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Integration by U-Substitution, Definite Integral

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Uploaded by on Apr 13, 2008

Integration by U-Substitution, Definite Integral

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Education

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Standard YouTube License

  • likes, 9 dislikes

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Uploader Comments (patrickJMT)

  • That was nice!

    I am just preparing for next term (which starts in a week). I was able to get an A in Calc 1 because of your videos. Figured I would start studying a little earlier because everyone is telling me that calc 2 is SO Much harder than calc 1.

    Do you feel this is true?

    Thanks,

    From Oregon

  • @wperlich integration is typically harder than differentiation, so yes, i think calc 2 is probably a bit more challenging

  • what happened to "+ C"

  • @eyeheartthekillers you do not need it for definite integrals. i just cancels out when we subtract so people do not bother including it

  • You're amazing, your videos are the only ones I've found (after searching for a good amount of time) that have really connected the topics for me. Thank you =)

  • @Sharylyn13 come back and watch any time ; ) 

Top Comments

  • Love your video, I like the HD ones more :) but everything great, even your voice is nice to listen to.. I got a calculus final tomorrow, and I feel a lot more comfortable after watching your videos.

    Thank you so much

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All Comments (161)

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  • this video is the oldest out of the ones i watched

  • How did he account for that first x^2 outside the quantity? From what I understand he only substituted the inside? Somebody must know.

  • @eyeheartthekillers

    It's a definite Integral, which is an exact value, so no +c is necessary.

  • your teaching is one of the clearest I've come across. Please become a math professor and teach at my university!!....we have some really shitty math profs where I am....

  • Ha, I actually spent a few minutes looking for this video! Seeing the giant white board (versus your smaller one or paper) threw me off, heh.

    I'm incredibly grateful for all you do though. Math has always been a bit difficult for me, but everything you've done has made it so much easier...and pleasant! Thank you so much =)

  • @TheSrrobinson Thanks for clearing that up, I was actually doing some problems today and kind of figured it out on my own. But im probably going to forget how it works anyway lol

  • @searching4somewhere i had the same question. If you do not change the bounds of integration then you would need to plug in x^3+1 for 'u'. However, since patrickJMT changed the bounds of integration with respect to 'u', we can simply plug in the new bounds for 'u'. Try doing it both ways, and you will get the same answer.

  • @TheSrrobinson nevermind I see how this works out. I see how changing the bounds of integration with respect to "u" allows us to not need to plug in the original value for "u". Also, I see that if you keep the bounds the same then one would need to plug in the x^3 + 1 value replacing "u" to finish solving. Eureka!

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