of the expression is X^3-2X^2+5X+C.
This was Great ....
What are the “little dx” things for? How come they’re there but seem to have nothing to do with the problem as they just disappear? And how did the 5 become 5x?
@4XLibelle
“the “little dx” things” simply indicate that the function being integrated, called “the integrand,” is the derivative of the function that results from the integration, is all. And that being the case, that is, and by the power rules for integration and differentiation being inverse processes one to the other, then, and therefore, the processes of differentiation and integration are inverses one to the other. That is, one process is used to find the original of the other process’ result. So, by integration, which is the inverse of differentiation, the integral of 5 becomes 5x, because the derivative of 5x = 5, that’s where 5 comes from; and, by differentiation, which is the inverse of integration, the derivative of 5x becomes 5, because the integral of 5 = 5x, that’s where 5x comes from… they’re inverse processes one to the other!
But bear in mind that the answer to an “indefinite integral,” such as the one here, is a “function, in some form f(x) + C. Whereas the answer to a “definite integral” is a “number,” some numerical value. A “definite integral” is so called because it’s “definite,” it’s.. “defined.” That is, it’s arbitrary such that when we’re told to find the “definite integral” of a function, say, the function f(x) = x^2, from x = 1 to x = 4 (1 and 4 are the lower and upper limits, or boundaries, of the integration and are called “the limits of integration”), what we’re being told to do is to find the “numerical” value, or sum, of the area beneath the curve of that particular function bounded by those points 1 and 4, the x-axis itself at the bottom of that area between and up to those points 1 and 4, inclusively, and the curve above the area beneath it between and up to those same points 1 and 4 for that one particular function f(x) = x^2 only, and not for any of the other functions that are family members with the integral of that particular function but differ some constant C and therefore have the form f(x) = x^3/3 + C because they too have the same derivative f’(x) = x^2 because the derivative of their constants is also 0. All other solutions, in other words, of the form f(x) = x^3/3 + C are absolutely excluded, arbitrarily, from consideration as a solution to a definite integral, or an answer to that one particular integrated function simply because it’s already “defined,” “definite;” we only want and therefore need, that is, the integral whose function is f(x) = x^3/3 as the one and only and final answer and solution, that is, we only want the numerical value, or integral of the function f(x) = x^2 and none of the other family members of the form f(x) = x^3/3 + C, period! In other words, we don’t want to know the numerical value of the area under the curve of the function f(x) = x^2 + 1, or the function f(x) = (x^2 - 7 from x = a to x = b, or the numerical values of the areas under the curves of any of the other functions of the family of functions having the form of f(x) = x^2 + C from x = 1 to x = 4 for this particular, “definite,” “defined,” exampled function. We only want the numerical value of the areas beneath the curve of the function f(x) = x^2 from 1 to 4, the “definite,” or “defined,” or, perhaps, arbitrary, or already chosen function, f(x) = x^2.
So, the “constant of integration” C, therefore, is there and must always be included along with the integral expression as an answer and solution for an indefinite because it allows and accounts for, and therefore accommodates the fact that an “indefinite integral” is “indefinite” and so can have a family of “functions” as its solutions, and not just one particular function as it’s solution. The derivative of the function y = 2x, for example, is 2. But 2x as the only solution for the indefinite integral of 2dx = 2x would be an incomplete answer since we’re looking for all “functions” that are solutions of the integrated function since, for instance, the derivative of the function y = 2x - 3, and a whole entire host and family of other functions in the form of 2x + C also = 2. So, the integral is “indefinite,” and is, F(x) = f(x) + C, F(x) simply being and meaning the antiderivative of the function being integrated. So the “constant of integration” “C” must be included with the expression for the integral of the function being integrated in acknowledgement and accommodation of that fact. That is, that there are also other “functions” that serve as solutions to the function being integrated since it’s “indefinite,” and as solutions differ from the integral of the integrated function, 2x, only by a constant C, since the derivative of a constant is 0 and therefore all have the exact same derivative …if that’s not confusing…?
But more specifically, and technically, integration is summation, it’s adding. That’s why the elongated S, which stands for and means summation. And as such, “the little dx things” also represent the infinitesimally small width of infinitesimally small rectangles beneath the curve of a function f(x) whose function values f(x) serve as the lengths of those rectangles. And the product of those two terms “f(x),” the length, and “dx,” the width, f(x)dx, the product of the areas of those rectangles, inherently and intuitively gives the areas of each of those infinitesimally small rectangles whose areas are then “summed” together, “added” together, to give by approximation the total area beneath the curve of the function f(x) as the approximations approach the total amount or area beneath the curve as the “limit” of the exact area beneath the curve! That method is a geometrical analysis long discovered and established centuries ago by Archimedes, and the crew of other geniuses following him up through the centuries to this day. So when we see the integral of 2dx, that is, the elongated S of 2dx, say, what we’re really looking at is a statement which expresses “the sum of the products of each of the areas of all of the infinitesimally small rectangles beneath the curve y = 2 whose lengths, or heights are 2, which is simply f(x), or the function values of the integrated function, and whose widths are dx, and so whose areas are the products 2•dx, or, 2dx. Circles, triangles, and rectangles, and the likes, have area formulas; but areas under quadratic and cubic functions, and the like, don’t have formulas. So the definite integral of f(x)dx serves as a formula for such functions since they represent and indeed are the length “f(x)” and the width “dx” of the rectangles of an infinitesimal number of infinitesimally small rectangles beneath a curve. And that works! Because remember, a rectangle does have a formula for its area, given by A = lw, that is, its length times its width! And the sum of the areas of all the rectangles under a curve given by the products of each of those infinitesimally small rectangles under that curve gives us the total area beneath that curve! Flippin simplistic genius, isn’t it! The very area itself beneath a curve becomes and indeed is the “limit” of the number of infinitesimally small rectangles that be be inscribed beneath it to give its exact area! The concept of the “limit” is the foundation of differential and integral calculus! In differential calculus, when we differentiate a function we reach a “limit” in approximating the slopes of secant lines of a function closer and closer down to a specific, instantaneous point on a function in finding the slope, or the “derivative” of that function at that specific point on the function, and in integral calculus, when we integrate a function we also reach a “limit” in approximating the area under a curve by inscribing an infinite number of infinitesimally small rectangles beneath it in finding the “ integral” of the function being integrated, or, that is, in finding the area beneath that curve.
I know… long-winded…?
at 6.55 we have the result of ∫ x^2 dx is (x^3)/3
Where is the constant? In the final answer we have a constant C, but this integral should have a constant in the answer as well.
@harpleblues he that. ∫ x^2 dx is (x^3)/3
If he thinks he is talking about intergrals that's wrong. ∫ x^2 dx = (x^3)/3. + C
If he is talking about antiderivatives he is also wrong because the antiderivatives of x^2 are (X^3)/3. + C
With the antiderivative we always mean all functions with the correct derivative, that's why we use the C . (X^3)/3 is just one of them.
There are two kind of integrals: indefinite and definite. He is teaching indefinite integrals by the way.
And the indefinite integral of a function is the set of all antiderivatives of that function, so that is why it includes the C
@harpleblues By the way: at 6:34 he making another mistake: ⎰x^2. does not have any meaning without a dx
Again one of his many errors in the videos.
@harpleblues
An antiderivative is the same thing as both an indefinite integral, and a definite integral. Saying “take the antiderivative of a function” is simply another way of saying, “take the integral of that function,” be that integral a definite integral or an indefinite integral! Integration (integrals) and differentiation (derivatives) are inverse processes of each other. That’s “ Part 1, or, The First Part of The Fundamental Theorem of Calculus”! The one undoes what the other does, mutually. If we took the derivative of the integrated function g(x) = x^3/3 + C, we’d get g’(x) = f(x) = x^2, bearing in mind that C is a constant, and that the derivative of a constant is 0. But we also get f(x) = x^2 when we take the derivative of g(x) = x^3/3, which doesn’t have the constant C in its expression. So that both expressions are solutions of the indefinite integral, except that one differs in value from the other by a constant of C. That being the case, the constant of integration C in the expression for the most general form of an indefinite integral [(x^(n + 1)]/(n + 1) ] + C allows for, and “must” allow for all solutions whose derivatives are the exact same. The very fact that the derivative of C is 0 makes both x^n and x^n + C both be solutions of an indefinite integral. So the constant of integration C in the expression for the most general form of the antiderivative allows for and “must” allow for that fact when taking the indefinite integral of a function. Therefore the constant of integration C “cannot” be omitted from the expression for the most general form of the indefinite integral. or the indefinite integral. And since definite integrals are “defined” by their “limits of integration,” that is, since the definite integral of some function g(x) is “defined” from a to b, or, in other words or terms, “defined” from 0 to 1, say, for example, the constant of integration C is not needed at the end of the integration because there’s only one solution for definite integrals simply because they’re already “defined.” It’s also there to find the particular solution as opposed to a general solution, of the antiderivative when we’re given information in terms of “conditions.” For example, we have the function f(x) = 3x^2. And we’re given an initial condition such that its indefinite integral g(x) = -1 when x = 1. So we first find the most general solution by taking the integral of f(x) = 3x^2, which equals g(x) = x^3 + C. Next, to find the particular solution of that general indefinite integral we substitute our given initial condition into the equation for that general indefinite integral f(x) = x^3 + C, which gives us, g(1) = (1)^3 + C = -1, which gives 1 + C = -1, which gives C = -2. So we’ve found that our C = -2. Then, Finally, we substitute -2 for C in the general solution given by g(x) = x^3 + C to get g(x) = x^3 -2. Which is the particular solution having that initial condition, as opposed to a general solution. Which all tells us that the particular solution we sought is f(x) = x^3 -2, that the derivative, f’(x) = 3x^2, is the slope of that curve, and finally, that the curve passes through the point (1, -1), which was our given initial condition given at the beginning of the problem. That’s the other use for and importance of the constant of integration C.
@@ndailorw5079 At the end you write "we know that x^2 is the anti-derivative of 2x". Why? If I start with f(x)=x^2 + 8. the derivative is f’(x) =2x.
So that means that x^2 + 8 is also "the" derivative of 2x? You are wrong in assuming that there is just one antiderivative ("the" anti-derivative" as you say). You can't write about calculating "the" anti-derivative as the inverse process of finding the derivative. That is because if you know the derivative you are not able to tell what your original function was.
@@henkhu100
And I absolutely agree, you’re absolutely correct. It’s incorrect to speak of antiderivatives absent of speaking of the constant of integration C that belongs to and is also crucial to that discussion… unless we’re discussing the definite integral, of course. So… I stand corrected.
I tried to mention as much about the constant of integration C around the middle section of my above post. But it’s true that I did a horrible job at it near the end of that post where I failed to make further mention of the crucial importance of adding the constant of integration C when discussing or doing the indefinite integral.
But, and at any rate, yeah, you’re absolutely correct! I stated above towards the middle section of my above post that since x^2 has 2x as its derivative, the general form of the expression for the antiderivative “must” reflect, allow and make accommodations for that fact by adding the constant of integration C to the expression [(x^n + 1)/n + 1] + C since an infinite family of x^2 (+/-) C’s also have 2x as their derivative since the derivative of any constant C is 0.
But I got sloppy and careless near the end of that post and incorrectly failed to continue and carry along in that portion of it any and all discussion of the constant C, since, as you correctly point out, the derivative of x^2 + C, for all C’s, has the exact same derivative as x^2, that derivative being 2x! So that for that very reason the general form of the antiderivative, again, must be expressed by adding C to the expression like so [(x^n+ 1)/n + 1] + C to allow for that truth.
But I’ll revisit my above comment and edit where needed and or delete where needed or delete altogether.
Thanks. Don’t want to confuse and mislead.
Elongated S is symbol for Summa =Sum in Latin .
To a lot of students who say Why do I need to take all this math algebra 1 and 2 Trigonometry And Geometry ?? You need to know ALL of that to do Calculus you will be using it all With calculus you can do and solve things that you as young math students don’t fully understand yet Calculus is truly mind blowing in sooo many ways Isaac Newton came up with this Thought of this .Calculus is not easy not by a long shot it wasn’t for me …Newton was a friken supper genius I can’t comprehend that kind of intelligence I have since forgotten a lot of what All the math I had once known In order to succeed in calculus you have to be an outstanding Algebra and Trigonometry student You simply can’t to calculus if you are not proficient in Algebra and Trigonometry Calculus is the key to unlocking the mathematical universe
I agree. I tried teaching my friends who aren’t taking calculus some things and they were completely lost so then I had to go back and show them OTHER rules and mathematical concepts and that just repeated.
also the reason you're made to learn things in school you'll never use is some people in the class will use it, but they must be exposed to it before they can do that...
I don’t get it so when using the power rule why in some instances you subtract one from the exponent but in other cases you add one to the exponent someone explain 😢
@huriznggraphics8200
The two processes of differentiation and integration are inverses of each other. For example, if we have a function y = x^2, say, then, its derivative, upon differentiating that function is, dy/dx = 2x, by the power rule for differentiation (x^n >> nx^(n - 1). But for integration, the power rule says do the inverse of what’s done with the power rule for differentiation, x^n >> x^(n + 1)/(n + 1). That is, when applying the power rule for differentiation and integration, and since the two processes are inverses of one another, where in differentiation we subtract 1 from the exponent then multiply the base of that exponent by the result we get from subtracting 1 from the exponent, in integration we do the inverse and add 1 to the exponent and divide the base by the result we get from adding 1 to the exponent. And “notice” that in doing so the operations of subtraction and addition, and the operations of multiplication and division are themselves inverse operations, which makes the whole thing and process work! So what good is all that?” If we needed to find the function that was differentiated and gave y = 2x as it’s result (remember that the function was y = x^2), we’d simply do the inverse of differentiation and integrate the new function y = 2x which we got by differentiating the function y = x^2. So we simply integrate y = 2x and we get (2x^2)/2 = x^2, by the power rule for integration, the inverse of differentiation, and get right back to the very original function y = x^2 which we started with at the very outset. That’s why you’re subtracting or adding 1, subtracting for differentiation, and adding for integration. Again, the two processes are inverses of each other. So, likewise, if I integrated a function I could differentiate the result and get back to the original function that was integrated.
A practical example, the equation for the law of falling bodies, wind is neglected, is the distance function s(t) = 1/2gt^2. The derivative of that function, the distance function, is the velocity function v(t) = gt and is called the first derivative of the distance function. The derivative of the velocity function is the acceleration function a(t) = g and is called the first derivative of the velocity function and the second derivative of the distance function. Now, suppose you only had the acceleration function but wanted to know how far a body falls in 8 seconds, but you only have the acceleration function a(t) = g. Well, you’d simply integrate the acceleration function a(t) = g to get v(t) = gt. But that’s just the velocity function, and it only shows how fast a body falls at some exact, specific instant of time, but you want to know how far a body falls in a given amount of time. No problem! You’d simply integrate once more for a second time by integrating the velocity function v(t) = gt to get the distance function for falling bodies, s(t) = 1/2gt^2, then plug in your numbers for time and the acceleration due to gravity and find how far a body falls within a given amount of time. And you’re done! Now in that entire process of getting back to the distance function from the acceleration function we were using the power rule for integration. For a(t) = g, to integrate back to the velocity function, g is a constant. But what function has a constant as a derivative? g times some variable. The only variable in this whole entire process is t! So the integral of a(t) = g is v(t) = gt, our velocity function! But we want the distance function. Ok. What function has as its derivative v(t) = gt? By the power rule for integration the integral of v(t) = gt is, s(t) =1/2gt^2! And that’s our distance function m. So that now the distance a body will fall in a given amount of time can be found! But notice at each integration the power rule for integration first added the variable t to the acceleration function to get us to the velocity function, and that same power rule for integration increased the exponent of the velocity function and divided it by 2 to gives us the distance function, which is the function we wanted! So we differentiated twice to get the acceleration function, then went backwards by two integrations to get back to the distance function, thanks to the power rule for both differentiation and integration!
So, if you were given the velocity function and you wanted to know a body’s acceleration at an exact time, you’d simply differentiate the velocity function because acceleration is the derivative of velocity. But if you were given that same velocity function and were told to find the distance a body falls over a given amount of time you’d integrate the velocity function because distance is the integral of velocity, or, velocity is the derivative of distance! So, for acceleration, you’d integrate once to get to velocity, and twice, by next integrating velocity, to get distance. For distance, you’d differentiate once to get to velocity, and twice, buy differentiating velocity, to get to acceleration because distance is an integral of velocity. And for velocity, you’d simply integrate once to get to distance since distance is the integral of velocity, and differentiate once to get to acceleration since acceleration is the derivative of velocity.
So, by the power rules for differentiation and integration, that’s what all the subtracting and adding the 1, and the multiplying and dividing by the results of that subtracting or adding 1 are all about.
Hope this helps ..and doesn’t confuse…
@@ndailorw5079 thanks man lol that was a whole book 😂
@chocolatecharley99 dope
This looks fun
Why did the dx go away?
It’s makes sense because you would distribute the elongated S to the rest. Now it makes sense
I’m watching cause i find calc a totally insane topic 😂
THANKYOU SIR
as a 9th grader really into math, i kinda dig calculus
thank you
I took precalc a couple years ago and it didn't even teach derivatives
Newton and Leibniz were two clever dudes.
im taking calc II in college watching this...
0:08 nope algebra 2
✨
❤
Goat
i love you
The integral is the squiggly line
Thank you but I'm poor in calculus ab 😂
Tysm! +1 sub. And +1 like
how am i so smart i am just a kid!!!!!!!!!?????????????????
First attribute a math teacher must have is Good English, good pronunciation, You are very deficient in that aspect, It is very distracting that you keep producing so much saliva in your mouth and you keep choking on it and swallowing it with difficulty. Also you say "prom" instead of "problem"
Quagen would go so hard rn