Your method is a bit like a CIRCULAR argument though because the derivatives of trig functions come from s=rtheta(length of circular arc) because with this and a unit circle considering the heights of small triangles we can conclude sin(theta) is approximately theta and thus limit sin(theta)/theta=1 and as a result d/dtheta(sin(theta))=cos(theta). S=rtheta can be derived from A=pir^2 in the following way A=integral C(r) dr and then using the Fundamental Theorem of Calculus to get C=2pir and thus s=rtheta as there are 2pi radians in a circle by definition.
Thank you for using plain and simple language to explain this. It makes it so much easier to deal with and understand. Maybe my Calc Prof. should take notes from you.
the integral of sqrt(1-x^2) does have a pretty intuitive geometric interpretation. if you go from 0 to 0.5 you have a cake slice bit and a right triangle bit. the triangle is easy and the cake slice bit has an area proportional to the angle of the slice, and that angle is just arcsin(x).
Going from, say, -5/2 to 0 is treatable as an area as well. A sector of a circle of radius 3 (need to calculate the angle) and a triangle of base=5/2 and height = sqrt(9-(5/4)^4).
You're the GOAT. The website my homework is on has a feature called "watch it" if you are stuck on a problem and their explanation made no sense. Would not have gotten the problem right if it weren't for you.
4:43 not necessary you have to use calc or use calculus 2. Because if you know the formula of area of circular segment you can calculate it by using simple formulas.
blackpenredpen The area of the circular segment is equal to the area of the circular sector minus the area of the triangular part. To prove it you don't have to know calculus 2) in your case if you consider integral from -2 to 0 , you just have to find the angle of circular sector. Then you have to subtract the area of circular segment from area of semicircle and divide by two)
By the way I frequently follow you, and your calculus is helping me very much. Although I solved it myself( very first question which you asked and I solved ) But what important for me was a method you applied to solve it.
Second one is quite easy with integration by parts To avoid trig subs we need two things 1. Integration by parts 2. Rewrite sqrt(9-x^2) as (9-x^2)/sqrt(9-x^2) Order does not matter We can rationalize this integral with Euler substitution sqrt(9-x^2)=(x-3)u
x^2=2^x. if i base 10 both sides, i get 10^(x^2)=10^(2^x). Using commutative property of exponents (I don't know what it is actually called), 10^(2^x )=10^(2^x) Therefore, x^2=2^x.
Is it possible to make a closed form for the integral of e^(-x^2)? I managed it for xe^(-x^2), the result being u = -x^2, f(x) = -.5 e^(-x^2), but it seems impossible otherwise...
If u square than sq root something, it becomes absolute value of that thing. So once u sq root both sides u get abs value of -i =abs value of I. This, of course, is true.
The problem is the action of taking the principal square roots. i^3=cos(-pi/2)+isin(-pi/2), (i^3)^2=cos(-pi)+isin(-pi)=cos(pi)+isin(pi) (changing to pi instead of -pi because of the definition of the principal argument of a complex number), so the principal square root of (i^3)^2 is cos(pi/2)+isin(pi/2), which is equal to i but not i^3. This is similar to the situation when you take square root of a negative real number, you get a positive number but not the original number.
![CAMPปลิ้น | EP.82[2/2] เมาท์มอยกับ 3 สาวเพื่อนซี้ ที่พร้อมขยี้พ่อหมีได้ทุกเมื่อ](http://i.ytimg.com/vi/Wwn7WnylNtQ/mqdefault.jpg)
thank you so much for this!! You taught me area integral interpretation in less than 5 minutes where my uni prof took 30!!!
I paused the vid at the beginning and calculated it with trig sub.
x = 3 sinθ (upper = 0 , lower = -π/2)
But I liked your method more.
I did polar coordinates in my head. ((r^2)/2 from 0 to 3)(pi/2)
Your method is the standard way of solving it.
Your method is a bit like a CIRCULAR argument though because the derivatives of trig functions come from s=rtheta(length of circular arc) because with this and a unit circle considering the heights of small triangles we can conclude sin(theta) is approximately theta and thus limit sin(theta)/theta=1 and as a result d/dtheta(sin(theta))=cos(theta).
S=rtheta can be derived from A=pir^2 in the following way A=integral C(r) dr and then using the Fundamental Theorem of Calculus to get C=2pir and thus s=rtheta as there are 2pi radians in a circle by definition.
Thank you for using plain and simple language to explain this. It makes it so much easier to deal with and understand. Maybe my Calc Prof. should take notes from you.
Thank you so much sir... You actually know the value of time... And even know how to make the students understand.. Thank you so much sir...
the integral of sqrt(1-x^2) does have a pretty intuitive geometric interpretation. if you go from 0 to 0.5 you have a cake slice bit and a right triangle bit. the triangle is easy and the cake slice bit has an area proportional to the angle of the slice, and that angle is just arcsin(x).
Going from, say, -5/2 to 0 is treatable as an area as well. A sector of a circle of radius 3 (need to calculate the angle) and a triangle of base=5/2 and height = sqrt(9-(5/4)^4).
Really helpful. I struggle with the hour long online lectures due to covid with my attention span and this helped me.
You're the GOAT. The website my homework is on has a feature called "watch it" if you are stuck on a problem and their explanation made no sense. Would not have gotten the problem right if it weren't for you.
Legend bro the explanation is much more simpler than other places
Ironic how this is the exact integral in one of our questions at university
Dozens of universities, actually.
How is that ironic?
@@JohnDoe-my8kj it's not
lol the question I’m struggling w and here we are lol
When have u became a thug😂😂😂 those sunglasses
thank you so much! i've been stuck on this question for the past 15 mins
thanks your better than most prof out there.
Integral of -2 to 0 of √(9-x^2) is actually doable without calculus as well
4:43 not necessary you have to use calc or use calculus 2. Because if you know the formula of area of circular segment you can calculate it by using simple formulas.
kevin51241 and how can we prove that formula?
blackpenredpen you meant : area of circular segment?
blackpenredpen The area of the circular segment is equal to the area of the circular sector minus the area of the triangular part. To prove it you don't have to know calculus 2) in your case if you consider integral from -2 to 0 , you just have to find the angle of circular sector. Then you have to subtract the area of circular segment from area of semicircle and divide by two)
kevin51241 ooh yea u r right! I forgot about geometry >_< lol
hey thanks man keep it up!!
Great video, This was the exact problem i was stuck on, and I just so happened to see this in my reccomendations
How is this the exact dame problem I needed help with...? =,=
y = sqrt(9-x^2) This is the arc of a circle, not a circle) because y >= 0 always.
Pls pls make a video on definition of limit in complex domain...
I LOVE UR HANDWRITING
The 2nd term is a nice subject of observation to an oral test.
By the way I frequently follow you, and your calculus is helping me very much.
Although I solved it myself( very first question which you asked and I solved )
But what important for me was a method you applied to solve it.
We have the same last name :)
Thank you sooo much!!
Hi! How about this? Evalueate definite integral from 0 to 1: sqrt(1+1/x)dx
Baby calculus is my favorite type of calculus!
Thank you
this was really good!!!
what about v(t)=4t-6 from t=1 to t=4 interpreting the integral in terms of areas?
Second one is quite easy with integration by parts
To avoid trig subs we need two things
1. Integration by parts
2. Rewrite sqrt(9-x^2) as (9-x^2)/sqrt(9-x^2)
Order does not matter
We can rationalize this integral with Euler substitution
sqrt(9-x^2)=(x-3)u
woah, damn, this was explained incredibly well :)
thanks so much!
Good job young man!
nice explanation
x^2=2^x.
if i base 10 both sides, i get 10^(x^2)=10^(2^x).
Using commutative property of exponents (I don't know what it is actually called), 10^(2^x )=10^(2^x)
Therefore, x^2=2^x.
Alan Armstrong you can't say a^(b^c) = a^(c^a)
eg 2^(3^4) != 2^(4^3)
2^81 != 2^64
edit
As of integers the only case when a^b=b^a & a != b is 2^4=4^2
Isnt the Area= 3+ integral of (sqrt(9-x^2))/4 -3, from -3 to 0.
I am thinking wrong or..?
Otherwise you use the area 3*1 two times?
Can you make a video on i factorial?
Anyone notice the sound at 2:54?
Is it possible to make a closed form for the integral of e^(-x^2)? I managed it for xe^(-x^2), the result being u = -x^2, f(x) = -.5 e^(-x^2), but it seems impossible otherwise...
Cool shades bro.
BROOO THIS IS THE EXACT PROBLEM ASSIGNED IN THE TEXTBOOK
Do the trig-sub !!
Looking cool
Lmao this questions on my test tmrw perfect
volumes?
Ok ,level completed ,level 2 double integral,level 3 triple integral ;)))
I love this method.... #yay
thanks
lots of sun in the classroom huh
trig identities. this is me right now in calc 2 :(((
isnt intergral from -3 -> 0 of 1 is 4? why he got 3??
3 * 1 =3 , how you got 4?
Pl solve my query
i^2=-1 and -i^2 = -1
So ....(-i)^2=i^2
((i)^3)^2=i^2
Square root both side...we get...
i^3 = i....pl solve my query
If u square than sq root something, it becomes absolute value of that thing. So once u sq root both sides u get abs value of -i =abs value of I. This, of course, is true.
The problem is the action of taking the principal square roots. i^3=cos(-pi/2)+isin(-pi/2), (i^3)^2=cos(-pi)+isin(-pi)=cos(pi)+isin(pi) (changing to pi instead of -pi because of the definition of the principal argument of a complex number), so the principal square root of (i^3)^2 is cos(pi/2)+isin(pi/2), which is equal to i but not i^3. This is similar to the situation when you take square root of a negative real number, you get a positive number but not the original number.
lettt goooo
Where is our lord and savior Justin Y.?
Bat shit crazy genius you are
Amazing
Brilliant
got to be honest didnt epect my homework to be the example
#YAY
👈😎👈 zoop
Hi
👓😎😎🤘
I'm gonna have to skip this one on the test. Thanks anyways.
Crack
Crack
He forgot +C 😋😋😋😋
No
It’s not an indefinite integral. It has limits so the family of functions is not needed
NooN
#YAY