Improper Integrals - Convergence and Divergence - Calculus 2
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- Опубликовано: 25 янв 2021
- This calculus 2 video tutorial explains how to evaluate improper integrals. It explains how to determine if the integral is convergent or divergent by expressing the limit as it approaches infinity. It also explains how to determine if the integral is improper by identifying any vertical asymptotes or infinite discontinuites. This video contains plenty of examples and practice problems.
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The (3x+1)² where did the squared go...I ddnt understand that part plz
@@paulineshiku1890 i know it's 6 months late but he substituted 3x+1 into "u" then raised it to the square afterwards (u^2). He did the integral of the full thing with "u" still in it and then replaced the "u" with 3x+1. So basically the squared just merged with the rest of the fraction
Great job , for those that interested in such problems , here my solution to 3 questions at the same time of the same book ( almost impossible integrals, sums and series) ruclips.net/video/dmrJCvXYcKo/видео.html
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Professor Organic Chemistry Tutor, thank you for another fantastic video/lecture on Improper Integrals in Calculus Two. This is a very important topic when it comes to Convergence and Divergence of Integrals in Calculus. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
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Thank you for the video. I'm going to learn improper integrals today in class. :) You videos have been so helpful!
13min video and I understood improper integrals way more than a hour and 40 min lecture. Thank you
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this guy is a legend, also how are u always covering what were doing in class. Like this was my first lecture in clac 2 this past month
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not me watching this 2 hours before my finals
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Can you please make a video on the type of improper integrals that have finite boundaries but infinite discontinuities?
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Sir, could you please do more video about improper integrals?
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if I was doing definite integrals and one of the x, y, or z components was divergent, what do I do?
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in problem 2 woudnt the intergal on -1/t be -ln(x)
5:42 why didn't you distribute the lim to the 1? why just apply lim to -1/t?
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