Amazing Solution of Difficult JEE Main Problem Based on Integration

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BHANNAT MATHS

BHANNAT MATHS

Күн бұрын

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In this video we will be solving a difficult integration problem from JEE Main 2022
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#bhannatmaths #jeepyq #integration #jee2024 #jee2023

Пікірлер: 126
@tanishqgoyal7180
@tanishqgoyal7180 Жыл бұрын
Being a JEE aspirant ,in Aman Sir's comment's section lot of students talk about concepts which we have not even heard, like Lambert W function etc.
@vibhashrivastava8846
@vibhashrivastava8846 Жыл бұрын
If u have any source or information about it, can u share it. I would like to learn too
@Balls99987
@Balls99987 Жыл бұрын
Inverse of x*e^x​@@vibhashrivastava8846
@Balls99987
@Balls99987 Жыл бұрын
@@vibhashrivastava8846 inverse of x*e^x
@TheHellBoy05
@TheHellBoy05 Жыл бұрын
@@vibhashrivastava8846 i talked about it once. here is a small explanation set f(x)=xe^x then f^-1(x)=W(x) so basically foW(x)=Wof(x)=x thus w(x)e^w(x)=x and W(xe^x)=x it is used to solve equations where the variable is exponentiated as well linear. like 2^a + a=5 inf series expansion is W(x)= ∑(-n)^(n-1)x^n/n! n=1
@adityajha2889
@adityajha2889 Жыл бұрын
Yeah
@MrAniruddhajog
@MrAniruddhajog 10 ай бұрын
I am a 31 year old doctor who has not solved mathematics since 2008. But this is putting fun in my otherwise boring professional life!
@karthikmalasani
@karthikmalasani Жыл бұрын
This can also be done with simple integration by observing that the messiest one to integrate is sqrt(1-y^2) which luckily is present in all options and in LHS and RHS. It is the same trick that led you to get a simple solution using geometry! Had that part not been common it would have been messy using geometry or otherwise! Left hand side: First part: integral(sqrt(2x)) dx|0 to 2 = 2sqrt(2)/3*x(3/2)|0 to 2 = 8/3. Second one: integral (sqrt(2x-x^2)) dx|0 to 2 = integral(sqrt(1-(x-1)^2)) dx|0 to 2 = integral(sqrt(1-y^2))dy|-1 to 1 We notice that the same integral is present on right side with limits 0 to 1. Add it to both sides makes LHS: 8/3 - integral(sqrt(1-y^2))dy|-1 to 0 = 8/3 - integral(sqrt(1-y^2)) dy | 0 to 1 (y = -x substitution followed by flipping limits and substitute x = y) Right hand side becomes: integral(1)dy|0 to 1 + integral(2)dy|1 to 2 - integral(y^2/2)dy|0 to 2 + I = 1 + 2 - 4/3 + I = 5/3 + I => I = 8/3 - 5/3 - integral(sqrt(1-y^2)) dy | 0 to 1 => I = integral(1 - sqrt(1-y^2)) dy | 0 to 1 Option C.
@shailnair2243
@shailnair2243 Жыл бұрын
Yeah thats something that will click in exam rather than going for graphs for me. Nice solution bro. Keep it up 👍👍👍
@aimanfatima6292
@aimanfatima6292 10 ай бұрын
Hello...Your method is amazing.. I have a small doubt .. You mentioned (in the 3rd paragraph )that integration -1to 0 √(1-y²)dy can be converted to integration 0 to 1√(1-y²)dy ... Can u please explain this step ..I mean how the limits are flipped without changing the expression...If it's some rule or property, please tell me it's name so that I can learn about it.
@JEE-oq1me
@JEE-oq1me 9 ай бұрын
im not reading all that but congratulations
@Balls99987
@Balls99987 3 ай бұрын
​@@aimanfatima6292Put y=-y
@kdhd100
@kdhd100 11 ай бұрын
The toughest part in this is to figure out areas along the y- axis. We usually don't do it along Y-axis, so understanding is difficult.
@LDR_Crafts
@LDR_Crafts Жыл бұрын
Problem: lim n infty 1 2^ n ( 1 sqrt(1 - 1/(2 ^ n)) + 1/(sqrt(1 - 2/(2 ^ n))) + 1/(sqrt(1 - 3/(2 ^ n))) +......+ 1 sqrt 1- 2^ n -1 2^ n ) is equal to -
@Balls99987
@Balls99987 Жыл бұрын
Put 2^n=t and proceed As 2^n->inf t->inf
@LDR_Crafts
@LDR_Crafts Жыл бұрын
@@Balls99987 thank you for suggesting a approach 🤗
@kdhd100
@kdhd100 11 ай бұрын
The problem is based on an old Russian book of Engineering Maths and the topic is double integral..... The same graph is shown.....
@ashis.di.sardar
@ashis.di.sardar 11 ай бұрын
The Way you teach Maths, wish you 1M Subs Fam soon!! Sir
@4m0d
@4m0d Жыл бұрын
beautiful observations sir, but its very difficult to think all this in exam
@danishhsable
@danishhsable Жыл бұрын
Tabhi toh maths ka difficulty sabse high rehta hai mere bhai
@shailnair2243
@shailnair2243 Жыл бұрын
Very very true
@vedantpatil8108
@vedantpatil8108 Жыл бұрын
Sir honestly bata ra hu mene bhi area ke concept se socha tha but me kuch miss kar raha tha mere soution me so answer match nahi kr raha tha aur sabhi books me standard meathod diya thai jisme bas integrate karke compare kiya hai, aaj finally jake mughe meri mistake ka pata chala jo me kr raha tha so THANK YOU SO MUCH SIR for this amazing solution😍
@shaileshs.3177
@shaileshs.3177 Жыл бұрын
Beautiful solution Aman Sir ! 👏👏👏
@AngrezLal-d5g
@AngrezLal-d5g Жыл бұрын
Gajab 🛐🛐🛐
@varshinilolla3090
@varshinilolla3090 Ай бұрын
8:00 Yaha pe the semi circle 1-√ (1-y^2) =x neeche tak extend hota hain(in the negative y region), lekin since our integral is bounded from [0,1] we didn't consider that section. Or to put it in other words, the equations of semi circles that we obtained by solving for x and y individually (-------> can be given as 4 different yet overlapping semi circles) is nothing but the equation of circle "restricted to a specific domain." And quite obviously, when we square it we are going to get the circle equation back. (algebraically, we get the "additional root"). Thanks for the video sir!!
@Shogun507
@Shogun507 11 ай бұрын
Absolutely genius! Only problem is, one mostly can't think in this way while giving the paper.
@shivamgupta9329
@shivamgupta9329 11 ай бұрын
Amazing explanation ❤❤
@IITIAN-f1z
@IITIAN-f1z 6 ай бұрын
gajab sir aapratim
@osowiecwalking9434
@osowiecwalking9434 Жыл бұрын
sir we can also transform root 2x-x2 to root 1-(x-1)^2 and by putting x-1= z it will be 2 * intreagtion 0 to 1 root 1-z^2 . now we can balance this above integral and match the constants on both side. I am getting C too
@omkumarsingh7
@omkumarsingh7 Жыл бұрын
Wow this is a very beautifully constructed Problem 🗿
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
Why u use only half curve can anyone explain
@omkumarsingh7
@omkumarsingh7 Жыл бұрын
​​@@Harshitkumar-cq1vb you mean y=√(2π) right ? If this is your problem then I will just say that since the values inside the sqrt should not be -ve that is 2π y also be +ve . The other half curve is y=-ve so that is why it doesn't exist in the coordinate plane it only will only exit in Argand Plane .
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
@@omkumarsingh7 y2= 4ax graph to negative mebhi jata hai coordinate plane pr
@omkumarsingh7
@omkumarsingh7 Жыл бұрын
@@Harshitkumar-cq1vb yes bro y² = 4ax jata hai negative but bro sawaal me hamme y = √(4ax) ke format me tha naaki y² = 4ax Agar abhi bhi clear na ho toh aap Google pe jaake ek yeh dono graph banana .
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
@@omkumarsingh7 sorry to disturb yoy but what is difference between y, 2 4ax and y =underroot 4ax
@Happy-uh7en
@Happy-uh7en Жыл бұрын
Sir you are the best in the world
@mgkkhan5891
@mgkkhan5891 5 ай бұрын
Sir please please make a video of this solving all these integral
@preetib6819
@preetib6819 Жыл бұрын
Good explanation
@omkamble5366
@omkamble5366 Жыл бұрын
Bhai dekh toh le
@abbeydudeuk
@abbeydudeuk 20 күн бұрын
Amazing
@ShivamBhalerao-i9j
@ShivamBhalerao-i9j Жыл бұрын
Really amazing question and solution
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
Why u use only half curve can anyone explain
@ShivamBhalerao-i9j
@ShivamBhalerao-i9j Жыл бұрын
@@Harshitkumar-cq1vb bhai x ki limits toh dekh
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
@@ShivamBhalerao-i9j bhai sun or reply jarur karna y2= 4ax ka graph x +ve axis ke upar bhi hota h Or niche bhi sir ne only upar wala curve liya or aap isla explanation dere ho ki limit dekho to bhai agar x + axis ke niche wale graph ko bhi liya jata tab limit kya hoti?
@ShivamBhalerao-i9j
@ShivamBhalerao-i9j Жыл бұрын
Bro my bad, original equation is y=root2x toh Y ka range toh 0 to infinity he Hoga na
@ShivamBhalerao-i9j
@ShivamBhalerao-i9j Жыл бұрын
Jab hum y^2=4ax me y=root4ax X-aixs ke upar wale portion ka equation Hota Hai and y=-root4ax X-axis ke niche wale portion ka equation
@aa01010
@aa01010 Жыл бұрын
Sir, amazing question and way to crack. i could never ever solve it this way. AMAZING
@mdasifeqbal2323
@mdasifeqbal2323 Жыл бұрын
Extremely complex question and tough solution.
@monujhembrom9279
@monujhembrom9279 Жыл бұрын
U r bsc student
@mdasifeqbal2323
@mdasifeqbal2323 Жыл бұрын
@@monujhembrom9279 I am a graduate engineer.
@itserfan07
@itserfan07 Жыл бұрын
amazing method sir ❤❤
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
Why u use only half curve can anyone explain
@shivanshnigam4015
@shivanshnigam4015 Жыл бұрын
Sirji itna analysis me 5min se jyada hi lagega pakka
@shailnair2243
@shailnair2243 Жыл бұрын
Sahi bole bhai ☺️☺️
@mathist-monster101
@mathist-monster101 Жыл бұрын
Wow sir, what an outstanding process of thinking !
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
Why u use only half curve can anyone explain
@smarter4163
@smarter4163 11 ай бұрын
​@@Harshitkumar-cq1vbbecause limit is from 1 to 2 😊 not from 0
@criticgamerz6382
@criticgamerz6382 Жыл бұрын
Indeed a good question ❤ , but had a different approach.
@kalindisingh2317
@kalindisingh2317 Жыл бұрын
Well anlyllisis sr ji
@afridbasha7547
@afridbasha7547 Жыл бұрын
Amazing explanation sir
@parthhooda3713
@parthhooda3713 3 ай бұрын
took me 3:06 to solve this question (by integrating) at home (in exam hall it would take much more)
@krrishrohilla2945
@krrishrohilla2945 Жыл бұрын
4:34 Sir yhi to problem hai Jab ham graph bnate hai to ye darr rheta hai kitni bar intersect karega aur kha par karega Aur aap phele se hi soch kar 1 sec mein bol gye Sir muje sabhi equations ke graph pta hai par aise questions par lga nhi pata Pls make a video on this problem🙏🙏🙏 Aur sir video mein ache se explain karna bhale hi 3 hr ki ho.😊
@noname-sl3to
@noname-sl3to 11 ай бұрын
Ek book aati play with graphs..... arihant ki shayad karlo tagdi hai
@BilalKhan-cq6mn
@BilalKhan-cq6mn 7 ай бұрын
Sir yaha pe sirf 3 inetegration karna hai Sqrt(2x) Sqrt(2x-x²) Sqrt(1-y²) and y²/2 Fir inki values ko har jagah daal dena hai. And not 7 integrals to be solved
@Rochak_Jain
@Rochak_Jain Жыл бұрын
Omg! Amazing problem and it's solution
@p.msaini7515
@p.msaini7515 Жыл бұрын
Sir pace series is best . I hope app shabi log ache hoge m bhi hu ekdham bhannat
@dhananjaykushwaha5783
@dhananjaykushwaha5783 Жыл бұрын
Sir may you create a video on maclaurin series expansion ?
@abcdWxyz-xm5ht
@abcdWxyz-xm5ht Жыл бұрын
Sir kya aap IIT jam ki preparation krate hai
@rishabhjain5849
@rishabhjain5849 Жыл бұрын
Phir bhi sir 4-5 min to lg hi jayenge scratch se aise soch kar solve krne mein kha 2.5 min mein hoga ye
@abhirupkundu2778
@abhirupkundu2778 Жыл бұрын
That is why skip such insane questions and attempt them at the end if any time is left.
@Siraj-123-q5p
@Siraj-123-q5p 5 ай бұрын
Sir is explaining thats why for sir it took so much time. In exam hall you dont have to explain everything you are doing, so your speed will be much faster.
@abhishekbaghel1979
@abhishekbaghel1979 6 күн бұрын
@imperfectness4
@imperfectness4 Жыл бұрын
Like for this LEGEND MATHEMAGICIAN ❤😊
@nikhilbth018
@nikhilbth018 Жыл бұрын
Sachin Sir ne kraya tha isse class me, Area se hi 😃
@Harshitkumar-cq1vb
@Harshitkumar-cq1vb Жыл бұрын
Why u use only half curve can anyone explain
@Retarc-me7ez
@Retarc-me7ez Жыл бұрын
Itne hard integral nhi hai agar dekha jae tho √1-y² ka integral y/2√1-y² + a²/2sin-1(y/a) se likha ja skta hai and jo limits di hai since wo kafi asan hai tho jab 0 and 1 put kroge tho simpy π/4 aajaega orr 2x-x2 wali ko bhi king rule lga kr 2×int(√1-x²)0to1 likha ja skta hai. Ya phir ek orr Kam kr skte hai 2x-x² wali term ko 2×int(√1-x²)0to1 likh kr isse L bol do orr uss trf wali 1-y² wali bhi L ho jaegi orr ab I ko L ke terms mein solve krke L mein hi answer compare krlo Graphs wala method unique hai and most probably or kisi easy ko bohot aasan bna skta hai but uss wale question ke liye yeh second wali approach (L assume krne walo) better hai
@5Stars49
@5Stars49 Жыл бұрын
🎉❤
@susmitdas771
@susmitdas771 11 ай бұрын
😮😮
@iconicsolos1234
@iconicsolos1234 11 ай бұрын
Sir mujhe lagta hai ye solution integrals solve karne se zyada complex way hai k
@BeltekTv
@BeltekTv Жыл бұрын
Sir. e^x² ka integration batana
@j.u.4.n620
@j.u.4.n620 Жыл бұрын
This is non integrable in indefinite But most of definite integrals can be solved by laplace transformation
@Balls99987
@Balls99987 Жыл бұрын
@@j.u.4.n620 can be solved only using polar coordinates
@shauryapratapdwiwedi8892
@shauryapratapdwiwedi8892 4 ай бұрын
i was one of them who got right ans.
@surendrarajpootrajpoot7399
@surendrarajpootrajpoot7399 4 ай бұрын
Bsc 1st year math mathod book integral calculus
@Raj-xt4fk
@Raj-xt4fk Жыл бұрын
Are bhai definite integration solve karo fir ans match karo Simple
@AvikGupta-ch7oh
@AvikGupta-ch7oh Жыл бұрын
Ha do fir 10 min
@Raj-xt4fk
@Raj-xt4fk Жыл бұрын
@@AvikGupta-ch7oh Bhai pagal hai kya, ye integration solve karne me 2 min bhi ni lagte. Agar tujhe 10 min lag raha hai, to sorry u need more practice
@shivanshnigam4015
@shivanshnigam4015 Жыл бұрын
@@Raj-xt4fk timer lagake 7 integrals solve kariye aap, 2 min se neeche ya uske aas paas bhi nahi ayega
@Raj-xt4fk
@Raj-xt4fk Жыл бұрын
​@@shivanshnigam4015 bhai tu meri baat sun. Wo jo root 2x - x2 wala integral hai, aur wo jo root 1- y2 wala hai, wo dono ekdam same hai. Rhs wale ko lhs mei bhej aur uska limit 0 se 1 ho jayga. Fir mentally trig sub karke kar le. Aur wo wala y2/2 wala to lulla hai. Ye sab to ho hi jaata hai.
@shivanshnigam4015
@shivanshnigam4015 Жыл бұрын
@@Raj-xt4fk trig sub ki jaruat nahi hai vo sqrt(1-y^2) wala sab jagah hai to usse side me rakh ke bhi kiya jaa sakta hai, lekin poora poora solve karne me toh thoda time lagega hi
@jayakchhag8161
@jayakchhag8161 Жыл бұрын
(SIR PLEASE READ THIS BY THIS WE COULD SOLVE IT IN 2MIN 40SEC) SIMPLEST WAY U WILL FIND IN THIS COMMENT SECTION Apply Queens property in LHS 2nd integral(splitting integral) in root 2x - x square Apply Kings propery in second integral of RHS ( splitting integral) in root 1-y square This can also be done with simple integration by observing that the messiest one to integrate is sqrt(1-y^2) which luckily is present in all options and in LHS and RHS. It is the same trick that led you to get a simple solution using geometry! Had that part not been common it would have been messy using geometry or otherwise! Left hand side: First part: integral (sqrt(2x)) dx|0 to 2=2sqrt(2)/3^ * * (3/2) / 0 to 2 = 8/3 . Second one: integral (sqrt(2x - x ^ 2)) dx|0 to 2 = integrate (sqrt(1 - (x - 1) ^ 2)) dx d * 10 to 2 = integrate (sqrt(1 - y ^ 2)) dy |-1 to 1 We notice that the same integral is present on right side with limits 0 to 1. Add it to both sides makes LHS: 8/3 - integrate (sqrt(1 - y ^ 2)) dy |-1 to c = 8/3 - integrate (sqrt(1 - y ^ 2)) dy dy | 0 to 1 ( y = - x substitution followed by flipping limits and substitute x = y) Right hand side becomes: integral(1)dy|0 to 1 + integrate (2) dy / 1 to 2- integral (y ^ 2 / 2) dy10 to 2 + 1 = 1 + 2 - 4/3 + 1 = 5/3 + 1 => I = 8/3 - 5/3 - integral (sqrt(1-y^2)) dy | 0 to 1 => 1 = integrate (1 - sqrt(1 - y ^ 2)) dy dy | 0 to 1 Option C.
@Balls99987
@Balls99987 Жыл бұрын
Paper khatam ,tan tan.
@shivanshnigam4015
@shivanshnigam4015 Жыл бұрын
Gajab Bhai, maine kiya tumhare method se 4 min se andar me hogaya 👌👌👍👍
@annapurnasarangi9999
@annapurnasarangi9999 Жыл бұрын
Did the same method, got the ans in 1min
@aimanfatima6292
@aimanfatima6292 10 ай бұрын
Hello...Your method is amazing.. I have a small doubt .. You mentioned (in the last 3rd paragraph )that integration -1to 0 √(1-y²)dy can be converted to integration 0 to 1√(1-y²)dy ... Can u please explain this step ..I mean how the limits are flipped without changing the expression...If it's some rule or property, please tell me it's name so that I can learn about it.
@abhinavrawat6632
@abhinavrawat6632 10 ай бұрын
​​@@aimanfatima6292being an even function it gives same area from -a to 0 as 0 to a
@Aman_Dixit.
@Aman_Dixit. Жыл бұрын
Differentiate karke check kar sakte kya
@SIVA_GANESH.B
@SIVA_GANESH.B Жыл бұрын
U can't differentiate becoz both are constants (area)
@DistinguishedGentleman1554
@DistinguishedGentleman1554 Жыл бұрын
Definite integral hai bhai differentiate karne pe zero aaega
@anvitmagadum7003
@anvitmagadum7003 7 ай бұрын
Tough question
@AlakMandal-di3ql
@AlakMandal-di3ql Жыл бұрын
Sir linear /linear ka graph ma ek video la aya.A sab student ka biniti ha please 🙏😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭
@debmalyadas7675
@debmalyadas7675 Жыл бұрын
Bhai asan hota hai Aod padhoge to khud banaloge
@ABH656
@ABH656 Жыл бұрын
Bhai simply itna yaad rakho range uska all real except coff of x in Nr / coff of x in Dr
@vedantsingh9156
@vedantsingh9156 Жыл бұрын
😶🫡🔥
@Poonamjidevi002
@Poonamjidevi002 Жыл бұрын
Eighth
@Mathematicsnepal05
@Mathematicsnepal05 Жыл бұрын
8th audience 😂
@harshitpandya3276
@harshitpandya3276 Жыл бұрын
Jaldi se comment kar deta hu first😂
@rudrathakur6253
@rudrathakur6253 Жыл бұрын
Matlab khud solve 11 min me karo aur bate 2-3 min me solve karne ki 😅
@navyaaru6295
@navyaaru6295 Жыл бұрын
Samjha bhi toh rahe hai hai bhai😅 aur starting ka samay bhi toh gaya tha iss question ke discussion mai related to how many students solved it and all. Tu karde isko part by part solve krke exam mai
@liopard8357
@liopard8357 Жыл бұрын
Aagar Bina intro aur Bina samjhaya hua solve karange to sir isko bus 2 minutes me hi kar denga
@iconicsolos1234
@iconicsolos1234 11 ай бұрын
Samjha bhi to rahe hai sir, waise sir ko ye question solve karne me in reality ek minute bhi nahi lagega
@Shogun507
@Shogun507 11 ай бұрын
He's also explaining. He could've solved it in less than 2 minutes since he knows the logic.
@Daksh-ek5ne
@Daksh-ek5ne 9 ай бұрын
​@@navyaaru62950:11
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