y = sqrt(x) is not a "function" by definition, because it doesn't pass the vertical line test. However, if you think about it, x= sqrt(y) is just the top half of a parabola opening to the right.
Because to make y= sqrt(x) a function we only take the principle sqrt(x). That is only using the positive values that square to x and not the negative values.
y = sqrt(x) does pass the vertical line test As for my original question, here's why: y = sqrt(x) has a domain of x >= 0 and a range of y >= 0. Thus, the squared form will also have the same restrictions (x = y^2, with x >= 0 and y >= 0)
I know this is 2 years later 😂😂 but it is 2-sqrt x because he was finding the area for the right side of the graph, which includes y=2 for the top function and y= sqrt x for the bottom function, building the subtraction: 2-sqrt x , hope this helped 😊
Why do we subtract the two functions? I'm completely lost because i thought integration would always represent the are under the curve until the axis. Wouldn't you have to add the area of the top and then the area at the bottom or am I missing something?
i have a question. if we want to find the area by integrating with respect to X, why do we have to make in into two pieces? can't we just integrate it from -2 to 4?
Great videos!! Could you please post one about (u,v) sustitution when finding areas using double integrals..?? You know.. Jacobian and all that stuff. Thanks!
hey lord of Calculus...would you like to give me an example in this topic(a Area Between Curves - Integrating with Respect to y) using trigonometric function?
12 years later still helping a stressed calc student thank you
i JUST needed this today for my upcoming exam in a week and was wondering if there's a video of it. I come home to see this. You sir, a are blessing!
did you ace your test?
@@jacklowe74Lol it was 12 years ago 😂😂
You explained this easily to someone who feels math is another language. Thank you!
I find your videos my favourite in every course I take of math courses ..Thanks
he smashes it Just non stop computations!! you make it look so easy very good man
JACK SPARROW those are flattering words from jack sparrow... or Isaac Newton
That sharpie sounds effing amazing
My upcoming exam is going to be a cakewalk thanks to you ¡Señor!
Brilliant ! So simply illustrated!
u are really cool.. u explain toics so smoothly.. thanks..
you are a great teacher
@mpatt79 glad i could help and THANKS FOR TELLING PEOPLE! that is the best thing one could do for me :)
I learn more from your videos than I ever do in class, thank you.
I just want to thank you so much for this video.
Question: when you square both sides of y = sqrt(x), you actually get a sideways parabola that is not like the original graph. What happened?
y = sqrt(x) is not a "function" by definition, because it doesn't pass the vertical line test. However, if you think about it, x= sqrt(y) is just the top half of a parabola opening to the right.
Because to make y= sqrt(x) a function we only take the principle sqrt(x). That is only using the positive values that square to x and not the negative values.
y = sqrt(x) does pass the vertical line test
As for my original question, here's why: y = sqrt(x) has a domain of x >= 0 and a range of y >= 0. Thus, the squared form will also have the same restrictions (x = y^2, with x >= 0 and y >= 0)
I'm from Mexico and in all of TH-cam in Latinoamérica there isn't video that explains this that easily
5 days without a video, i was beginning to die!!
@VictorAndScience yes, yes they should. you should write them.
Great video, thanks for the help
why is it 2- sqrt x (1:26)
I know this is 2 years later 😂😂 but it is 2-sqrt x because he was finding the area for the right side of the graph, which includes y=2 for the top function and y= sqrt x for the bottom function, building the subtraction: 2-sqrt x , hope this helped 😊
Why do we subtract the two functions? I'm completely lost because i thought integration would always represent the are under the curve until the axis. Wouldn't you have to add the area of the top and then the area at the bottom or am I missing something?
You are the man, always managing to help me out :)
Do you think you could do a video about finding area under a curve using the n-rectangles method?
Where are you now, 10 years later?
thank you so much. really useful video.Just understand how to do my homework right away.
i have a question. if we want to find the area by integrating with respect to X, why do we have to make in into two pieces? can't we just integrate it from -2 to 4?
Thank you so much good sir!
Ik this is 9 years old, but thanks for teaching me this better than my prof
@Spiceman333 this is not the only one!
Really appreciate the videos bro
ever messed around with a Schrodinger equation?
thanks a lot. now i;m ready for the exam
thanks heaps, very useful for the upcoming hsc maths test :)
Great videos!! Could you please post one about (u,v) sustitution when finding areas using double integrals..?? You know.. Jacobian and all that stuff.
Thanks!
Thank you for explaining like I'm 5, as stupid as it sounds it makes it so easy to understand.
You are amazing bud !
hey lord of Calculus...would you like to give me an example in this topic(a Area Between Curves - Integrating with Respect to y) using trigonometric function?
@superbubbleman i already have
Another excellent video
Thanks a lot ..
You really helped me ..
Fantaaaastic
u just saved my butt thank u !!!
Thank you.
You have to split it up because it crosses the x axis.
very clear thanks!
what year do we take this?
Ki this is calculus 2
Sound on this video is barely audible....
@ArsonistInUrFirewall perfect : )
Thanks boss
Why didn't he write +C?
because it's gets cancelled when we do the definite integration
Sharpie should be paying you or all these videos.
Thumbs up if u came here through Mr. Brose's given link from Cerritos HIGH
whos watching this in 2021?
Thank you.