Відео

Thank you so very much Professor for covering this topic! Keep up the sharing!

Nice that one was fun. U-sub trig sub an natural logs all stuff I enjoy lol

Note: at 53:04 forgot to write “prime” in the denominator, it should be |T’|.

Spice it up with chain rule of the day

Im having trouble understanding the examples that involve x=_ not being or being in the domain. In example 3 you say x=1 is not in the domain. now is that because we put 1 in for x and it equaled 0? because then in question four you say that h=0 is not in the domain but if we put 0 in for h it is???

yes Prof.V thank you so much for the new trick! I let cos(t) = u because I was afraid of the integral of 1/sin(t) but now you have opened my eyes I won't run away from such again. thank you so much.

Dear Prof.V I hope you are doing well thank you so much for the integral of the day. i let u = ln(x) and du = 1/x*dx and the was in the form sqrt( 1 - (u)^2 )/(u) and did trig sub by letting cost = u and -sint*dt=du and then simplify and break it up into two integral 1: integral sect and 2: integral of cost . the endpoint was -ln(sect + tant ) + sint + C by changing back to x it is -ln( (sqrt(1-(lnx)^2 ) + 1 ) /(lnx) ) + sqrt( 1 - (lnx)^2 ) + C let me enjoy the video

Hi Professor V, l used u=sq root1-t^2 for my second sub. And I have something different.

Yay I did it!!! 😊😊

First try!

¡¡Super bonita integral!! 🙂

is ln(sqrt(10)-1) also an acceptable answer? I was able to simplify ln(9/sqrt(10)-1) down to ln(9(sqrt(10)-1)/10-1)) which is just ln(9(sqrt(10)-1)/9) which then simplifies to ln(sqrt(10)-1)

Outstanding example

I can't seem to get the right answer to 3. I simplified and found the antiderivative but when I calculate it does not give me the right answer. Do you have to split the integrals?

Cosecant is prettier to look at but secant is a better conversationalist...Who said math can't be romantic....?

Another excellent video! This was a fun one. Just one question. You’re allowed to use the limit comp. test when the terms of your series aren’t always positive? Cuz the first term of the series you used the test on is zero, which is obviously neither positive or negative. I guess it’s not that big of a deal, but it’s still something I just wanna clarify lol

Thanks more dear!

in part 1 why would e^x be squared in youre following the formula uv - integral of vdu (by parts)

Woo! I was pleased to get the same conclusion. Thank you, Professor V!

It was so useful to understand thx

Fantastic videos and outstanding teaching! One question: what textbook are you using for these lectures?

Thank you so very much Professor V! This video is so helpful to understand the concepts in this section!

Thank you so much for this video!! You don't even know how helpful this is!! :)

Could we factorize and say the limit isequal to ln(6).ln(x-3) as x->3from the right?

thanks prof v. teaching is definitely a skill you have honed to perfection. with all your little tips nd neat little tricks, man im a huge fan lol.

Professor, why did we choose the negative at 55:13?

You’re amazing ❤ thank you so much

Dear Prof.V good luck with your summer school ! At least I have something to work on. you did an outstanding work with creating videos it does help a lot in university . anyways please enjoy the rest of your day

Dear Prof.V I hope you are doing well, for this one I let u = e^t and du= e^t*dt ,u(ln(4)) = 4 and u(0) = 1 and get a integral of 1/sqrt( u^2 +9) from 1 to 4. and apply u-sub by letting u = 3*tan(x) and du = 3*(sec(x))^2dx when I put everything back to the original from I got integral of sec(x) which is equal to ln|tan(x) + sec(x)| but now change from x to u which was equal to ln( (sqrt(9+u^2)/3 + u/3 ) and the deference was ln(9) - ln(sqrt(10) + 1) hopefully it is right. let me enjoy the video.

Easy peasy😄

melika khanzadeh said thank you very much for your help.

What a incredible teacher 😮❤

doing this and then watching the video makes me so grateful for hyperbolics

Thank you so much, Professor V. I can't wait for the day's integral.

This was great review! Thank you for the video. :)

I am happy to see daily integration and series videos through the summer!

thanks

Thank you :))))

just to confirm for the last problem, is there any specific reason why we pulled out the x from the 6 and 7? Just wanted to see whether or not it really mattered if we kept the top as just 6x^2 + 7x

Nice! I just used nth term test cuz the limit equals infinity cuz e^n obviously grows faster than n^2 in the long run.

Could we use the test for divergence? We can show, as a limit, as n-> infinity the nth term does not go to zero using L’Hôpital’s Rule twice. Is this correct?

Nice. Ratio tests are one of my favorite type of series problems. Now I can look forward to integral and series problems each day.

Thank you Professor V for your kindness and commitment, your seed will sprout a knowledgeable future...! 😊❤️

I was stressing so bad about this because I didn't understand it at all and I have a test on it tomorrow. Thank you so much this genuinely got me from not understanding it at all to fully understanding it. THANK YOU

I really enjoy your integral of the day problems. Is there any chance of getting a "series of the day" ?

Very helpful videos, thank you! I'm glad i just stumbled upon your channel while reviewing for Calc II final. Another great teacher to add to my lineup of online tutors!

That was a good one. So satisfying when my answer matches yours. I have been doing series problems and watching your series videos but trig sub is my favorite integral technique so when that comes up I have to give it a try lol.

I think if you rationalize the denominator at the end, it does help simplify. Great vid like always! Enjoyed it!

id love series of the day and testing for convergence! And related rates are also one of my favorites, lol.