Good job; I learned Calculus in college and loved it, but never used it in my career as a Chemist, so the knowledge departed. I think it may be fun to relearn.
One has a function f(x) = 2x^2 - 3x + 2 where the limits on the x - axis is x=0 to x=2. To find the area of this function, one takes the integral of f(x) and one has a = 2/3*x^3 - 3/2x^2 + 2x with the limits of x=0 to x=2. Once one calculates the limits, one has the result a = 10/3
If anyone knows of a good video or page that teaches slow people like me "how to determine a function of any curve", I'd be much appreciative. And thank you TabletClass for your useful videos! IOWs, this helpful video from TabletClass has taught me how to find the area size if I already know the curve's function. But what do I do if the function still needs to be determined?
To determine the function that matches a given curve, you'd usually match the curve with the family of functions that have a similar shape. You usually also look for a physical basis for why this curve is a given shape, if the curve represents real-world data. Such as matching a temperature vs time graph to an exponential decay. Then, you'd establish known points on the curve, and set up a system of equations to solve for the parameters that define the curve. For instance, consider a polynomial function that passes through the points (0, 7), (1, 0), (5, 32), and (7, 0). You'd set up a system of equations based on the general form of a cubic, y=a*x^3 + b*x^2 + c*x + d, and apply each of these known points to x and y. You'll then solve this system for a, b, c, and d. So your equations would be: 7 = d 0=a + b + c + d 32 = 125*a + 25*b + 5*c + d 0 = 343*a + 49*b + 7*c + d Trivially, we know d = 6, because the y-intercept was given as one of the points and all terms containing x disappear. So we can simplify the system accordingly. We can also divide the middle equation by 5, and the final equation by 7. -7 =a + b + c 25 = 125*a + 25*b + 5*c, simplifies to: 5 = 25*a + 5*b + c -7 = 343*a + 49*b + 7*c, simplifies to: -1 = 49*a + 7*b + c From this point forward, we'd set up an augmented matrix, and use Gauss-Jordan elimination (or other linear algebra methods of our choice) to solve for a, b, & c. The solutions are: a = -1 b = 9 c = -15 d = 7 And thus the equation is: y = -x^3 + 9*x^2 - 15*x + 7
@carultch great explanation, but is still doesn't make sense to me. Do you know of any videos that could basically explain what you just wrote? Thanks
The math is not hard but not all teachers do a good job of explaining all the different symbols. For example Capitol F is the integral so Capitol F prime is the derivative of the integral which equals little f our original function. Small dx means infinitesimal change in x. Teachers often jus say thing without explaining well what they are. After a few weeks of calculus all students should know what capitol F prime and small dx means without explanation but in the beginning math teachers need to explain this and explain it over and over and over to drive the point home for a couple weeks
The capital and lowercase letters do not necessarily mean derivative and antiderivative. In fact, in Vector Calculus, it was common to use F and f in the other way around. The prime usually represents differentiation, but it can also just represent a variant on the original letter. The point is, don't get too attached to notation. It can be different in a different application. The d notation an extension of Δ notation, where Δx means change in x. The dx you can think of as "dinky x", since it is a little tiny change in x. If a problem is expecting you to know that f'(x) = F(x), and f(x) = integral F(x) dx, then the problem needs to define its terms. Like "find a function f(x), such that the derivative of f(x) equals the function given below as F(x)".
Excellent tutorial! Just a question; what would the unit of measurement be for the calculated "area under the curve"; I know it all depends on the units for X and Y. OK! Let's say X is time in years and Y is the number of newly diagnosed cases of a certain type of cancer; what is the unit for the calculated figure (in this case 10/3 what?). Hope the question is clear.
Area is measured in square units. It could be any unit; In other words, units could be m², cm², inch², ft², etc. In this case he didn't specify what the units are, but the answer is (10/3)units², or 3.333units². In your case if x is time, and y is diagnosed cases, then if the unit isn't officially defined, you can make up your own unit. You could have 3.333omid² 😀.
seems like we take the x portion of the DX and multiply it against the 3 terms in the equation, x *x2 and x*x and 2*x and of course apply the rule to the denominators..
That reasoning only works for polynomial terms. It doesn't work in general. Locally, integration is a form of multiplication, because you are adding up each value of the function, each multiplied by a tiny change in the function (dx). So each term we are adding up is f(x) * dx, and we have an infinite number of terms to add up along the way. This is why for constants, it is as simple as just multiplying by x. For terms like x, it isn't as simple as just multiplying by x, because we also have to account for the fact that x and x^2 are different sizes along the way. For integrating a term like k*x where k is a constant, it is like the area of a triangle. Its height is k*x, and its base is x. It's area is thus 1/2*k*x^2. For higher ordered polynomial terms, we need to account for the fact that these terms are a lot slower to grow, and spend a lot more "time" closer to zero. It grows slowly near zero, and grows faster and faster, the farther away from zero you get. The higher the exponent, the more extreme this behavior becomes. This is why rather than dividing by 2 like we did for the linear term, we divide by the new exponent (3 for x^2, which integrates to 1/3*x^3).
When it is an indefinite integral, you always add the +C. The reason is that there are an infinite number of functions that only differ by an added constant from each other, all of which have the same derivative. So to collectively represent this infinite family of functions, we add an arbitrary constant, of +C, when asked for the indefinite integral, or general solution to the antiderivative of a given function. If it were a definite integral, you can keep it simple, and let C = 0. No matter what you make C equal to, as long as it is the same for evaluating both limits, it will cancel itself out anyway. So you don't need to think about +C for indefinite integrals. To give some context on what it means in an application, suppose you were given a speed vs time function for a car, and wanted to know what mile marker it ends its journey after a given amount of time. Suppose its speed vs time is constant at 40 mph, and it travels for 2 hours, and it takes a negligible amount of time for it to speed up and slow down. Do you have enough information to know what mile marker it ended upon? No, you don't. You know it travelled 80 miles, but 80 miles from where? Since there are an infinite number of possible starting locations, there are an infinite number of possible solutions. So in this case C represents the initial position. C in general, doesn't necessarily represent the initial position, but it is directly determined by it.
@@marknascimento5037that doesn't really help. Let's say you had a word shaped room with lots of curves and you wanted to know the floor area exactly. How would you do it using calculus?
@@MrDirkles You'd have to have a mathematical definition of all the letters in that word. For instance, let's suppose a room shaped like the letter D is defined by a vertical line through the origin, and the equation x = 1 - (y - 1)^4. Plot this, and confirm that it looks like the letter D. Observe that the vertical line intersects the curve at y=0 and y=2 To calculate the area, we need to integrate the function relative to y, to find the area between the vertical line, and the quartic polynomial defining the rest of the letter D. You could expand it, but you don't really need to. So we are now integrating: integral (1 - (y - 1)^4) dy Let u = y-1. Observe that du/dy = 1, since this is a simple linear term with a slope of 1. Thus, we can rewrite it as: integral (1 - u^4) du Increase each exponent by 1, and divide by the new exponent: integral of 1 du = u integral of -u^4 du = -1/5*u^5 Thus we have: u - 1/5*u^5 Translate back to the y-world: y - 1 - 1/5*(y-1)^4 Evaluate this at y=2, and subtract the evaluation of this at y=0: At y=2, we have .8 At y=0, we have -1.8 Thus, the area inside this D-shaped room, is 1.6 square units.
Why add 1 to the power? Why not add 2, or 3? How did you get the equation for y to begin with? I appreciate the lesson, but I would advise that if you do this agin, explain each and every equation and why/how it works.
I am 74 years old and found your site a few weeks ago....I am really interested in learning just for fun....you make it easy...my question is this: I see where the 2, 0 come from....but can you have a parabola be on the positive side and the negative side as well?.....example: 3, -2.......how would you find the are under that using the same equation? would you just follow the same process but now have to do it for 3 and -1? then, would you have 2 answers...or just use the positive answer/ ...thanks........OKU
@okupchurchill7850 You’d still use the same process, but you’d simply go from -2 to 3, instead. And no, you’d get one answer, or one solution. Here you’re only substituting values for x, namely, -2 and 3, in order to reveal the answer. And area, length, time, etc., can never be negative, in the strictest sense and meaning of the existence of those things… there’s no such thing as a negative amount of land, or a negative amount of length, or a negative amount of time, etc., say, for example! But, nonetheless, acres, feet, minutes, etc., may and can be added to or subtracted from one another, respectively. So that caution with regards to your algebra must be taken when calculating the integral. For an example and for a trial run, try taking the integral of y = x^2, or, f(x) = x^2, if you prefer that notation, from -2 to 3, or perhaps, and perhaps somewhat clearer, from -3 to 3. When you get to that part where the expression becomes (x^3/3) - (x^3/3) from -2 to 3 (or from -3 to 3, if you choose) you must remember and bear in mind that where -3 is substituted for x you’re then actually multiplying (x^3/3) by (-1), by the algebraic rule for subtraction, like so, (-1)[(-3)^3/3)] = (-1)(-27/3) = 27/3! Which is all to say that we must first change the sign of operation (in this case subtraction) to its inverse or opposite operation (addition), then change the sign of the term(s) that follow after the sign of operation. For example, 4 - (-4) >> 4 + (+4) >> 4 + 4 = 8. So, and with that in mind, taking the integral of f(x) = x^2 from -3 to 3, for example, we get x^3/3 - x^3/3 from -3 to 3, which itself becomes, and again, by the rule, [(x^3/3) + ((-1)(x^3/3))] from -3 to 3 = 54/3 = 18sq. units. Most of the mistakes in calculus are algebraic! Therefore, strength in algebra, and trigonometry, makes life in calculus easier and much much less of a headache! But although the sheer intellectual genius and beauty both essential to and fundamental and foundational to calculus make it relatively easy, that doesn’t mean nor should it be inferred, necessarily, that it’s a walk in the park…
Ten thirds makes me think 🤔 🧐 about the area underneath the parabola “y = 2x^2 - 3x + 2” and enclosed by the x-axis, the y-axis, and the line y = 2; the four sides of the yellow region.
@ Bradley Martinez The instructor’s final answer makes it pretty obvious that “ten thirds” is the area beneath the curve (that is, the area In square units, whatever we choose those units to be, whether we choose for our units feet, inches, meters, or whatever we choose for our units) so much so that it becomes unnecessary and redundant to say “ten thirds makes me think about the area underneath the parabola “y = 2x^2 - 3x + 2”…” And the area beneath the curve is bounded by the three sides, or the three lines, x = 0 (which itself is the y-axis!) from y = 0 all the way up to y = 2 along, again, the y-axis (not the entire y-axis, mind you), and by the line y = 0 (which is itself the x-axis!) all the way from x = 0 to x = 2 along, again, the x-axis (again, not the entire x-axis, mind you), and by the line x = 2 all the way from y = 0 to y = 4, and finally by the curve y = 2x^2 - 3x + 2 itself bounded by those three lines are indeed the four sides of the yellow region! And what, if anything, does the line y = 2 have to do with this particular graph and finding it’s area?
Nope, there’s no explanation of how the original formula describes the curve. There’s no explanation of how you got from 6 & 4 to 18/3 & 12/3 (although I think I can guess). And the area is 10/3 what? I can’t cut a board to 10/3 without knowing the unit of measure and even the wouldn’t it be 3.333? At 58 years old I’ve discovered the math I didn’t think I would use in high school is actually applicable in Machinist trades. But I’m a very literal thinker and math teachers do not explain problems literally which is why we fail.
@@DeerAssassin And your reply to me tells us all we need to know about you, and why you posted your first comment regarding this video in the first place… 58 years old, eh…?
@@ndailorw5079 what it tells you is your explanation was uppity and lengthy, and somewhat insulting. Like 40 years ago in high school math class, there’s always an assumption that the student should know more. I can explain algebra and most geometry to my kid because I know what the placeholders are for and what they represent. I watched about a dozen of your videos and learned several things but there is some basic information missing. My response to your reply shows I’m old enough to get to the point, it wasn’t helpful. Helpful would be pointing me to the video explaining what basic information I’m missing to help the kid. Thanks.
@@DeerAssassin Fair enough. My apologies for the insult. But you also say that you “watched all my videos..” …these aren’t my videos. I’m simply a viewer commenting on these videos along with everyone else here. And I suppose that you’re right about my assumptions, at least to some extent. It was just that in my thinking, by my assumptions, that is, (16/3 - 6 + 4) going to (16/3 - 18/3 + 12/3) was obvious to anyone with even a basic knowledge of working with fractions, just a simple matter of finding a common denominator for the three terms inside the parenthesis, is all, to my mind at least. …it’s no longer calculus at that point. And again, I assumed that anyone familiar with basic geometry knows that area problems are given in square units… just as volume problems are given in cubic units. But again, my apologies for my lengthiness and my uppitiness!
Why do you feel the need to go on and on about why we need calculus, your life's story, how important it is to take notes, why it is that people fail, how many mistakes you've made. Then you go into the whole linear path that leads to calculus. WHY? JUST GET ON WITH IT!!!!!! My god, no real teaching occurred until about 6 minutes into the video! I had calculus, and more math beyond that about 40 years ago so was curious what your video would say. But I just couldn't stomach it.
Cool... so this is where I can find all the geeks and nerds I used to beat up! And before you say "but now I rich" sorry I am too. And had many girlfriends back in the day. Isn't life outside the basement great?!
🤣 I was a jock in HS playing football, baseball, basketball in advanced calculus, chem n physics. I was friends with everyone cause didn’t matter and many girlfriends. Really wish was around you back then, it would’ve been fun!! Edit: advance to advanced
@@ICLight412 I was playing football ⚽️ at the time. It was the 80s in England and those were rough times. Frankly we used to get in a lot of fights and there was a lot of drinking. After uni, I went into comedy writing and made a good living. A mate of mine from football was smart like you and started a medical billing company with a £250,000 investment that I loaned him. He came up with some computer software algorithm thing and sold the company for £3.2 billion. I guess that nine figure check I received is due to maths boffins like you. And get this.. he wasn't even Asian! 🤑
@@rileybuchan9813 not "your". It's "you're". I may have barely made my maths minimums, but I did graduate from Thames Valley University with a first in English.
I love your intro to Calculus videos! You make it look really easy (and I know C gets much more complex), almost like magic!
Good job; I learned Calculus in college and loved it, but never used it in my career as a Chemist, so the knowledge departed. I think it may be fun to relearn.
Thanks for explanation. Also "took" calculus and prerequisites. Relearning opens a lot of doors to scientific discovery.
@@steveolmore6091like for puzzle solving sake?
I struggle to see any joy in this.
I am still trying though
finally, somebody who used something other than 0-1 range THANK YOU
I'm in 8th grade and I actually understand this,
I watched from 6:38 to 14:09 that's all you really need to watch to understand the video
Calculus starts at 6:38
Hey, I’m a 13 year old and I completely understood this video! Good job!
Where are you from
@@Rannaghor-sonali uk, why?
@@e3lord856 just like that
@@e3lord856 cuz I am also 13
Doing the same
Wow!!! It does not get any easier than this, it truly doesn't.👏
A = [ 2/3 . x³ - 3/2 . x² + 2x] from 2 to 0 = 2/3 . 8 - 3/2 . 4 + 2 . 2 - (0 - 0 + 0) = 16/3 - 12/2 + 4 =
32/6 - 36/6 + 24/6 = 20/6 = 10/3 units²
One has a function f(x) = 2x^2 - 3x + 2 where the limits on the x - axis is x=0 to x=2. To find the area of this function, one takes the integral of f(x) and one has a = 2/3*x^3 - 3/2x^2 + 2x with the limits of x=0 to x=2. Once one calculates the limits, one has the result a = 10/3
If anyone knows of a good video or page that teaches slow people like me "how to determine a function of any curve", I'd be much appreciative. And thank you TabletClass for your useful videos!
IOWs, this helpful video from TabletClass has taught me how to find the area size if I already know the curve's function. But what do I do if the function still needs to be determined?
To determine the function that matches a given curve, you'd usually match the curve with the family of functions that have a similar shape. You usually also look for a physical basis for why this curve is a given shape, if the curve represents real-world data. Such as matching a temperature vs time graph to an exponential decay. Then, you'd establish known points on the curve, and set up a system of equations to solve for the parameters that define the curve.
For instance, consider a polynomial function that passes through the points (0, 7), (1, 0), (5, 32), and (7, 0). You'd set up a system of equations based on the general form of a cubic, y=a*x^3 + b*x^2 + c*x + d, and apply each of these known points to x and y. You'll then solve this system for a, b, c, and d.
So your equations would be:
7 = d
0=a + b + c + d
32 = 125*a + 25*b + 5*c + d
0 = 343*a + 49*b + 7*c + d
Trivially, we know d = 6, because the y-intercept was given as one of the points and all terms containing x disappear. So we can simplify the system accordingly. We can also divide the middle equation by 5, and the final equation by 7.
-7 =a + b + c
25 = 125*a + 25*b + 5*c, simplifies to: 5 = 25*a + 5*b + c
-7 = 343*a + 49*b + 7*c, simplifies to: -1 = 49*a + 7*b + c
From this point forward, we'd set up an augmented matrix, and use Gauss-Jordan elimination (or other linear algebra methods of our choice) to solve for a, b, & c. The solutions are:
a = -1
b = 9
c = -15
d = 7
And thus the equation is:
y = -x^3 + 9*x^2 - 15*x + 7
@carultch great explanation, but is still doesn't make sense to me. Do you know of any videos that could basically explain what you just wrote? Thanks
The math is not hard but not all teachers do a good job of explaining all the different symbols. For example Capitol F is the integral so Capitol F prime is the derivative of the integral which equals little f our original function. Small dx means infinitesimal change in x. Teachers often jus say thing without explaining well what they are. After a few weeks of calculus all students should know what capitol F prime and small dx means without explanation but in the beginning math teachers need to explain this and explain it over and over and over to drive the point home for a couple weeks
The capital and lowercase letters do not necessarily mean derivative and antiderivative. In fact, in Vector Calculus, it was common to use F and f in the other way around. The prime usually represents differentiation, but it can also just represent a variant on the original letter. The point is, don't get too attached to notation. It can be different in a different application. The d notation an extension of Δ notation, where Δx means change in x. The dx you can think of as "dinky x", since it is a little tiny change in x.
If a problem is expecting you to know that f'(x) = F(x), and f(x) = integral F(x) dx, then the problem needs to define its terms. Like "find a function f(x), such that the derivative of f(x) equals the function given below as F(x)".
Very good, thanks a lot. 👍👍👍🙏🙏🙏😀😀😀
Awesome !!!!! Precalculus was soooo long ago. Do I have to start over,, is there an Algebra for Calculus ?
Following the steps is no problem. However, I have no idea why the steps work.
Or how they came to be really
For sure do not be intimidated. The fun gets better and more interesting!
Wonderful explanation. You make things easier to understand. By any do you cover calculus for IB Diploma
How to reach "discription" ?
Cool stuff, in engineering we had to work out volumes on weird forms using triple integration.
I enjoyed it. (49 year-old coding student.)
Hello teacher I liked your videos on math different courses. Can you in the future have note taking techniques on calculus.
Excellent tutorial! Just a question; what would the unit of measurement be for the calculated "area under the curve"; I know it all depends on the units for X and Y.
OK! Let's say X is time in years and Y is the number of newly diagnosed cases of a certain type of cancer; what is the unit for the calculated figure (in this case 10/3 what?).
Hope the question is clear.
Area is measured in square units. It could be any unit; In other words, units could be m², cm², inch², ft², etc. In this case he didn't specify what the units are, but the answer is (10/3)units², or 3.333units². In your case if x is time, and y is diagnosed cases, then if the unit isn't officially defined, you can make up your own unit. You could have 3.333omid² 😀.
where does 3x come from???
seems like we take the x portion of the DX and multiply it against the 3 terms in the equation, x *x2 and x*x and 2*x and of course apply the rule to the denominators..
That reasoning only works for polynomial terms. It doesn't work in general.
Locally, integration is a form of multiplication, because you are adding up each value of the function, each multiplied by a tiny change in the function (dx). So each term we are adding up is f(x) * dx, and we have an infinite number of terms to add up along the way. This is why for constants, it is as simple as just multiplying by x.
For terms like x, it isn't as simple as just multiplying by x, because we also have to account for the fact that x and x^2 are different sizes along the way. For integrating a term like k*x where k is a constant, it is like the area of a triangle. Its height is k*x, and its base is x. It's area is thus 1/2*k*x^2.
For higher ordered polynomial terms, we need to account for the fact that these terms are a lot slower to grow, and spend a lot more "time" closer to zero. It grows slowly near zero, and grows faster and faster, the farther away from zero you get. The higher the exponent, the more extreme this behavior becomes. This is why rather than dividing by 2 like we did for the linear term, we divide by the new exponent (3 for x^2, which integrates to 1/3*x^3).
When is c, the integration constant important?
When it is an indefinite integral, you always add the +C.
The reason is that there are an infinite number of functions that only differ by an added constant from each other, all of which have the same derivative. So to collectively represent this infinite family of functions, we add an arbitrary constant, of +C, when asked for the indefinite integral, or general solution to the antiderivative of a given function.
If it were a definite integral, you can keep it simple, and let C = 0. No matter what you make C equal to, as long as it is the same for evaluating both limits, it will cancel itself out anyway. So you don't need to think about +C for indefinite integrals.
To give some context on what it means in an application, suppose you were given a speed vs time function for a car, and wanted to know what mile marker it ends its journey after a given amount of time. Suppose its speed vs time is constant at 40 mph, and it travels for 2 hours, and it takes a negligible amount of time for it to speed up and slow down. Do you have enough information to know what mile marker it ended upon? No, you don't. You know it travelled 80 miles, but 80 miles from where? Since there are an infinite number of possible starting locations, there are an infinite number of possible solutions. So in this case C represents the initial position.
C in general, doesn't necessarily represent the initial position, but it is directly determined by it.
@@carultch thank you! Much appreciated
Really great
I am in grade 9 & I understood , thanks man
calculus Area starts at 7:36
Thanks!
It would really help if it was indicated how this pertains to real world situations
Everywhere we look there's calculus
@@marknascimento5037that doesn't really help. Let's say you had a word shaped room with lots of curves and you wanted to know the floor area exactly. How would you do it using calculus?
@@MrDirkles You'd have to have a mathematical definition of all the letters in that word.
For instance, let's suppose a room shaped like the letter D is defined by a vertical line through the origin, and the equation x = 1 - (y - 1)^4. Plot this, and confirm that it looks like the letter D. Observe that the vertical line intersects the curve at y=0 and y=2
To calculate the area, we need to integrate the function relative to y, to find the area between the vertical line, and the quartic polynomial defining the rest of the letter D. You could expand it, but you don't really need to.
So we are now integrating:
integral (1 - (y - 1)^4) dy
Let u = y-1. Observe that du/dy = 1, since this is a simple linear term with a slope of 1. Thus, we can rewrite it as:
integral (1 - u^4) du
Increase each exponent by 1, and divide by the new exponent:
integral of 1 du = u
integral of -u^4 du = -1/5*u^5
Thus we have:
u - 1/5*u^5
Translate back to the y-world:
y - 1 - 1/5*(y-1)^4
Evaluate this at y=2, and subtract the evaluation of this at y=0:
At y=2, we have .8
At y=0, we have -1.8
Thus, the area inside this D-shaped room, is 1.6 square units.
That's really cool thanks @@carultch
Why add 1 to the power? Why not add 2, or 3? How did you get the equation for y to begin with? I appreciate the lesson, but I would advise that if you do this agin, explain each and every equation and why/how it works.
At the end of the problem why did you give everything a 3 denominator I'm lost after that
its basic fraction addition. the rule is to add/substract fractions together, you have to make them have the same denominator
Great
I am in class 7 and I am copying all the things and then showing to my classmates and flexing there that I calculate this myself.😂😂😂
I am 74 years old and found your site a few weeks ago....I am really interested in learning just for fun....you make it easy...my question is this: I see where the 2, 0 come from....but can you have a parabola be on the positive side and the negative side as well?.....example: 3, -2.......how would you find the are under that using the same equation? would you just follow the same process but now have to do it for 3 and -1? then, would you have 2 answers...or just use the positive answer/
...thanks........OKU
@okupchurchill7850
You’d still use the same process, but you’d simply go from -2 to 3, instead. And no, you’d get one answer, or one solution. Here you’re only substituting values for x, namely, -2 and 3, in order to reveal the answer. And area, length, time, etc., can never be negative, in the strictest sense and meaning of the existence of those things… there’s no such thing as a negative amount of land, or a negative amount of length, or a negative amount of time, etc., say, for example! But, nonetheless, acres, feet, minutes, etc., may and can be added to or subtracted from one another, respectively. So that caution with regards to your algebra must be taken when calculating the integral. For an example and for a trial run, try taking the integral of y = x^2, or, f(x) = x^2, if you prefer that notation, from -2 to 3, or perhaps, and perhaps somewhat clearer, from -3 to 3. When you get to that part where the expression becomes (x^3/3) - (x^3/3) from -2 to 3 (or from -3 to 3, if you choose) you must remember and bear in mind that where -3 is substituted for x you’re then actually multiplying (x^3/3) by (-1), by the algebraic rule for subtraction, like so, (-1)[(-3)^3/3)] = (-1)(-27/3) = 27/3! Which is all to say that we must first change the sign of operation (in this case subtraction) to its inverse or opposite operation (addition), then change the sign of the term(s) that follow after the sign of operation. For example, 4 - (-4) >> 4 + (+4) >> 4 + 4 = 8. So, and with that in mind, taking the integral of f(x) = x^2 from -3 to 3, for example, we get x^3/3 - x^3/3 from -3 to 3, which itself becomes, and again, by the rule, [(x^3/3) + ((-1)(x^3/3))] from -3 to 3 = 54/3 = 18sq. units. Most of the mistakes in calculus are algebraic! Therefore, strength in algebra, and trigonometry, makes life in calculus easier and much much less of a headache! But although the sheer intellectual genius and beauty both essential to and fundamental and foundational to calculus make it relatively easy, that doesn’t mean nor should it be inferred, necessarily, that it’s a walk in the park…
Immensely helpful
Love this
That's good!!!
The hardest calculus is Two calculus
Ten thirds makes me think 🤔 🧐 about the area underneath the parabola “y = 2x^2 - 3x + 2” and enclosed by the x-axis, the y-axis, and the line y = 2; the four sides of the yellow region.
@ Bradley Martinez
The instructor’s final answer makes it pretty obvious that “ten thirds” is the area beneath the curve (that is, the area In square units, whatever we choose those units to be, whether we choose for our units feet, inches, meters, or whatever we choose for our units) so much so that it becomes unnecessary and redundant to say “ten thirds makes me think about the area underneath the parabola “y = 2x^2 - 3x + 2”…” And the area beneath the curve is bounded by the three sides, or the three lines, x = 0 (which itself is the y-axis!) from y = 0 all the way up to y = 2 along, again, the y-axis (not the entire y-axis, mind you), and by the line y = 0 (which is itself the x-axis!) all the way from x = 0 to x = 2 along, again, the x-axis (again, not the entire x-axis, mind you), and by the line x = 2 all the way from y = 0 to y = 4, and finally by the curve y = 2x^2 - 3x + 2 itself bounded by those three lines are indeed the four sides of the yellow region! And what, if anything, does the line y = 2 have to do with this particular graph and finding it’s area?
Great!
The calculus is easy. most people have problems with algebra, not calculus
What does the area under a curve tell you? Why find area under a curve?
you don't explain the creation of the 2 rules
Nope, there’s no explanation of how the original formula describes the curve. There’s no explanation of how you got from 6 & 4 to 18/3 & 12/3 (although I think I can guess). And the area is 10/3 what? I can’t cut a board to 10/3 without knowing the unit of measure and even the wouldn’t it be 3.333? At 58 years old I’ve discovered the math I didn’t think I would use in high school is actually applicable in Machinist trades. But I’m a very literal thinker and math teachers do not explain problems literally which is why we fail.
@@ndailorw5079 you’re about as helpful as an ingrown hair on a ballsack. 😆👉🏻
@@DeerAssassin
And your reply to me tells us all we need to know about you, and why you posted your first comment regarding this video in the first place… 58 years old, eh…?
@@ndailorw5079 what it tells you is your explanation was uppity and lengthy, and somewhat insulting. Like 40 years ago in high school math class, there’s always an assumption that the student should know more. I can explain algebra and most geometry to my kid because I know what the placeholders are for and what they represent. I watched about a dozen of your videos and learned several things but there is some basic information missing. My response to your reply shows I’m old enough to get to the point, it wasn’t helpful. Helpful would be pointing me to the video explaining what basic information I’m missing to help the kid. Thanks.
@@DeerAssassin
Fair enough. My apologies for the insult. But you also say that you “watched all my videos..” …these aren’t my videos. I’m simply a viewer commenting on these videos along with everyone else here. And I suppose that you’re right about my assumptions, at least to some extent. It was just that in my thinking, by my assumptions, that is, (16/3 - 6 + 4) going to (16/3 - 18/3 + 12/3) was obvious to anyone with even a basic knowledge of working with fractions, just a simple matter of finding a common denominator for the three terms inside the parenthesis, is all, to my mind at least. …it’s no longer calculus at that point. And again, I assumed that anyone familiar with basic geometry knows that area problems are given in square units… just as volume problems are given in cubic units. But again, my apologies for my lengthiness and my uppitiness!
@@DeerAssassin
I see you deleted my post…
✨👌🏻
Get to the subject matter already!
Why does this man always say nothing in the first 3-4 minutes of each video?
Why do you feel the need to go on and on about why we need calculus, your life's story, how important it is to take notes, why it is that people fail, how many mistakes you've made. Then you go into the whole linear path that leads to calculus. WHY? JUST GET ON WITH IT!!!!!! My god, no real teaching occurred until about 6 minutes into the video! I had calculus, and more math beyond that about 40 years ago so was curious what your video would say. But I just couldn't stomach it.
Too much talk at the beginning without explaining actual math. Boring!
The intro is waaaaay too long. 5 mins in and still not started the problem....
This guys to wordy
Although a very nice tutorial, WAY too much talking
His explanations are too wordy with sidetrack information and sales pitches for his other classes resulting in mental diversion from the main topic.
You should teach self promotion. You waffle on for more than 6min about yourself.
Dear God, I clicked off because I couldn't stand your lengthy, shameless rambling ,and repetitive pitch for your courses.
I don't really know if it gets better I meaannn, it's calculus, if you can point me in a better direction I'll take it
Your loss. I skipped it and learned some calculus.
Cool... so this is where I can find all the geeks and nerds I used to beat up! And before you say "but now I rich" sorry I am too. And had many girlfriends back in the day. Isn't life outside the basement great?!
🤣 I was a jock in HS playing football, baseball, basketball in advanced calculus, chem n physics. I was friends with everyone cause didn’t matter and many girlfriends. Really wish was around you back then, it would’ve been fun!!
Edit: advance to advanced
@@ICLight412 I was playing football ⚽️ at the time. It was the 80s in England and those were rough times. Frankly we used to get in a lot of fights and there was a lot of drinking. After uni, I went into comedy writing and made a good living. A mate of mine from football was smart like you and started a medical billing company with a £250,000 investment that I loaned him. He came up with some computer software algorithm thing and sold the company for £3.2 billion. I guess that nine figure check I received is due to maths boffins like you. And get this.. he wasn't even Asian! 🤑
@@felixleiter5092 your strange
@@rileybuchan9813 not "your". It's "you're". I may have barely made my maths minimums, but I did graduate from Thames Valley University with a first in English.
@@felixleiter5092 hahaha