Imagine this question being on your final exam. You go through all of these steps and calculations just to forget the +C at the end which single-handedly causes you to fail the exam.
This is completely off topic, but I think I came up with an interesting problem. Given the recursive functions: f(n) = af(n-1) + bg(n-1) g(n) = ag(n-1) + bf(n-1) Where a=0.5 and a+b=φ Show that both f and g are fibbonacci sequences. I'll put my solution at the end. Stumbled upon this fact while trying to solve for g for any a or b... idk how others may attempt to solve it but I wrote them as a discrete time convolution. I got this equation for g(n): g(n) = (b^2-a^2)g(n-2)+2ag(n-1) Was trying to solve for f or g to write sin(nt) in terms of sin(t) and cos(t)... but accidentally made a mistake, one of the equations were supposed to add and the other was supposed to subtract the b term.... was doing this in the first place because, I was hoping the equation for sin(nt) would help solve the integral for (sin(Nt)cot(t/2)+cos(Nt))cos(nt) where N and n are two non-negative integers. Which I plugged into a couple integral calculators... and it didn't give me a solution. When you evaluate that integral in respect to t from -π to π, and divide by 2π, its supposed to be an inverse discrete-time fourier transform that should equal 1 for all |n|
I’m a sophomore in high school and have no idea what he’s doing but I still watch a bunch of his videos because they are interesting and relaxing. Keep up the good work👍
i was in your position a couple years ago. i’m now in calc 2, watched this whole video and actually understood all of it, and i’m very proud of myself. keep studying and you’ll get there💪
It is necessary to replace the variable in time to avoid radicals. ∫6dx/(x^4-6)=6^(1/4)∫ dt/(t^4-1). t=x/6^(1/4). Here the fraction under the integral sign is so simple that it can be decomposed into elementary fractions without the method of indefinite coefficients. 1/(t^4-1)=(1/2) *{1/(t^2-1)-1/(t^2+1)}=(1/2)*{(1/2)[1/(t-1)-1/(t+1)]-1/(t^2+1)}= = (1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)
I've always wanted to be a teacher, but after watching you, it made me feel so dumb that I'm no longer sure I'm capable of it. I hope I can be as intelligent as you are one day.
STUDYING HARD NOW, HE'S GOTTA RECLAIM ALL OF THE MATH THAT ESCAPED FROM HIS BRAIN INTEGRALS, DERIVATIVES AND PARTIAL FRACTIONS HE'S MASTERED THEM ALL NOW HE'S READY FOR ACTIONNNN MATH-E-MATIC! AUT-O-MATIC! FEELS LIKE MAGIC WHEN YOU DERIVE JUST TO STAY ALIVEEEEE! MATH-E-MATIC! AUT-O-MATIC! MATH-E-MATIC! AUT-O-MATIC! *music fades out as you ride off on a motorcycle*
Just took my Calculus CLEP exam and got the score I needed; 64 ain't great, but i get my credit so it's cool. I've never been confident in Math, and quite frankly I went into the exam expecting to fail- I'm quite happy that wasn't the case. Im very grateful for all the videos you make, and passing this exam was definitely thanks to you lol. Keep it up!
Can you please make a video that introduces hyperbolic functions and some sample problems with them? I’ve never known anything about them and would love to learn more about them!
here is a challenge: find f(x) such that f'(x) = f(x+1) oh and i forgot something: there is a x such that f(x) = 1 (just multiply the solution by some number to get f(x) = 1 if you don't have it already). This is just to exclude f(x) = 0
The partial fraction decomposition can be done without a system of equations if you consider that 1 = 1/(2*sqrt(6))*[(x^2+sqrt(6))-(x^2-sqrt(6))]. This improved memory efficiency allowed me do the problem in my head reasonably fast. Also, BPRP was running out of room and the video was already long so he used hyperbolic trig functions. But you can decompose 1/(x^2-sqrt(6)) with the same trick or Heaviside cover up.
I’m 41 and just learning calculus. I just sat there with wide eyes and all shocked at how much work it’s needed. Damn. I think my brain almost literally exploded. Lol.
bro hes getting serious with these problems. I love your videos BPRP. Keep up these amazing videos. Im a high school student in AP Calculus AB. You help me alot with studying! Thanks man!
The final answer can be further simplified into: -[(∜6)/2]coth⁻¹[x/(∜6)] - [(∜6)/2]tan⁻¹[x/(∜6)] + C however, what if the question is: (Without the restrictions/without the previous infinite series question) ∫[6/(x⁴-6)]dx for -∜6 < x < ∜6 Since coth⁻¹[x/(∜6)] is not defined for -∜6 < x < ∜6, what expression would it be? !
A young man as cool as a Shaolin monk with a Kung Fu Master beard and has a Pokemon ball in his hand explaining math problems... This combination made my yung hollow mind literally explode.
You could have added the dx in the original problem. It was probably written in white. Excellent problem. I can solve it today, thanks to what I’ve learned from you over the past few years. I may have told you before that as a physicist, my first year of calculus was an “honors” class where all we did was proofs. The only think I really loved about that class was that we used Apostle and I fell in love with proofs by induction.
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 how I can solve this Question
At first, this question seemed pretty daunting but when I started solving I thought this question was really simple. I solved inside equations pretty quickly but when I got to the integration I was like "bruh f*ck this shit". In my opinion, the integration isn't difficult, it's just very lengthy.
Hello sir. A friend of mine got an equation he couldn't respond in an exam and i'm sure the solution isn't that hard. It's cos(z^2)=-4 Thank you if you can respond this one 😁
Wolfram actually gives me a solution: x = -2 + exp( ( log( t /(t^2 +2) ) *( 1 - t^2 ))/(t^2+1) + log(2)*(1 - t^2) /(t^2 +1) ) where t:=tan(x/2) and a numeric value of x=2.66536 + 2*n*pi .... maybe because of this video Wolfram updated their algorithm. (it's 6th December now and vid was posted 26th Nov )
I may be wrong with the following thought, but here goes: if inside a fraction, both numerator and denominator have the same sign - in our case, both are positive - its absolute value is the fraction itself. Which means that |6/(x^4)|sqrt_4(6), and not |x|>sqrt_4(6).
@@Freddy-pp4nc because they are saying to. That is the structure of the question. If it is in first principles form, you need to do it as such. If it's in d/dx form, you can just derive it. With first principles at least you know where you're working to, so it isn't completely abstract algebra.
I think there is an error in the last expression. The red part in factor of coth-1 and tan-1 should be (root4(6))^2 instead of just root4(6). So after simplification you get (-1/2)coth-1(x/root4(6)) - (1/2)tan-1(x/root4(6)) + c
I feel he did the partial integral in a way that was more work, but it could be the only way. I thought we could just do partial and take g’ = 6, g = 6x, h= (x^4 - 6)^-1, h’ = 0.25ln(x^4 - 6) and the do hg -integral(h’g) for the answer. Why wouldn’t this work?
For the first step, the inside of the limit, Couldn't you have used l'hopital's rule to do it quicker? The outcome matches what you got, but I was unsure if the methodology holds up the same
Why do americans/english people always want to use l'hôpital's rule ??? It's not that efficient, and here, recognizing that you have the definition of the derivative is much quicker. In general, taylor expansions or equivalents are much more useful than l'hôpital's rule, which is quite outdated. I really don't understand why some people only learn to do limits using this rule.
Hey blackpenredpen, i have a problem ive tried to solve, but no answer yet. It goes like this LOGa(x) = a^x , you need to solve for an a value so tangency occurs (both equations kiss eachother) Ive got the value to be about 1.45, but its not satisfying to estimate it lol Thanks:)
It's okay. This was basically an entire calculus final exam crammed into a single question. You don't normally get this kind of thing in an actual class. Even extra credit questions usually aren't this hard.
I don't really know what your issue is, but I hope this helps. The series is a geometric series (of the form a^r, where |a| should be less than 1 for the series to converge. If the series starts from n=1, then the value of the series is a/(1-a). If you're confused about the inequality, then it's 6/|x^4|< 1, 61). Hope this helps.
I did the last integral by my own and I dunno how u get the secont part(arccoth). I used a formula for 1/2a*ln(x+a/x-a). if u still read this comment few months after answer please, am I right or I did a mistace i didn't noticed. Love U
Do you like my videos edited or unedited?
unedited
Sir do you conduct zoom meetings also.. please share to me also 😞😞😢😋😋
Un
Unedited
Unedited but u should speed up erasing and writing way too long things
You know things go serious if you see bprp using 5 colors on a question
Now he’s BPRPBPRPPP
@@Claymore_II i had a stroke guessing it xD
1. that was actually funny
2. you somehow have about twice as many likes as THE PINNED COMMENT BPRP MADE
HOW
@@ananas-nevermind As you can see, many other comments are hearted by bprp. I exchanged the heart for 500 likes.
jk lol 😂
@@Claymore_II bprpbpppgp
bp¹=black pen
rp=red pen
bp²=blue pen
pp=purple pen
gp=green pen
Imagine this question being on your final exam. You go through all of these steps and calculations just to forget the +C at the end which single-handedly causes you to fail the exam.
Bruh you forgor the dx
Bro if I am forgetting only the c part in this question...... Then I am sure.... I will be able to solve other questions easily 🤣🤣🤣🤣
It's called stuying ET (Elektrotechnic) and IT (Informationtechnic)...
Then you have a garbage math teacher/professor if your only mistake of forgetting the +C at the end causes you to fail the entire exam.
@@user-yr3uj6go8i all or nothing
Oh yeah it's just the inverse hyperbolic cotangent of x over the fourth root of 6, I could have guessed that
For sure🤣🤣❤️
It’s crazy the shit they make you remember in calc but then in any science they give formula sheets longer than the test
You forgot the plus c
This is completely off topic, but I think I came up with an interesting problem.
Given the recursive functions:
f(n) = af(n-1) + bg(n-1)
g(n) = ag(n-1) + bf(n-1)
Where a=0.5 and a+b=φ
Show that both f and g are fibbonacci sequences.
I'll put my solution at the end.
Stumbled upon this fact while trying to solve for g for any a or b... idk how others may attempt to solve it but I wrote them as a discrete time convolution. I got this equation for g(n):
g(n) = (b^2-a^2)g(n-2)+2ag(n-1)
Was trying to solve for f or g to write sin(nt) in terms of sin(t) and cos(t)... but accidentally made a mistake, one of the equations were supposed to add and the other was supposed to subtract the b term.... was doing this in the first place because, I was hoping the equation for sin(nt) would help solve the integral for (sin(Nt)cot(t/2)+cos(Nt))cos(nt) where N and n are two non-negative integers. Which I plugged into a couple integral calculators... and it didn't give me a solution.
When you evaluate that integral in respect to t from -π to π, and divide by 2π, its supposed to be an inverse discrete-time fourier transform that should equal 1 for all |n|
Challenge in Maths 🏆🏆🔥🔥
th-cam.com/video/VoZZetC5GfU/w-d-xo.html
All that effort for 12 likes lol
@@dizzyd7315 lol
@@dizzyd7315 Other people are illiterate.
He’s always energetic and I love it.
you can see the passion through his eyes
I’m a sophomore in high school and have no idea what he’s doing but I still watch a bunch of his videos because they are interesting and relaxing. Keep up the good work👍
Same, I really enjoy these kind of videos :D
i was in your position a couple years ago. i’m now in calc 2, watched this whole video and actually understood all of it, and i’m very proud of myself. keep studying and you’ll get there💪
I did the same thing, been learning from him for a while
13:19 LMFAO I FELT THAT “why am I doing this” 💀
yeah it sounds so genuine lol
You know it's gonna be hard when he doesn't recommend calc teachers to use this in a test at the start of the video
lol
@blackpenredpen ngl I'd pass out if I saw that question 😳😳
@@blackpenredpen Challenge in Maths 🏆🏆🔥🔥
th-cam.com/video/VoZZetC5GfU/w-d-xo.html
Wait. Where's the dx?
That's -100 score for you.
My first thought also.
@@xinpingdonohoe3978 same
Fermat : "I have discovered a truly marvelous proof of this, which this board is too narrow to contain."
blackpenredpen : "hold my beer"
"blackpenredpen"->"blackpenredpenbluepenpurplepengreenpen"
Peak character development ladies and gentlemen
then comes penbrow
Limits, derivatives and integration in one question. It should be called fundamental question of the calculus...Complete package.. Amazing
Don’t forget the sigma!
5 minutes for the limit, differentiation and series; and 13 minutes for integration... Well...
more like 5 minutes to do partial fractions and 3 minutes to do the integral
Idk if it’s easier but I factored out the 1/6 on the top and bottom from the beginning to lessen the burden of roots to get 1/((x^4/6)-1).
Heyy I solved this on my own before I went further into the video. I'm proud of myself 😍
It is necessary to replace the variable in time to avoid radicals.
∫6dx/(x^4-6)=6^(1/4)∫ dt/(t^4-1). t=x/6^(1/4).
Here the fraction under the integral sign is so simple that it can be decomposed into elementary
fractions without the method of indefinite coefficients.
1/(t^4-1)=(1/2) *{1/(t^2-1)-1/(t^2+1)}=(1/2)*{(1/2)[1/(t-1)-1/(t+1)]-1/(t^2+1)}=
= (1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)
6^(1/4)* ∫ [(1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)]dx=
= 6^(1/4)* {(1/4)*ln|t-1|-(1/4)*ln|t+1|-(1/2)* atan(t)} +C=
=(6^(1/4)/4)* ln|(t-1)/(t+1)|-(6^(1/4)/2)*atan (t) +C =
=(a/4)* ln|(x-a)/(x+a)|-(a/2)*atan (x/a) +C . a= 6^(1/4).
I've always wanted to be a teacher, but after watching you, it made me feel so dumb that I'm no longer sure I'm capable of it. I hope I can be as intelligent as you are one day.
Just keep practicing man. Everything comes on point to the ones who really know what they want. Just giving you some strength ❤️!
Remember that he’s been teaching for about a decade at a massive institution
Watching your videos, I actually started loving math !
Partial fractions could be used in a different way by adding and subtracting x^2 (x squared) and 6^(1/2) (square root of 6)
This made me realize how much I've forgotten about partial fraction decomposition 🥲 time for a study montage
I got all the way to the integral on my own and said "I do math for fun, but I draw the line at PFD."
STUDYING HARD NOW, HE'S GOTTA RECLAIM
ALL OF THE MATH THAT ESCAPED FROM HIS BRAIN
INTEGRALS, DERIVATIVES AND PARTIAL FRACTIONS
HE'S MASTERED THEM ALL NOW HE'S READY FOR ACTIONNNN
MATH-E-MATIC!
AUT-O-MATIC!
FEELS LIKE MAGIC WHEN YOU DERIVE
JUST TO STAY ALIVEEEEE!
MATH-E-MATIC!
AUT-O-MATIC!
MATH-E-MATIC!
AUT-O-MATIC!
*music fades out as you ride off on a motorcycle*
First time in history I'm seeing this guy forget the integration dx in the question itself.
I like these mega problems! They're great review.
Glad to hear : )
Just took my Calculus CLEP exam and got the score I needed; 64 ain't great, but i get my credit so it's cool. I've never been confident in Math, and quite frankly I went into the exam expecting to fail- I'm quite happy that wasn't the case. Im very grateful for all the videos you make, and passing this exam was definitely thanks to you lol. Keep it up!
Never heard of it until looking it up. Seems very silly they wouldn’t just give you the AP Exam.
Can you please make a video that introduces hyperbolic functions and some sample problems with them? I’ve never known anything about them and would love to learn more about them!
An old video here th-cam.com/video/aC5cYc7XhIs/w-d-xo.html
@@blackpenredpen thanks!!!
Wait until you hear about elliptic functions.
here is a challenge: find f(x) such that
f'(x) = f(x+1)
oh and i forgot something: there is a x such that f(x) = 1 (just multiply the solution by some number to get f(x) = 1 if you don't have it already). This is just to exclude f(x) = 0
Integration of 1/sin^7x + cos^7x possible
It makes me so happy to see someone this exited about maths! Keep up the entertaining content😁👍🏼!
The partial fraction decomposition can be done without a system of equations if you consider that 1 = 1/(2*sqrt(6))*[(x^2+sqrt(6))-(x^2-sqrt(6))]. This improved memory efficiency allowed me do the problem in my head reasonably fast. Also, BPRP was running out of room and the video was already long so he used hyperbolic trig functions. But you can decompose 1/(x^2-sqrt(6)) with the same trick or Heaviside cover up.
My thoughts exactly.
All i know is f(x)=x² so f'(x)=2x
Ah yes, blackpenredpenbluepengreenpenpurplepen
coefficients can be simplify to -(6)^(1/4)
I’m 41 and just learning calculus. I just sat there with wide eyes and all shocked at how much work it’s needed. Damn. I think my brain almost literally exploded. Lol.
bro hes getting serious with these problems. I love your videos BPRP. Keep up these amazing videos. Im a high school student in AP Calculus AB. You help me alot with studying! Thanks man!
Glad to hear it, thank you : )
@@blackpenredpen yessirrrr
I lost it at the integration part
Until then i was really happy that things were going smooth
Erase part of the board!!! Also at the end, you could simplify square root of 6 / (2 * 4th root of 6) to just (4th root of 6) / 2.
The final answer can be further simplified into:
-[(∜6)/2]coth⁻¹[x/(∜6)] - [(∜6)/2]tan⁻¹[x/(∜6)] + C
however, what if the question is: (Without the restrictions/without the previous infinite series question)
∫[6/(x⁴-6)]dx for -∜6 < x < ∜6
Since coth⁻¹[x/(∜6)] is not defined for -∜6 < x < ∜6, what expression would it be?
!
A young man as cool as a Shaolin monk with a Kung Fu Master beard and has a Pokemon ball in his hand explaining math problems... This combination made my yung hollow mind literally explode.
I really felt the "Why am I doing this" at 13:20 on deeper levels than is comprehensible.
Thank you I have tears of joy now
You could have added the dx in the original problem. It was probably written in white.
Excellent problem. I can solve it today, thanks to what I’ve learned from you over the past few years.
I may have told you before that as a physicist, my first year of calculus was an “honors” class where all we did was proofs. The only think I really loved about that class was that we used Apostle and I fell in love with proofs by induction.
Congratulations from a French student. 🇫🇷❤️
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 how I can solve this Question
These are the kinds of questions my General Chemistry I and II professor was fond of - the entire final exam in one question.
An absolute monster
This guy is gonna change my career!
Your videos are everything
13:26 “why am i doing this”. that’s how i feel when i do implicit differentiation
Love your content helps with my love of mathematics, regards
just learned what's the derivative of coth^-1. Great video !
At first, this question seemed pretty daunting but when I started solving I thought this question was really simple. I solved inside equations pretty quickly but when I got to the integration I was like "bruh f*ck this shit". In my opinion, the integration isn't difficult, it's just very lengthy.
Blackpenredpen Collab with purplepen greenpen and bluepen
I just realized how much more I have to learn
Hello sir.
A friend of mine got an equation he couldn't respond in an exam and i'm sure the solution isn't that hard.
It's cos(z^2)=-4
Thank you if you can respond this one 😁
The students: "don't worry he said he would put only one ex. on the exam"
The exercise: are you sure about that 😂😂🤣🤣
Hahaha!
More all one calculus questions thanks)
There's no dx at the end of the integral at the top :( xD
You're right.
It is incomplete
Wolfram actually gives me a solution: x = -2 + exp( ( log( t /(t^2 +2) ) *( 1 - t^2 ))/(t^2+1) + log(2)*(1 - t^2) /(t^2 +1) )
where t:=tan(x/2) and a numeric value of x=2.66536 + 2*n*pi .... maybe because of this video Wolfram updated their algorithm. (it's 6th December now and vid was posted 26th Nov )
Solved by me ❤
Date 17 April 9:29 AM
All trivial, the hardest part is the integral and that is taught in elementary
sqrt6/4thrt6 can be simplified to 4thrt6 but it’s trivial I guess
i never thought i would be so satisfied by an math equation :D
I am glad to hear that : )
I may be wrong with the following thought, but here goes: if inside a fraction, both numerator and denominator have the same sign - in our case, both are positive - its absolute value is the fraction itself. Which means that |6/(x^4)|sqrt_4(6), and not |x|>sqrt_4(6).
At 10:00 it would be better 6/√6 = √6 and then B=√6/2
the very first limit: isn't the definition of derivative?
derivative of x^(-2) is -2/x^3
Yes u are right. He mentioned it in the video
But when it's in first principles form, you should do it as an actual algebra question.
@@xinpingdonohoe3978 why, unless they say you have to why not save time
@@Freddy-pp4nc because they are saying to. That is the structure of the question. If it is in first principles form, you need to do it as such. If it's in d/dx form, you can just derive it. With first principles at least you know where you're working to, so it isn't completely abstract algebra.
Had to take calculus 1, 2 and 3 for my computer science degree (plus other maths), I Barely remember how to do derivatives.
its amazing sir.. your videos always enlighten me.....
I love how you look at it so proudly when you're done, like it's your child. 😂
😆
I think there is an error in the last expression. The red part in factor of coth-1 and tan-1 should be (root4(6))^2 instead of just root4(6).
So after simplification you get (-1/2)coth-1(x/root4(6)) - (1/2)tan-1(x/root4(6)) + c
Great content man. As always
06:32 It also can be nice way to use Heavyside cover up method, I guess. :)
I love this!
@blackpenredpen
I do this in my mind.
*Just swap integral symbol with sum symbol*
I feel he did the partial integral in a way that was more work, but it could be the only way. I thought we could just do partial and take g’ = 6, g = 6x, h= (x^4 - 6)^-1, h’ = 0.25ln(x^4 - 6) and the do hg -integral(h’g) for the answer. Why wouldn’t this work?
I would just chose c and hope the next quistion won't be hard
13:19 - My reaction when it's 4 am and I'm only watching these videos
If you have 1/(0/1), would that be undefined, or would it be 0 because (a/b)/(c/d)=(a/b)×(d/c) meaning 1×(0/1)=0
Couldn't you have simplified the answer since √6 / ∜6 = 6^(1/2) / 6^(1/4) = 6^(1/2 - 1/4) = 6^(1/4) = ∜6 ?
I think this should be on your other channel called blackpenbluepengreenpenpurplepenredpen.
😆
You are a madman!
I was following along till integration (partial fractions) and then I lost you when you were talking about hyberbolic functions 🤯😅
For the first step, the inside of the limit, Couldn't you have used l'hopital's rule to do it quicker? The outcome matches what you got, but I was unsure if the methodology holds up the same
Why do americans/english people always want to use l'hôpital's rule ??? It's not that efficient, and here, recognizing that you have the definition of the derivative is much quicker. In general, taylor expansions or equivalents are much more useful than l'hôpital's rule, which is quite outdated. I really don't understand why some people only learn to do limits using this rule.
Somebody get this man a bigger whiteboard! xD
Your awsome Professor)
Is this ASMR? I get the feels just by watching this
Hey blackpenredpen, i have a problem ive tried to solve, but no answer yet. It goes like this
LOGa(x) = a^x , you need to solve for an a value so tangency occurs (both equations kiss eachother)
Ive got the value to be about 1.45, but its not satisfying to estimate it lol
Thanks:)
You're missing a dx after the integral as soon as it starts
13:19
6/x^3
Math is my current favorite and best subject, I’m currently in precalc on track to take Ap calc bc next year. I am now very scared of calc lol
It's okay. This was basically an entire calculus final exam crammed into a single question. You don't normally get this kind of thing in an actual class. Even extra credit questions usually aren't this hard.
Wait, when solving the Σ, how did x⁴ leave since the fraction is (6/x⁴)'? Shouldn't it be something else?
he says it's raised to the FIRST POWER, so it's just ^1
I don't understand your issue with the series. What's wrong with him answer?
I don't really know what your issue is, but I hope this helps.
The series is a geometric series (of the form a^r, where |a| should be less than 1 for the series to converge. If the series starts from n=1, then the value of the series is a/(1-a).
If you're confused about the inequality, then it's 6/|x^4|< 1,
61).
Hope this helps.
apart from remembering the +c, the dx is important too.
next time incorporate Reimann sum or Trapezoid sum and FWD difference operator
Ah yes, my favorite episode of blackpenredpenbluepengreenpenpurplepen
There are two kind of people in this world: 1) +-3/sqrt(6) & 2) +-sqrt(6)/2.
on todays episode of blackpenredpenbluepengreenpenpurplepen
I did the last integral by my own and I dunno how u get the secont part(arccoth). I used a formula for 1/2a*ln(x+a/x-a). if u still read this comment few months after answer please, am I right or I did a mistace i didn't noticed. Love U
It's nearly 2am and I've been up for over an hour trying to do this. I lay restless in bed thinking about this question
Awosome
the "why am i doing this" is so real
wait wait wait wait .......what, problem is solved .....It was looking terrific