1 minute integral vs. 9 minutes integral, trig sub, calculus 2 tutorial
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- čas přidán 13. 02. 2019
- Learn trig substitution with this calculus tutorial! We will go over the integral of sqrt(9-x^2) and the integral of sqrt(9-x^2) from -3 to 3.
Check out my 100 integrals for more integration practice for your Calculus 1 or calculus 2 class. • 100 integrals (world r...
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I looked at the second one geometrically and it turns out that the first part (the arcsin) is the area of a circular sector and the second one (basically 1/2xy) is the area of a triangle. The sum of these two areas gives the value of the integral
SkiFire13
Wow!!! This is extremely cool!!
I actually trying to do the second one geometrically. It's fun, but it took me more than 10 minutes.
So, basically, for me, it's better to do the integral using standard way
Mindblown!!! Years doing and teaching these integrals and never spotted this amazing fact!! Much easier and understandable! Thanks mate!!
So θr²/2 is there area of a circular sector? I didn't know that... That's a useful trick
WhoistheJC? yes simplest to understand as a fraction (θ/2π) of a circle (πr^2)
Area of semi-circle is a great technique for these. Calc 1 (before trig substitution) professors sometimes throw this on tests!
I like the fact that this video is 11 minutes, so it indeed is a 1 minute vs 10 minute integral 🤣
You also have to consider that he is moving and a discussing. However, if he's just going to solve this without minding the audience, it would be less than a minute and less that 9 minutes.
Both can be done geometrically; i.e., without calculus. The definite integral is just a semicircle of radius 3, and so, is:
∫₋₋₃³ √(9-x²) dx = ½π·3² = 9π/2
The indefinite integral can be evaluated (starting from x=0 and adding a constant of integration at the end) as a right triangle + circular sector:
r = 3; y = √(r²-x²); θ = sin⁻⁻¹(x/r)
∫ √(9-x²) dx = ½xy + ½·r²θ + C = ½x√(9-x²) + (9/2)sin⁻⁻¹(x/3) + C
Aha! I see now that I've been scooped by SkiFire13. Kudos to him/her!
I promise I didn't see that before I answered; I've done this sort of integral before.
Fred
Wew sir
For the indefinite integral, it is a squarish-bottomed slice of a circle bounded by a diameter and then two lines perpendicular to it. Such a figure can be decomposed into a "circular cap" (area bound by a chord) and a trapezium.
Again, outstanding presentation ... I wish I had instructors like you in my college years back in the Stone Age :)
I am from Nicaragua, your canal is very great
Thanks bro. That double angle identity coming back to haunt me a decade later!
I like how my course book has this as it's second test problem on integral substitutions when you're just trying to understand the very basics on even how to get started.
We have a formula for root(a square-x square) but I never bothered to learn how it was derived. Thanks to you, now I know and won’t forget the formula :)
Thank you for using the Doraemon theme song. Brought back memories :')
Thanks dude! I've been learning integrals from you, because i think it's interesting. I'm still 14 years old, hopefully i can master all of these stuffs.
Hey ,you are really amazing . You make me love mathematics
you're awesome dude.
channel name = blackpenredpen
him: uses blue pen
me: ✌️
Hi, It would be nice if you make a video explaining the issue of why does everyone prefers "positive stuff", like positive roots, or cancelling square roots with seconds powers, being the second power inside of the square root. All that things.
Whoever looked at an integral like that and realized you could solve it by substituting a trig function was a straight-up genius. I can't even imagine making that connection.
thank you so much!!!
It's cool ..
On another topic, I found a quadratic equation that can find the levies in the right triangle when raising height to the rest:
c²x²-c³x² + (ab) ².
You can record ab = ch and save a torch in two disappearances.
The square beacon yields two results, the two levies that stand on the rest.
You explained this better than my professor, homework, and Isaac Newton
Doraemon music at start😎😎😎
Ahhh the nostalgia
Now we can proof that 9/2*arcsin(1) + 0 - 9/2*arcsin(-1) - 0 = 9/2*pi, and so arcsin(1) - arcsin(-1) = pi. How cool is that. : )
Why aren't you doing potions? You seem more interested in muggle math😑
Yes, and since arcsin is an odd function, this proves arcsin(1) = π/2
@@angelmendez-rivera351 More easily:
Take a right triangle with angle theta. As theta approaches 90° i.e. pi/2, that means you'd approach an isosceles triangle with two right angles. The legs (one opposite theta, the other the hypotenuse) would have to be equal length(isosceles triangle symmetry), so sin(pi/2) = 1, and thus arcsin(1) = pi/2
I used to do potions for a period of 3 years... I'm taking some rest right now. : )
Interesting you upload this just as we are studying trigonometric substitution in calc II
What a cool way to spend Valentine's Day.
Is nice to look that the integral is well defined in [-3,3] and the argument in arcsin is ×/3 (domain of arsin is [-1,1]) solving any problem with the value of x
i really dont know why i keep seeing integral resolutions and all this stuff when i barely know pre-calculus lol
i love you so much
Integral challenge:
sinx/(1+e^x) dx from pi/2 to -pi/2
Have been stuck in this problem myself.
Like your videos btw. They serve as a great source of problems whenever I almost end up wasting my time on CZcams.
Can you do it with trigonomteric sub?
I think so.
I am from moroco, Your canal is a very nice .
AIMAD ELOUARDACHI thanks!
Your a great teacher my friend
Thank you.
Superb
Plz make video on" volume of certain 3d solids "
Thank you ☺️
Well in India, you're directly taught the formula for √(a^2-x^2)
that was fun to watch
Does this generalize to a semicircle of radius r? Something like:
(r^2/2)arcsin(x/r)+(1/2)x(sqrt(r^2-x^2))+C
awesome video
The 10 min can also be solved in 1 min , using a formula which can be derived by geometry or algebraically.
True however in my calculus class you have to have finished the question with calculus or they take away Mark's
Why can we assume cos(θ) is positive?
1:29 Bob from Incredibles: "PI IS PI"
Sir i think there is a formula for integration of rt of a^2-x^2 which is given by x/2 rt (a^2-x^2) +a^2/2 sin inverse of x/a
+c
..where c is a constant
Don't memorise it as a formula, it's actually the area of a sector plus the area of a triangle
i like how the right one is super simple but the left one is pretty complicated
Great video as always!
It would be nice if you could help me with that promlem though.
f(x)=ax^2+bx+c and there is a pair of values of x that f(q)=w and f(w)=q prove that this function cannot do the same thing with a pair of numbers other than q and w
Try solving for q using that equation.
But how can you get the area of a circle without first doing the calculus?
Why is it that when you use trig sub you don't have to use absolute value when taking the square root of something squared?
What happened to the 9-9sin^2 theta why/what did he mean when he said he factored out the 9?
How will we know when to sub a trig value in for x
What’s this integral talking about outside of -3 to 3, Maple says it is close to 5*i from 3 to 5
hello as a beginner im wondering why cant we use sqrt(9-x^2) as (9-x^2)^(1/2) and use chain rule to solve this
The beard's looking good!
There is a formula with which we can get the answer in 10 Seconds. It's a direct formula for these kind of sums. There is no need to use area under a curve.
Hi, how different or same would the solution be if we were to substitute x=3cos(@) ?
Aditya Iyer Not much different.
Can you not do the arcs in directly without trig substitution?
The integral of something in the form dx/[root(a^2-u^2)] is arcsin(u/a) +C. A is a constant and u is a variable expression.
In this case the answer would be arcsin(x/3) +C.
NICE !!
Good job here.
I have a question towards the first one, how can you be sure this is a half circle? How do you know the function is round (it is symmetrical I know but how do you that it's also round and not curvy in other form).
the function was y= sqrt(9-x^2)
rearranging you get:
y^2=9-x^2
x^2 + y^2 = 9
this is the equation of a circle thats how he knew it was a full semi-circle
@@chronyx685 thanks alot bro
need to state explicitly θ is acute?
6:54 Now do this integral geometrically. It's really satisfying.
HelloItsMe There we go again with that nonsense.
@@angelmendez-rivera351 can you stop please, you just don't know what I mean but then just don't comment on my comment. You can do this geometrically by dividing it up into a circle sector and a triangle
Hi both of my long-term viewers!
I think I know what Helloitsme meant as someone else commented it too. I will pin that comment now.
HelloItsMe Yes, that is called evaluating a definite integral. Quite different from getting an antiderivative. If you think I do not know what you are talking about, then bother to prove it and explain it.
YAY!!!!!!!
Snejpu : ))))
i am abotut to lose my damn mind.
Why would we use the formula of area I wanna know why this particular method?
That was not really a 10minute integral, in our school we are taught general formulae for a lot of integrals
√(a^2 - x^2) being one of them.
Taking the bounds off = taking the training wheels off and it's time for the rubber to hit the road!
Blackpenbluepenredpen
But how do you know what the area of a circle is without going through the other integral first?
Because the circle is a shape whose area we know, long before we learn what integration is.
Why can't you use a u substitution to solve the integral? Am I missing something?
You can call your new variable anything you want, you just do it the opposite way of what traditional u-substitution is.
Usually, when people talk about u-substitution, they mean identifying a composition of functions originally in the integrand, and recognizing the derivative of the inner function, multiplied by it. Or a constant multiple thereof. You use a u-substitution, to eliminate the inner function, and just integrate the outer function. You are looking for something that could've been made by the chain rule, and undoing that process.
With trig sub, you do the reverse. You introduce an additional inner level of functions, so that it can be simplified through a trig identity.
@@carultch Thanks man, I've been stuck on that question for the past 4 years, I can finally pass my Calc final!
Seriously though I respect the detailed explanation, I'm a Math major in university now and compared to what i'm doing now, I'd kill to get back the days of just doing trig and u-subs...
10 sec integral (if you know formulas)vs 1 min integral
just integrate by substitution makes it way easier.
There are many errors here. 1. Can not substitute x=3 sinx unless|x|≤3. 2. Can not set √cos²x=cosx. That is only true for cosx≥0, which it not always true. The errors goes on and on. Please repost.
That's why we have Properties in 12th Maths NCERT in India!
Sorry I will have the life time integral
Thank you
Thnaks. sqrt(cos^2(x)) should be |cos(x)| , why do you omit modules?
Cos^2(x) is always positive for real values of x therefore sqrt(cos^2(x)) is also positive.
Ethan Bottomley-Mason No, that is not how that worls. By definition, sqrt(x^2) = |x|. x^2 is always positive if x is real, but from this, it does not follow that x is positive.
Petruschko Ukropovich You can omit the modules because the anti-derivative usually is only well-defined and only makes sense in real-numbers for an interval of values in which cosine happens to be nonnegative, making the absolute value unnecessary.
@@angelmendez-rivera351 can you explain it again?
@@99Albileo For every x in [-3,3] , theta is in [-pi/2 , pi/2], so cos(theta) is positive, and we can drop the absolute value.
Amazing!
Was pretty sure we would use complex integration. Strange that we end up with arcsines whereas some parts of the domain are clearly in the imaginary world...
black pen red pen BLUE PEN?
A less neat way to write sqrt(9-x^(2))/3 would have been cos(arcsin(x/3)), which was my initial response. I don't know why this baffles me.
Doraemon intro tune?
By the way, your videos are very good.
I like the challenges mathematics.
AIMAD ELOUARDACHI me too!!!
@@blackpenredpen est ce que je peux faire l intégration par partie pour résoudre cette intégrale?
Ah someone just changed their CZcams profile photo
Just got an A in Calc 3 :)
i spend literally 1 hour to find the second one, I am the dumbest person on this earth
Be nice to your people 😂😂😂 5:00
What else can one say about it but just great 🙂
Challenge from India . What is the average distance of all possible random points within a square of length 1 unit
: )
Can someone solve this: square root of x square minus 4
I’ll just put theta = arcsin (x/3) and call it a day
👏👏👏👏
Sin inverse >>>> arc sin
I like arcsin because sometimes I use sin^(-1)x instead of csc(x)
Nah, arcsin is better because it is 1. Easier to type and write in most situations, and 2. Unambiguous notation is objectively always better than ambiguous notation. No matter how much you want to argue from context, sin^(-1) will never be unambiguous.
I don’t see why the indefinite takes 10 minutes.
I thought that was √q-3sin(theta) lol
After ward I realized it was 9
You are doing it soo long it have formula then it is only 1min
Easy...
Srinidhi A V good job!!!!
Thanks!
good job!
+c
thinks
Doremon back
Nice to watch but I didnt understand 90% of it
Nice beard!
Thank you!!!
3rd
What if we dont use the double angle identity?
1st here
Stop boasting and lying. You spend hours to solve it, and then put it on video for one second.
Abed Bob
Ok : )