Timothy.Siu
Prophet 9
made a mistake on the third last line shud be 2u^43unitz said:
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integration help (2 Viewers)
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Timothy.Siu
Prophet 9
because u left it in terms of u, and secondly its wrong anyway.P.T.F.E said:
lyounamu
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Just follow the freaking question.
You do that and you will get it wrong. When people post up question, they are more interested in the solution than answer.
x = u^2 -1
dx/du = 2u
x sqrt(x+1) dx = (u^2-1)u dx
= (u^2-1)u . 2u du
= 2u^4 - 2u^2 du
Integrate so you get:
2u^5/5 - 2u^3/3 + c
But u^2 = x+1
so u^5 = (x+1)^2.5
and u^3 = (x+1)^1.5
so
answer = 2(x+1)^2.5/5 - 2(x+1)^1.5/3 + c
you will obviously see that using x = u^2 -1 isn't much harder. It's like easier.
You do that and you will get it wrong. When people post up question, they are more interested in the solution than answer.
x = u^2 -1
dx/du = 2u
x sqrt(x+1) dx = (u^2-1)u dx
= (u^2-1)u . 2u du
= 2u^4 - 2u^2 du
Integrate so you get:
2u^5/5 - 2u^3/3 + c
But u^2 = x+1
so u^5 = (x+1)^2.5
and u^3 = (x+1)^1.5
so
answer = 2(x+1)^2.5/5 - 2(x+1)^1.5/3 + c
you will obviously see that using x = u^2 -1 isn't much harder. It's like easier.
jellybelly59
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umm... continuum was on the spot with the answerP.T.F.E said:
Pwnage101
Moderator
i'm not namu, but can i answer the Q? lolContinuum said:Namu, do you know if there's a difference between the reverse chain rule and integration by substitution ?
well they are basically the same
ull neva be asked to 'use the reveers chain rule
but basically, with the reverse chain rule technically u do use a substitution in ur head, but when u sub in du it turns out 'nicely' ie:
INT (sinxcosx).dx
now u could use a substituion like u=sinx, but by using 'the reverse chain rule' u notice that this will yield an answer that is (1/2)(sinx)^2 + C - ie when u sub in du=cosx.dx it turns out as INT u.du, which is a very simple function, so technically they are the same thing - we call it 'ther reverse chain rule' because when u derive something via teh chain rule, u multiply teh function by it';s derivative at the end, and so if we have an integration with a function multiplied by its derivatrive, its sorta the revers, and that derivcative disappears
if u dont get it, just use a substitution, its the SAME THING, ur just doing more owrking out on the page, which is a good thing!
BUT dont worry about it
if u do 4U u'll get the hang of things and itll be much smoother in the integration chapter
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