I had example 3 on my homework and i was confused about it until you broke it down and explained it PERFECTLY. Thank you this helped a lot with overlapping curves 👍
the lower bound should be less than the upper bound. If we set 3pi/2 as the lower bound then the upper bound would be 2pi, other wise, we could rewrite the expression as negative 0 to 3pi/2
I had example 3 on my homework and i was confused about it until you broke it down and explained it PERFECTLY. Thank you this helped a lot with overlapping curves 👍
out of all the videos ive watched, this one helped me the most; tyvm.
The absolute best way to learn calc, thanks algebros you guys are the best !!!!
OMG thank you so much I finally understand this, I was so confused😮💨
Oh my god thank you so much I finally understand this
This is an excellent video.
For problem number 2, why did you go from -pi/2 to 0 and not 3pi/2 to 0?
because i think it gives negative value i dont know why but we always have to use bigger upper limit than lower limit
the lower bound should be less than the upper bound. If we set 3pi/2 as the lower bound then the upper bound would be 2pi, other wise, we could rewrite the expression as negative 0 to 3pi/2
10:30? wtf happened, how is it so you just decided to set zero to be in this point
I think maybe because the function -6cosø whenø=0, r=-6?
well idk but if you decide to set zero in the original way it has no difference. the integral is still pi/2 to 2pi/3
W mans
Fr
I failed
you philled
😂
Too easy.