How to use Calculus to solve a basic math problem
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- Опубликовано: 8 сен 2024
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How to use calculus to find the area of a triangle. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at tcmathacademy....
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It is a 2 second job. Consider it a 3x4 rectangle the area is 12, divide by 2 and the triangle area is 6.
But like... What if I wanna overcomplicate little things!?
@@analog8163
In this case your life will be a mess.
I must say you are so talented in how you speak to really make things simple…!…Thx…!!!
Finally an explaination that is clear! I have spent years reading and studying in an effeort to find an explaination that makes the subject of integration understandable. Thank you. I would like to call attention to what I found to be confusing. There is a prectice of drawing a squiggly line and then refer to the "x" axis as representing the boundary to the irregulare line. In fact the irregular line is a "function" that is integrated. It is NOT just a squiggly line!
No video where showing with an example but not it is clear how to use the calculas formula
It's unbelievable how good you are at simplifying and teaching such a complex topic
Your simple to learn and understanding the basics had exposed the problems of over 25 yes on how to understand this process of calculus. If I had had you earlier in my schooling,I would have been great in mathematics. Tanks
I actually had calculus 1 & 2 way back when - I know this is a simple integral but when you did it I found out if you don't use it you lose it. I can't wait to see more calculus from you..... I may have to get myself a graphing calculator.....TI - 84 hint hint to follow along. The other graphing calculators I are half the price but colleges use this one the most so students can follow along.... even if they switch schools.
It's an easy problem to solve. Take the integral of the function f(x) = 3/4x with the integral from 0 to 4. the function is 3/8x^2. The solution is substitute the limits o and 4 and the solution is 3*(4)^2/8 - 3*(0)^2/8 = 6
Thanks for enthusiasm towards math. And I love your 80's references. Math and Physics were my favorite courses in high school. Went on to college for Computer Science, but also got a math minor. I love watching the videos!
starts at 5:22
Lol 😂
Explain the integration mechanism not just the rules
The next extension to this problem is: y=3/4x + 1. It can be solved using geometry formulas and calculus too. It is a little obvious. It is useful to show the equivalance of using geometry and calculus for these special cases.
I was always impressed when I took calculus how the basic algebra equations were proved and/or derived.
I would make that right triangle into a rectangle to compute the area and then divide by two
…same thing, that’s just 1/2ab, or, ab/2, the area of a triangle, which is just the same thing as the product ab multiplied by 1/2, or, ab/2…?
What you didn't explain in the calculus portion is the relationship to the thin rectangles. The height of the rectangles is the 3/4 X and the width is the an extremely small portion called dx. You kind of jumped past this.
It's amazing that if I would have had you tube in HS, I would not have flunked out of Algebra 2, and I would have passed basic algebra with more then a C, anyway mathematics are so very fascinating when taught properly.
I agree, with a proper tutor/teacher, Mathematics can be the most fun subject.
@@mikhail6746 I guess if you like it, it could be fun, but the application of this stuff is more fun then just solving basic equations. And you can kind of see where and how it is never ending.
(Base x height)/2
(4*3)/2=6
Took a lot longer to type this than to figure it out. The 90⁰ angle kept it simple.
I think that you may have missed the point. Calculus is not used to do basic algebra problems, but rather to find solutions (as he indicated) for finding the area under irregular curves. Also, you can rotate the curve to determine the volume of a solid object such as a flower vase. The solutions can't be found with algebra unless you want to add up an infinite number of infinitely small rectangles. Also, BH/2 doesn't require a 90 degree angle. It works for any triangle.
Good video but why does this actually give you the area. What's the reasoning behind this. L x 1/2 h is obvious. Integration is not. Please explain?
It should be noted that your calculation of base x height is of a right triangle. In non-right tirangles we say base times perpendicular height.
Wow!! you are such an amazing teacher. I had no prior calculus exp or knowledge whatsoever and I clearly followed along the entire way. Thank you sir for your free content. Masterclass coming soon? 🤣.
Choosen example is excellent.
Currently taking highschool pre calc! Cant wait to take AP calc AB next year
I didn’t really appreciate mathematics until I was an adult and my interest has only increased. Then the internet, and now I can sit, listen and learn to my hearts content….. I have been retired for years now and my interest has stayed the same.
Integral 4 to 0 of 3/4 x dx = [ 3/8 x² ] 4 to 0 = 3/8 . (4)² - 3/8 . (0)² = 3 . 16 / 8 - 0 = 6 units²
Check: Area = ½ . 3 . 4 = 6 units²
…0 to 4, not, 4 to 0. …x2 - x1. Your correct answer proves as much. 4 to 0 would give -6 square units… which is impossible for the area under this curve… it’s the area from a to b, not, the area from b to a; (a) being 0, and (b) being 4. The integration is b minus a, which you correctly did…
Thank you
6:09 Just to keep the answer Koscher, the area of the triangle is 6 'square units'.
3*4/2 = 6
Base*height/2
Wake me up when you get started on the solution. This is not really a critical comment, just saying that less rambling and repetitive talking and more directness in approaching the solution using calculus techniques. Math is a beautiful topic, and yes I am a mathematician and have used it in the real world for years. Area finding using calculus is a way for finding areas under "curves" that are not linear. Also, calculus can be used to derive formulas, for example, the volume of a sphere and even the equation for finding the area of a triangle.
Thanks!
Same way if you want to know the area inside a parabola (which have no formula by itself), you just have to find the equation of the parabola and then integrate it. Just like that!! So if you want to find the area inside of that TMJ, which you have a couple of parabolas, you just want to add them together and integrate it. Just like thatt!! I never become so happy with a math like this😄
thanks!
Thank you and thank you for your video
It does makes sense for highschool students but not in elementary students.
Calculus is a really fun topic I believe!!
very nice!
Area equals 3 times 4 divided by 2 Ans 6
…correct answer, but that’s not calculus… which is the point here. Did you listen to this instructor at the very beginning? What do you do with the area under either side of a parabola, say, or some other geometrical figure not in the shape of a square or triangle and doesn’t have a ready-made formula for solving its area? He used this triangular figure to show that calculus can be used to solve not only areas of this shape that have ready-made formulas for solving their areas, but can also be used to solve much more complex area problems of shape that have no formulas at all for solving their areas… that’s the point here. He’s teaching calculus, not basic mathematics…
He’s not looking and asking for your method… besides, he explained your method at the outset… so he obviously understands it and knows about it.
Love your math videos. They are a wonderful review of lessons learned decades ago. Thank you.
This would be a nice class but I can't stay awake while he talks about something other than the problem!
All his other videos I get, but these calculus ones, I really don't get!
…could you “get” changing a fraction into a decimal if you didn’t understand and know division and multiplication…? Same here… you must know and understand all the mathematical disciplines that prepare you to “get” calculus. …just can’t do fractions without understanding and knowing the division and multiplication necessary to do them. Same here. You must understand and know the mathematics, geometry, algebra, trigonometry, etc. that’s before and lead up to calculus and prerequisite and necessary to understanding and doing it, is all.
@@ndailorw5079 Yes 3/8 x 16 = 6 3/8 = 0.375 and times that by 16 is 6
@@almklit
…no offense, but, what’s your point?
I was only trying to make a point to the poster above, who started this thread. I wasn’t asking for an answer to the math problem I presented. I was simply being rhetorical, somewhat, is all.
The poster above says they didn’t “get,” didn’t understand calculus, so I simply tried to explain that they must first understand and know all the prerequisite math that prepares the student for doing calculus, and that leads to and guides them in understanding and knowing it, is all.
@@ndailorw5079 To be honest I don't know myself, I was just reflecting.
Omg TY!!! I was stuck relearning on a physics class problem. The professor online was going over it, I knew it involved calculus and I was stuck. I’ve been watching your videos, knew I’ll find the one video to get what I needed. Today seen this and WALLAH!!! 😁 so happy it came back n figured it out with your help. I was frustrated couldn’t remember but it’s been 20 plus years.
Force Distribution in a straight line:
Sum 0-L can’t put that stretched S lol S sum symbol for this.
Xr = S(B/L x^2 dx)/S(B/L x dx)
B/L a constant (base length can bring out)
Xr = (B/L S x^2dx)/(B/L S x dx)
Next integrate where I got lost
(B/L x^3/3 S)/(B/L x^2/2 S)=
2/3L.
I was thinking, oh boy, what’s going on. I looked online and couldn’t find it. I did but wasn’t explained clear enough as you did. After seeing this, I get it. Sorry for being to long or if don’t understand but again, thanks again for turning the light bulb on again in my brain 😁😁😁😁😁
Haven’t been in HS since 68 and got this in about 10 seconds in my head. This guy wants to impress people on how complicated he can make a simple problem!
@@rondouglas5147 happy simple for you, I remember the area of a triangle but calculus got stuck on dividing x^2dx/xdx. Just froze n this helped.
THANKS...
So, if the range is -4 to 4, the area is 12, but that doesn't work if you plug in 4 and -4. Your answer comes out to 0. If you use 0 and -4 for the lower half, you come up with -6. Knowing it is area, do you just work out the absolute value of the two areas and add them together?
@ Larry Patterson
For this particular problem, the limits of integration are 0 and 4. Therefore, 0 to 4 is the domain for this problem. Only that portion of the graph is considered here so that the domain not concerned about the entire length of the graph or any other portion of it. Here, the problem only considers the domain and range that forms the triangle, that is, the limits of integration which include and are from 0 to 4. This is the integral of y = 3/4x, the range, from 0 to 4, the domain. It’s not the integral of y = 3/4x with a domain from -4 to 4.
I think he missed the step of why it was y = 3/4x in the first place.. obviously that wasn't the main point of this. But that is the gradient of that straight line of the triangle or hypotenuse.
5:05 go back 10 seconds. He started then
What about the constant of integration? Did you forget to mention it?
@ Walter G
This is a “definite” integral, not an “indefinite” integral.
The definite integral is a “number”! And It’s a number strictly because it has “limits of integration.” The limits of integration here are 0 and 4. The “limits of integration” “define” the integral, they make the integral of the function being integrated “definite,” and the only and final solution or answer to the integrated function because they give the integral itself an actual, definite, and final numerical value when they’re substituted for its variable x.
The indefinite integral, on the other hand, and by and for comparison, is a “function,” and not a number. The indefinite integral has no “limits of integration” that can be finally plugged into it to give it a final numerical value and so is and will remain merely a function. That is, again, the indefinite integral, [(x^(n + 1)/(n + 1)] + C, is a function, not a number, simply put. The area problem for this triangle could never be solved definitively and numerically and finally, had it been presented instead as an indefinite integral. Because it would then have no limits of integration that could be plugged into its x variable to give a definite and final and only numerical answer for the function being integrated. Which is unlike this particular case of a definite integral where the limits of integration are 0 and 4, for x = 0 to x = 4 along the x-axis. So the answer and solution to the indefinite integral version of this same definite integral is 3x^2/8 + C, a “function,” and not a number! And since the solution to an indefinite integral is a function, and not a number, the “constant of integration” C must also and always be included as a term of the indefinite integral expression because it’s the most general integral because allows for all functions of the form F(x) + C to be solutions to the integrated function since they all have the same derivative. That is, the derivative of y = 2x, which is 2, is the same as the derivative of y = 2x - 3, of y = 2x + 6, of y = 2x + 13, of y = 2x - 11…..etc.! So the indefinite integral of y’ = 2, the derivative of y = 2x, which is the indefinite integral of 2dx, must be written as 2x + C as opposed to simply 2x because all the above functions are solutions to the integrated function. If the integral was written simply as 2x, it would exclude all other solutions to the function being integrated that also have 2 as their derivative.
If I would’ve had this knowledge 55 years ago I would’ve sweat a lot less blood rote memory got me pass several concepts that very obvious now.
Nice one
Love your videos! You are a great teacher.
If you stop the too much explanation, you will be a good tutor
6
I meant 25 yrs
What's the name of the software that you use to illustrate the problem?
The dx needs to be defined.
When you say "M'kay" a lot, you sound like southpark...
The sides are 3x4 = 12 for a rectangle, but it's a triangle which is 1/2 of a rectangle, so 1/2 of 12 is 6 - If you have to figure one that's not a right angle, break it into 2 triangles with right angles? ....probably why I barely scraped by in every math class I ever took...
@ A P2
He knows that. But he’s teaching calculus here, and he’s showing us how to solve the problem using calculus because there are more complicated functions than this one that has to be done using calculus. He knows that the area of a triangle is A = 1/2bh, but he’s not concerned with that here and is only concerned about finding areas under curves using calculus because there are functions that don’t have formulas like squares, rectangles, and triangles do, to name just a few, and so, in effect, sort of speak, integration produces a new function that allows us to find the area under the curves of those functions. And he used this function to prove just that point. Some functions don’t have established formulas like the triangle here has, so you have integrate the given function and produce a new function that allows you to find the area using it. He teaching how to use integral calculus in order to teach us how to integrate a function such as this one, as well as how to integrate much more complicated functions than this one… he’s teaching how to integrate! He’s simply using the triangle as a proof of that fact! It’s a simple function that’s being used to show the very same basic principles and techniques that will be used later in the study of calculus on more complicated functions. He more than knows how to and that he can find the area of the triangle using the equation A = 1/2bh if he wanted to, but he’s not teaching algebra and geometry here, he’s not teaching that, he’s teaching calculus, so he’s using calculus to solve the problem because calculus will be tool needed later on in the study of calculus to solve much more complicated problems than the problem of simply finding the area of a triangle!
Masha Allah 😇😇🥰🥰💯💯
no prattles on for far too long, blocked in future
Sir.good day.
Your kind guidance required.
Let we have a shape like some of that part is circular and some part is triangular and rest of that is straight line.
I need to find the area.of that shape using calculus....
How would I found the function or equation for that shape.
Kindly reply.
Thanks
10 ?
Thumbs up
Can't you just use the formula for a triangle? If you do you come up with the same answer.
Why complicate something simple? 1/2 x 4 x 3 = 6.
@Michael43713
…you’re right. The formula for this right triangle is the much simpler and faster route to take for this particular problem. But… you missed the point, here. He’s teaching calculus, here. He uses this simple problem that has a simple formula for solving this simple, particular problem to show and prove and assure his audience that calculus can be used to solve problems that have formulas for their solutions as well as used for solving much more complicated and even complex problems that don’t have formulas at all for their solutions. That’s the point here; the point here is learning calculus, and how to apply it…? In solving for the area beneath the curve of a parabola, for instance, there is no formula to come to our aid… except calculus itself ..so then… what do we do..? Well… that’s when our good ole’ friend calculus comes to our aid, as well.
So in using calculus here to solve this simple problem by treating it as though it doesn’t already have a specific formula for solving it, the instructor here shows and proves that calculus can be used not only to solve problems that already have specific formulas for solving them, but that calculus can also be used to solve problems that don’t have specific formulas for solving them! He well knows that 1/2ab is the simplest method here, but that’s not his point, here…?
He is not an instructor. He has a serious problem of diarrhrea of the mouth!
This video is very confusing and most of the comments are confused too. You start with 5th grade math to calculate the area of a triangle that almost everyone know of A=1/2*B*H and you took half your video to do the obvious. Most of us are already bored out of our minds. Then you jump to 9th grade algebra 1 to show the formula of the hypotenuse of the triangle as Y=3/4*X and yes that is obvious too to a lot of people. Then you back tracked and poorly explained about Y=MX+B and what that meant and a lot of people didn't understand and the rest were again bored to death. We were not here for a simplistic refresher in triangles or algebra 1 and the definition of what a line is all about. We a here for how to use Calculus. You talked about doing an infinite sum of little rectangles under the line. This is where you went off the rails. You never did tell us how to come up with the formula of the rectangle. It is A=H*B for the rectangle. Then you really made me laugh when you said "Oh we have is notation of dx that we use". Wrong!!!! Show the notation of how the rectangle is calculated by A=Y*delta(x), then y=3/4x and the formula is 3/4x*delta(x). Then they would know that it is the sum of all the little rectangles of 3/4x*delta(x) and in calculus the notation is the intregal(3/4x dx). You didn't teach anything after that. All you did was us some magic to do the simple integration and didn't explain any of it. You left your students confused and bored to death. This video is a total failure. Half of it was explaining 5th grade math and the important parts you skipped over and then used magic to come up with an answer. You took something simple and turned it into a confused mess. Real math is not mumbo-jumbo and hand waving like you did in this video. You wasted the time of your students. My calculus professors in college makes you look home-schooled.
Thank you but talks too much before going into to solve the problem. I'm not even what when he stared
Simple things but talk for a excessive long period.
Easy for you to say that but not easy if you don't have the concept. Be patient.
@@kwaananse6424
Elaboration can be simple and quick if the contents have been fully absorbed by the speaker! For example, E=MC square is simple enough to describe the complex issue. It is a matter of ability instead of a matter of patience.
I like to think of every factor of an equation having a descending power of x.
i.e. x^n +/-….x^2 +/- x^1 +/-x^0. Also, dx is there, but is ignored and is not part of the solution, why? Could it be that, in this case the magnitude of dx approaches 1/infinity?
Sounds like u have PhD or MD
Have some respect for the time and effort the guy is spending on you
You forgot the constant of integration.
You don't need a constant of integration on a definite integral
@harpleblues
When we take the indefinite integral of a function we’re essentially and effectually taking the antiderivative of that function, and vice versa. The terms are interchangeable… they both mean and do the same thing.
But the phrase ‘take the “antiderivative” of…’ better describes and gives a clearer picture of the process of integration than does the phrase ‘take the “indefinite integral” of….’
For instance, for the function f(x) = x^2, f’(x) = x^2 = 2x… the derivative is 2x. So, the antiderivative of the derivative 2x gives us x^3/3 + C. We can see that the antiderivative of the derivative reversed the entire process of differentiation carried out by the derivative, since integration (the indefinite integral, in this particular case, the other is the definite integral) and differentiation (the derivative) are inverse processes; they mutually undo what the other does.
For example, in taking the derivative of f(x) = x^2 we get f’(x) = 2x. We then can see that taking the derivative of a function multiplies the function by the value of the function’s exponent while at the same time decreases the value of the function’s exponent by a value of 1 less than the value of the function’s exponent to give, in this particular case, 2x. Now, taking the antiderivative of the derivative 2x we get F’(x) = x^3/3 + C by which we can also see that the antiderivative reverses the entire process carried out by the derivative in that it “ANTIDERIVES” the derivative, sort of speak; the antiderivative does the exact opposite to the derivative function of what the derivative did to the original function f(x) = x^2 by dividing, as opposed to multiplying the derivative function by a value 1 greater than the value of the derivative function’s exponent while at the same time increasing, as opposed to decreasing the derivative function’s exponent by a value of 1 greater than the derivative function’s exponent and thus the antiderivative does the inverse and exact opposite to the derivative function that the derivative did to the original function and in doing so brings us right back to the original function f(x) = x^2! So we can see that the antiderivative describes to us and shows us that it’s anti derivatives… lol! And as such, the phrase ‘take the antiderivative of…’ describes the process of integration much better than the phrase ‘take the indefinite integral of…,’ or ‘take the definite integral of…’
And because the two processes of taking the derivative and taking the antiderivative are mutually inverse processes the derivative is also anti antiderivatives… lol!
✨👌🏻
I am there's for class and class miss you ,as well and I am doing more math as well I more math class, I am going back to school a agree
Lot of work to get A = 6 ??????
6 what? 6 square what?
It doesn't specify the units, which can be inches, feet, yards or light years.
Crikey, you really love to hear yourself speak.
Tangent times square of highest point? Always?
P
Where were you 55 years ago? LOL.
Skip the first 5 min of all his videos to get to the actual learning...good lesson though
OMG, can you make this any more complicated than you already did? I cot the answer in about ten seconds in my head. You have a right triangle with sides of 3 and 4. Make it a rectangle with 3 and 4 as the sides, that equals an area of 12. Devide by 2 for the area of triangle!!!!!! You are way too complicated in all of your explanations.
Sure. But the point is that one can use calculus to find the area. It illustrates the equivalance in this special circumstance.
Ron, you didn’t get the point of the video, which is not finding out the area of the triangle, but explaining in the most simple way how does calculus work. And John nailed it.
It becomes very boring when it takes you too long to make an introduction.
Too much warbling....just get on with it.
Drag it out boring like math classes
unbelievably boring - just get on with it!
Waffle waffle waffle.
Yet more yada, yada, yada... blah, blah, blah.. It's all ultra verbose and prolix - just totally unnecessary padding - and I suspect for no more reason than to extend the time taken to get the real message over to the listener/observer. However, the m o does allow more time for pop-up adverts. Had I been sitting in a class and had all of this 'prattling guff' directed at me, I'd have unceremoniously walked out!
Or you could just use basic math 🙄
You talk a lot
Quick solution, this is a right triangle. You've given two sides, it's basically half of a rectangle. 4x3/2=6, as Lee Thompson states below. Why Why Why make things so bloody complicated. It does not teach anything in doing so. You are playing math games, not teaching.
bruh he's j showing a way to do it w calculus bro literally said to do it the easy way in the vid too lowk the calc version w integrals light asf too.
He is illustrating that calculus delivers the same answer in this special case, as it should.
It is a waste of your valuable time sir
talk too much about him. And then repeating simple phrases and steps over and over again...not good.