In Desmos, if you click on either the lower or upper bounds (in this case -10 and 10) you can change the range of the exponents and even the step (i.e. going from 1.0 to 1.0 instead of 0.1 to 0.1).
I imagine that if you were competing you would want to have the basic elements of these integrals well memorized because they could really slow you down otherwise.
Hi Mike. Very true! Time is a big factor in the competition. Not sure I said it in the video but I was thinking this particular problem wouldn’t work in a competition because of the approximate solution. 🤔
@@owlsmath So, tending to zero won't cut it? I am "doubly stupid" being both self taught and from a physics/engineering background, so I am not averse to fudging things a bit to get (close enough to!) the answer. LOL
@@MikeMagTech 😆 I was just thinking about how would you express the answer. Sometimes they can say something like "round your answer to 3 decimal places" but that seems kind of awkward :)
i think it would be interesting to perhaps see a reduction formula for a certain function or two you've reviewed before either x^m ln^n (x) or (xln(x))^n whichever is easier
@@the.lemon.linguist ok. I know I did the problem multiple ways and one time i think I said I'd "never do that one again" 🤣 I think one of them I used IBP but not sure
I think you can do better than saying that it is approximately 0. You could have gotten that by recognizing that you are raising something under 1 to a large power. The error of the truncation is roughly half of the first omitted term, so the integral is approximately 1/4050. The true value is 1/4048.000988.
@@slavinojunepri7648 It's true for many alternating series where the ratio between the terms goes to -1. You can check it by considering the remainder of the series grouped in pairs in two ways.
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These types of integrals are pretty common for competitive exam but hard to solve
Hey Skyfall. Yes, I think it would be a real problem without the reduction formula. 👍👍👍
In Desmos, if you click on either the lower or upper bounds (in this case -10 and 10) you can change the range of the exponents and even the step (i.e. going from 1.0 to 1.0 instead of 0.1 to 0.1).
Wow thanks for that! I was thinking there should be a way. 😆
I imagine that if you were competing you would want to have the basic elements of these integrals well memorized because they could really slow you down otherwise.
Hi Mike. Very true! Time is a big factor in the competition. Not sure I said it in the video but I was thinking this particular problem wouldn’t work in a competition because of the approximate solution. 🤔
@@owlsmath So, tending to zero won't cut it? I am "doubly stupid" being both self taught and from a physics/engineering background, so I am not averse to fudging things a bit to get (close enough to!) the answer. LOL
@@MikeMagTech 😆 I was just thinking about how would you express the answer. Sometimes they can say something like "round your answer to 3 decimal places" but that seems kind of awkward :)
Great video! Love the arctan approximation😊
Thanks! Yeah at first I was thinking it’s an ugly solution but then arctan saved the day 😂
i think it would be interesting to perhaps see a reduction formula for a certain function or two you've reviewed before
either x^m ln^n (x)
or (xln(x))^n
whichever is easier
Hi Lemon. I think I did it before but I"m not sure on that.
@@owlsmath oh, alright!
i'll look to see
@@the.lemon.linguist ok. I know I did the problem multiple ways and one time i think I said I'd "never do that one again" 🤣 I think one of them I used IBP but not sure
I think you can do better than saying that it is approximately 0. You could have gotten that by recognizing that you are raising something under 1 to a large power. The error of the truncation is roughly half of the first omitted term, so the integral is approximately 1/4050. The true value is 1/4048.000988.
How did you come up with this assessment on the error? Is this true for every series?
@@slavinojunepri7648 It's true for many alternating series where the ratio between the terms goes to -1. You can check it by considering the remainder of the series grouped in pairs in two ways.