Comparison Theorem for improper integral, calculus 2 tutorial

  Рет қаралды 9,173

blackpenredpen

blackpenredpen

5 жыл бұрын

Calculus 2 tutorial. Learn and understand the comparison theorem for improper integrals.
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Пікірлер: 34
@quahntasy
@quahntasy 5 жыл бұрын
New channel name - White chalk
@rot6015
@rot6015 5 жыл бұрын
Whitechalkredchalk
@johnmin3821
@johnmin3821 5 жыл бұрын
You can't just conclude that's old. *Don't do that.* Your explanations are the best haha
@AlRoderick
@AlRoderick 5 жыл бұрын
I think he wants us to be sure that we put down the correct inequality right there.
@Gold161803
@Gold161803 5 жыл бұрын
But don't worry if they're negative even though that would make the inequality wrong... we still love him though
@qmzp2
@qmzp2 5 жыл бұрын
I like how your recent releases are perfectly in sync with what I'm studying in calc 2, haha. Thanks for the videos!
@futfan9092
@futfan9092 5 жыл бұрын
I figured that the denominator was really close to the sqrt((x^2+sqrt2)^2). The sqrt and square cancel out, so you are left with the intgeral of 1/(x^2+sqrt2), which is 1/sqrt(sqrt2)*arctan(x/sqrt(sqrt2)), which definitely converges, so this one probably does too
@himanshnegi
@himanshnegi 5 жыл бұрын
Sir its my humble request to please checkout the "jee advance" integration sums. Believe me they are really complicated so u should make videos on them, as most of the sums blow up my mind. P.S.- sorry for my bad english.
@mrdeep7694
@mrdeep7694 5 жыл бұрын
thank you
@bipinmaurya3824
@bipinmaurya3824 5 жыл бұрын
Thanks sir..i am viewing right now
@snejpu2508
@snejpu2508 5 жыл бұрын
You and Peyam uploading exactly at the same time. : )
@chaoticoli09
@chaoticoli09 5 жыл бұрын
I think it's unnecessary to prove that 1/x^2 >= 1/(sqrt(x^4 +... even terms + positive constant)). 1/sqrt(x^4) = 1/x^2, so adding more positive things to the denominator makes the entire expression smaller.
@arvindkumar531
@arvindkumar531 5 жыл бұрын
We can make this inequality stronger by noting that 1/(x**2 + 1) = 1/sqrt(x**4 + 2*(x**2) + 1) > 1/sqrt(x**4 + 3*(x**2) + 2 ) > 1/sqrt(x**4 + 4*(x**2) + 4 ) = 1/(x**2 + 2) Where ' ** ' is the power function Now integrating both the extreme sides we find that the required value lies between ( pi/2 , pi/(2*sqrt(2)) - arctan(1/sqrt(2)) / sqrt(2) )
@ingilizcehazrlk9134
@ingilizcehazrlk9134 5 жыл бұрын
Great
@matefixfix1338
@matefixfix1338 5 жыл бұрын
Eres un crack
@pauljackson3491
@pauljackson3491 5 жыл бұрын
-2 < 1 (-2)^2 < (1)^2 4 < 1 wait... When you square the numbers you lose any conditional direction. It's like when you multiply be a negative number you must change the direction; squaring multiplies the left by -2 and the right by +1, does the greater-than become a less-than or not! Regardless, I thumbs upped since that still works and can help find out if the integral even converges. Of course I always thumbs up your videos anyway. We need more teachers/professors like yourself. I like using the word (well non-word) un-ir-regardless-ness-ful-ly. Sure the affixes may not make sense but it sounds cool/stupid.
@ananyapathak8701
@ananyapathak8701 5 жыл бұрын
Missed your color switch style!
@3P141592651
@3P141592651 5 жыл бұрын
積分1/2 (secx)^(1/2) ,從pi/3到pi/2。與此題一樣
@azamingxzc9570
@azamingxzc9570 5 жыл бұрын
Nice suit ',:D
@Ispeaktrollish
@Ispeaktrollish 5 жыл бұрын
Could you do the graph of e^(-e^x))? It's pretty interesting!
@Sup3rlum
@Sup3rlum 5 жыл бұрын
#YAY
@sihle5671
@sihle5671 3 жыл бұрын
"the big balls"😋😂😂😂
@82rah
@82rah 5 жыл бұрын
Can't you make the stronger statement that definite integral 1/sqrt(x^4 +3x^2 + 2) < 1 ?
@jacksonsingh855
@jacksonsingh855 5 жыл бұрын
whats the value of integral though??
@user-cq6pd3nw3d
@user-cq6pd3nw3d 5 жыл бұрын
0.7269
@arfanm9440
@arfanm9440 5 жыл бұрын
How did you know I’m taking Calc 2 and I have the test which will cover Comparison Theorem coming Monday
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Arfan M : )
@MarkMcDaniel
@MarkMcDaniel 5 жыл бұрын
I'm guessing that chalk costs much less than dry erase markers.
@linearealgebra4006
@linearealgebra4006 5 жыл бұрын
Soo low Can you sometimes do some multivariable Analysis
@bipinmaurya3824
@bipinmaurya3824 5 жыл бұрын
I think It's special for jee mains
@himanshnegi
@himanshnegi 5 жыл бұрын
Are u preparing for it.
@bipinmaurya3824
@bipinmaurya3824 5 жыл бұрын
Yes i'm going to be attending the examination of jee mains 1st batch exam2018
@gradecracker
@gradecracker 5 жыл бұрын
Are you a teacher? If so what school
@Gold161803
@Gold161803 5 жыл бұрын
UC Berkeley I think. Don't quote me :p
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