Sorry everyone! At 2:21 I overlooked parts as a possible solution method. You can solve with parts using the recursive method.

@mardinjordan8642

Ай бұрын

Can you please make a video on how to do it am extremely struggling on it

@devsehgal7432

25 күн бұрын

@@mardinjordan8642 Don't worry about it too much this is an extreme example

you can do it by parts, it gives that the initial integral is x(3^lnx) -(ln3)(the integral) you just add it to the other side and then divide by 1+ln3

adw1z's simplification is best, but requires some down-field vision. the usub in the video is the best obvious approach, but e^u*3^u you want to convert the 3 to e^(log3), and then you can combine. technically you get another u-type substitution, but a linear one so you can just "reverse chain rule" it

Learning maths from you from Sydney Australia 🇦🇺 Great work!

beautiful question

@MrAstburyMaths

Ай бұрын

Thanks

the better method is that x^ln3 = 3^lnx. So its mere instant, obviously its not allowed in the q but yeah.

@adw1z

Ай бұрын

u can still cheese it by doing the u = ln(x) sub, x^(ln3) dx = e^(u(1 + ln(3))) du, its just as easy and saves lots of time

A very quick trick is to notice: 3^ln(x) = exp(ln3)^ln(x) = exp(ln(x))^ln(3) = x^(ln3), which integrates to: (ln(3) + 1)^-1 x^(ln(3) + 1) + C = x* 3^ln(x) / (ln(3) + 1) + C Edit: I just read now, u need a ln(x) sub, which is very annoying. But it makes life much easier using the above method still: u = lnx so e^u = x, e^u du = dx, and then integrate e^(uln(3))e^u du = e^(u(1+ln3)) du, which becomes: 1/(1+ln3) e^(u(1+ln3)) + C = x* 3^ln(x) / (ln(3) + 1) + C Avoids any sort of parts or partial differentiation for 7 marks :)

@memelord_699

Ай бұрын

Yep, did it this way pretty quickly.

@sikf

Ай бұрын

That’s genius 👏🏿

@spencergee6948

22 күн бұрын

Very good but the question asks us to use the substitution u = lnx. If we didn't do that would we be marked down?

@adw1z

22 күн бұрын

@@spencergee6948 I wouldn't risk it, so I'd try and use a u = ln(x) sub wherever possible. Even in the shortcut method u can use it, which saves a lot of time and then u are guaranteed all the marks

where do you find these questions? I saw some people mention madasmaths but I can't find these questions. Can you please send me the link?

i got up to (3e)^u and got ln(3e) * (3e)^u. from there i ended up with 3^(ln(x)) * x(1 + ln(3)) + c, but the 1+ln(3) is meant to be the denominator so i just divided everything by (1 + ln(3))^2 and got (3^(ln(x)))/(1 + ln(3)) + c/((1 + ln(3))^2) and since c/((1+ln(3))^2) is just a constant i just renamed it and turned it into + C on its own. that got me the thing u had to show. would that work or is that the wrong method?

recursive method is perfect for this

Sir at 4:42 why did you consider to differentiate 1/lna. Where did this came from?

@fried_noodless

Ай бұрын

I think it's because integration and differentiation are inverse functions, and when you integrate a function you multiply by 1 over the function. Not entirely sure though.

I was astounded by your knowledge in the missing xi shorts, so here’s a harder one: Which club links: -Dean Moxey -Adam Randell -Mike Williamson

@MrAstburyMaths

Ай бұрын

That’s mighty tough! I don’t even know who the 2 top two are. I know Williamson played for Newcastle so I’ll guess them!

@CounterThePress

Ай бұрын

@@MrAstburyMaths Dean Moxey played at Crystal Palace while Randell plays for Plymouth after a loan spell at this club. Mike Williamson came through their academy and Neil Warnock now acts as an advisor. It was Torquay united, tough one to be fair, had to research a couple of them

Isn’t 3^lnx, equal to (e^ln3)^lnx, = (e^lnx)^ln3 = x^ln3 So does it not just integrate to (x^(ln3 + 1))/(ln3 + 1)

@ishaandas5273

Ай бұрын

Genius 😳

@Chakamatics

26 күн бұрын

That's correct but the question said u need to use the substitution method

Easy🤣love how easy high school is compared to stuff later down the road

By parts work well

@Fate212

Ай бұрын

yeah you jus gotta rearrange the parts

@SmellyPoop-mm4ov

Ай бұрын

@@Fate212 fr why is bro tripping its easy

@jamal8917

Ай бұрын

fr its quicker than this way

@MrAstburyMaths

Ай бұрын

Yes sorry my mistake, parts does work well here

@MrAstburyMaths

Ай бұрын

Sorry for tripping, my mistake, parts does work well here

Why can't we use e = pi = 3 ? Or is integration by engineers not allowed?

where is this question from? Im trying to find out how to do this by parts thanks 😃

@MrAstburyMaths

Ай бұрын

Madas SYN Paper W

How did you know to use e^u?

@afernz4999

Ай бұрын

After differentiating, try sub in x*du = dx, you will get get 3^u * x du which has the integral in terms of u and x, so you would need to get x in terms of u, which would be e^u. Sir here has just instantly put it to e^u before x as usually these types of problems require you to do that, but not all of course.

Use reduction formula

May i ask, how do you know that by part swill be trickier?

@MrAstburyMaths

Ай бұрын

Sorry my mistake, parts does work well here.

@trqpurpz7402

Ай бұрын

You can’t do this by parts, I tried it and it puts you into an infinite loop- to show what I mean, Try using parts on ∫3^u e^u du , you should find that within the very first part, the repeated integral step in the formula is the exact same; hence, an infinite loop

@MrAstburyMaths

Ай бұрын

@@trqpurpz7402 You need to know the recursive step for integration by parts in order for it to work. This type of step is not on the Edexcel Spec. That's why I didn't think about it during the video. It does work though!

@s6leh

Ай бұрын

@@trqpurpz7402 do it once then add it on i think

@trqpurpz7402

Ай бұрын

@@MrAstburyMaths wow, maths is so interesting! I just watched a recursive integration step video and I understand it. If I was to use this in the exam, would I lose marks is my question?

hi could anyone please explain how to do this question by parts?

@Hoopy_

Ай бұрын

u=e^u, dv/dx=3^u substitute into parts formula du/dx=e^u, v=3^uln3

@evanpoole7829

Ай бұрын

I havent tried this but maybe 3^lnx x one, choose one as the value to integrate. Kinda similar to how when yoi integrate lnx you integrate lnx x 1.

@ronakkumar3295

Ай бұрын

For parts this is how I did it: So if you sub u = lnx you get the integral of 3^u e^u as shown in the video Now parts: z = 3^u so dz/du = ln(3)*3^u (ofc u normally say u = something then du/dx = something but in this case we have already used the letter u so that would make things confusing) dv/du = e^u so v = e^u integral of e^u*3^u = e^u*3^u - integral of ln(3)*3^u*e^u (Using by parts) Now notice that when u do it by parts you have the same integral again. So u can say I = integral of e^u*3^u So: I = e^u*3^u -ln(3)*I Hence I = e^u*3^u/(ln3+1) Then sub back in u = lnx, then u get the thing they want, let me know if i explained something badly

@Hippogan

Ай бұрын

@@ronakkumar3295 Thank you very much I did it again and now it works!

@areen1215

Ай бұрын

@@ronakkumar3295 hey the first part is all good but i dont understand how the( ln3 I )became (ln3 + 1) ? and the denominator

i cant believe i actually managed to do it first try tbf it just all flowed well but I'm most worried about the modelling quadratic questions that come up ahha I never know how to do them even though they're meant to be the easier marks

@ravjayakodi2746

Ай бұрын

Do U mean the stuff when they give U the minimum or maximum point and then figure it out ?

@anmty-fk2dp

Ай бұрын

bro no the exponential modelling qs shake me to my core

@MrAstburyMaths

Ай бұрын

Nice work 👌

that was so fun omfg

whats wrong with saying xdu=dx?

@medievalrat324

18 күн бұрын

because you don't want any "x" terms in your integral when you're using u-substitution- you just want it in terms of "u" so you have to convert all "x" terms to "u" terms

I solved this integral with parts and sub very easily the easiest integral ever why did u say parts wont get u far?? u integrate 3^u and differentiate e^u then add the integral to the other side and simplify so easy?

@MrAstburyMaths

Ай бұрын

Hi sorry you found the integral too easy. It’s been hard trying to find super hard questions everyday and record, edit and post them. Yes, you are right parts does work well, sorry for my mistake there. I think both methods are equally acceptable and efficient.

@Mathiscool-jx9mv

Ай бұрын

@@MrAstburyMaths its alr G, ill forgive ur incompetence just this time (￣y▽,￣)╭

@jamzfym

Ай бұрын

how did you integrate the integral when doing it by parts

@Mathiscool-jx9mv

Ай бұрын

@@jamzfym u need to treat it like an equation and "add or subtract" the rhs to lhs then simplfy

@mardinjordan8642

Ай бұрын

@@Mathiscool-jx9mvwhy did you make the u= e^u

i think this question is harder than alevel

May I ask what exam board this question is from?

@julesbourlot5684

Ай бұрын

Edexcel

@adailyfact

Ай бұрын

@@julesbourlot5684 international or normal?

@ZeroWinz

Ай бұрын

this isnt off any of the actual edexcel papers so far this is from a MADASMATHS synoptic paper ( W I think ) and this paper is a question of a paper rated above the a level standard difficulty

@adailyfact

Ай бұрын

@@ZeroWinz Ok thank you so much mate.

lightwork 😀

@MrAstburyMaths

Ай бұрын

👍