Properties of Integrals and Evaluating Definite Integrals
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- Опубликовано: 24 апр 2018
- Now that we know that integration simply requires evaluating an antiderivative, we don't have to look at rectangles anymore! But integration can also be a very difficult technique. Let's start simple by learning some properties of integrals, and getting some practice with evaluating simple definite integrals.
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I never knew someone can teach calculus like this.. you taught very smartly so that anybody can understand it and also love it.. you're a math-magician and mathematician for sure. If you people reading this comment agree, then watch his every video.
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Dave, you took my math teacher's job, literally plays your video in class instead of actually teaching.
I am a college student at CUNY taking Calculus I. My professor is terrible. I spent 4+ hours scouring the internet trying to learn how to solve an Integrals problem that was included in my homework. In the first two minutes, this video made me understand how to go about tackling the problem. Thank you so much. Making education free and accessible to all is just ... all I can say is thank you. Thank you.
I hope your college studies went well!
2 years later and this video has taught me more than my three classes last week
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Please do a video on complex integrals.
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I watched it and I took notes.
Thanks
This is enough for my CA Foundation Level.
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wow. gamechanger. Got a business intro. calc final tomorrow. wish me luck
How did it go?
@@user-bs7mv5rh1u well….. not so well😭
However i did so well on the first few exams so i was able to pass the class 😁
@@user-bs7mv5rh1u thx for askin
How can I contract you
what if the area is defined but but you have to find the value of an defined integral
Good sir
I want to ask a question that in reaction of sodium and chlorine to make salt how can chlorine react with sodium when it already cl2
it gets reduced!
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binge watching before my exam haha. You are the GOAT for making this playlist.
How did it go?
@@user-bs7mv5rh1u unfortunately i procrastinated a lot so it was bad lol..But i am catching up again these days...
@@slowfern
May I know if you're a university student?
the best
why is the integ of 1 is x/1 , it is x why should is divide 4:36 ?
AWESOME! you teach math by teaching it as algorithms instead of concepts. that is you give us math as an opaque black box ,we dont see or understand how it works inside th hood.you just teach us how to use math with no understanding or proofs. i dont "understand " calculus , I DONT CARE. i just know how to use it the same way i drive a car without seeing the engine. i just want to use math to bulid cool circuits. i just wanna pass the exam. i dont givea sh!!t about perfection or understanding . this is the way math should be taught to engineers and normal people. let the philosophers worry about elegance and perfection and derivations and proofs. more power to you sir.😄 keep up the good work.
Hey professor Dave, why in the 3rd example did you put the x^2 from the denominator in the spot of the 1 as x^-2? isn't the whole function being divided by x^2? wish me luck on my calc final tomorrow!
7 months late, but this is because according to exponent laws, 1/(x^n) = x^(-n). So 1/x^2 = x^(-2). Just basic simplification....cheers!
@@srisaipatnaik2776 lol you're good. When the pandemic hit, I flunked calc and chem so I'm no longer in a stem major XD oh well life is easier now
@@SilverDragonGaming Well, I hope you're all fine now...
the power formula is like the simplest formula possible for integration
ok this is simple I tried the integral from 0 to 1 of x and got 0.5 which I also calculated as the area of a triangle and got 1/2(1) = 0.5
Dave, When you intergrate something don't you have to put a constant C? or is it only if the range is not specified?
yes that's for indefinite integrals! coming soon.
From what I got, C is also there for definite integrals, but when you do `F(b) - F(a)` and the constant is on both sides, it just cancels out.
@@bvolpato Yes, it is there, it just cancels out, and it becomes unnecessary to write. We can make the constant of integration anything we want, when evaluating definite integrals, and to keep it simple, we just make it zero.
Some classes will give you problems such as "find a function F(x) whose derivative is f(x)." By virtue of stating the problem with the indefinite article "a", no matter what you make the constant of integration, or even if you omit it entirely, you've technically answered the question correctly.
@@bvolpato One example where it there is value in keeping track of the constants of integration, is in the theory of structural beams. When finding the deflection curve of beams, you often need to evaluate double integrals of integrands such as "k*x * (L-x)", where L is the length between supports and k is a constant from the problem specifics, and its details are a topic for another day. This is for a simple supported beam with a uniformly distributed load across two supports length L apart.
Evaluating the double integral, we get y=-1/12*k*x^4 + 1/6*k*L*x^3 + C*x + D, where C and D are arbitrary constants of integration. We want y to equal zero at x=0 and at x=L, because the supports fix the beam in place at these locations. We then solve our two equations for the unknowns, and get D=0 and C = -1/12*k*L^3, which allows us to construct the equation of the elastic curve as y=-1/12*k*x^4 + 1/6*k*L*x^3 - 1/12*k*L^3.
Done.
Is the upper limit b always greater than the lower limit a? What is the upper limit is less than the lower limit?
Is it just okay? Thanks.
The upper limit is most commonly greater than the lower limit of integration, but it doesn't need to be. If the upper limit is less than the lower limit, then the integral outcome is simply the negative of what it ordinarily would be. When the upper limit is greater than the lower limit, the integral indicates positive area above the curve, and negative area below the curve. Vice-versa when the limits are switched.
An application of reversed limits of integration, is work done by a gas (positive) and work done on a gas (negative) in an expansion or compression process, like you see in the theory of operation of a car's engine. The process is represented as a path on a pressure (vertical) vs volume (horizontal) graph, and the area under the path is the work done by the gas.
If we always put the limits of integration in increasing order, then we would always get a positive area on this plot. This is consistent with the gas expanding and doing positive work. But, when the gas is compressed, there is work done on the gas, so the work needs to be negative. And it is consistent with the way the process happens, since we integrate from a big volume to a small volume to make it negative.
@@carultch Thank you. This is so helpful!
Imagine a RUclipsr teaching you calculus quicker than your college professor :))
Hey prof! What'll be Antiderivative of x^-1
ln(x)
@@ProfessorDaveExplains but prof isn't it x^(-1 + 1) /( -1 + 1) which is just 0/0..
right so that algorithm doesn't work for this exception, i know i derived this fact in one of these calculus videos i just don't remember which one, perhaps there is one on integrals of trig functions or logarithmic functions or something?
@@ProfessorDaveExplains thanks prof
You're special
4:48 why he takes 1/3 instead of 8/3 ?!!!
That's a way to simplify the equation. Some people write it as 1/3 times 2^3 and some write it like 2^3 over 3
Both result in the same answer
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Hi Professor Dave.
I have a request that you may find interesting or not, it’s a mathematical one. I have always found it hard to understand “Inter-universal Teichmüller theory”, “The Collatz Conjecture”, and “The Birch and Swinnerton-Dyer Conjecture”. If you would make a video on one of these topics that would really help 😊
I’m studying a Master’s in mathematics in Denmark. I’m sorry for my English, I’m not mothertongue in English.
Btw, I love your content.
4:04
*gets few answers right* *my confidence "tym to rise "*
professor dave: " don't get cocky"
my confidence: "tym to go back to my hole *pyeonnnnnng*
I don’t know if the variable is x or t
The d-term will specify what the variable of integration is. As long as you don't write what I call "disrespectful integrals" like ∫cos(x) instead of ∫cos(x) dx, you'll see the intended variable of integration in the notation.
8:15 / 9:47 i got 31 over 4.....
or 15/2.... some one help
For the compeehension checking who else got different answers?
now I am thinking of how infinitely thin a straight line ought to be.......
By definition, it has no thickness. You could say that it has an infinitesimal thickness.
I feel like I'm relying on you so heavily for first year I don't know how I'm going to cope with second year 🙈🤣 I hope there's second year content here somewhere, I don't even know what it is yet though
I’ve got what you need.
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69th comment
You skip a lot of steps.
Still being held back 😂
Smaller steps are explained in previous Videos