integration help (3 Viewers)

[jellybelly59]

integrate the following using the given substitution

x. [root(x+1)] when x = (u^2)-1

 

[Timothy.Siu]

2/5(x+1)^5/2 -2/3(x+1)^3/2+c

imo, that substitution is stupid, shud jsut do u=x+1

 

[shaon0]

Let u=x+1

S (u-1)sqrt(u) du
= S u.sqrt(u) du - S sqrt(u) du
= 2/5 (u)^(5/2) - 2/3 (u)^(3/2) +C
= 2/5 (x+1)^(5/2) - 2/3 (x+1)^(3/2) +C
Best way to do this is let u= x+1

 

[Continuum]

What's the difference between integration by substitution and reverse chain rule ? I'm getting conflicting answers - some people say its the same thing but under a different name, while others say that they're abit different.

 

[Timothy.Siu]

What's the difference between integration by substitution and reverse chain rule ? I'm getting conflicting answers - some people say its the same thing but under a different name, while others say that they're abit different.

u cant reverse chain rule this, well not that i know of because its product rule as well.

 

[shaon0]

What's the difference between integration by substitution and reverse chain rule ? I'm getting conflicting answers - some people say its the same thing but under a different name, while others say that they're abit different.

Reverse chain rule uses u and does a reverse chain rule using it. For this q you can only use u=x+1 where you can't reverse chain rule this (as far as i know).

 

[jellybelly59]

but the textbook is telling me to use that substitution.

 

[Continuum]

Reverse chain rule uses u and does a reverse chain rule using it. For this q you can only use u=x+1 where you can't reverse chain rule this (as far as i know).

That doesn't really answer the question.....

Plus, you use u for integration by substitution as well.

 

[shaon0]

but the textbook is telling me to use that substitution.

then u'd have to go; u = sqrt(x+1) and use chain rule to find the derivative so you can get rid of the dx.

 

[shaon0]

That doesn't really answer the question.....

Plus, you use u for integration by substitution as well.

reverse chain rule is usually used for composite functions.
Where as substitution in integration can be used in non-composite function integration as well as composite function integration.

 

[P.T.F.E]

x.√(x+1 ) when x= u^2-1
u^2-1√(u^2-1+1)
u^2-1√(u^2 )
(u^2-1)u
u^3-u
Then integrate this
∫〖u^3- u〗
=u^4/4- u^2/2
HOPE THIS HELPS U.

 

[P.T.F.E]

Let u=x+1

S (u-1)sqrt(u) du
= S u.sqrt(u) du - S sqrt(u) du
= 2/5 (u)^(5/2) - 2/3 (u)^(3/2) +C
= 2/5 (x+1)^(5/2) - 2/3 (x+1)^(3/2) +C
Best way to do this is let u= x+1

YOU CANT DO THAT IF IT ASKS U TO USE THE SUBSTITUTUION U CAN'T JUST MAKE UR OWN RULES UP!!!

 

[Timothy.Siu]

x.√(x+1 ) when x= u^2-1
u^2-1√(u^2-1+1)
u^2-1√(u^2 )
(u^2-1)u
u^3-u
Then integrate this
∫〖u^3- u〗
=u^4/4- u^2/2
HOPE THIS HELPS U.

wrong

 

[shaon0]

YOU CANT DO THAT IF IT ASKS U TO USE THE SUBSTITUTUION U CAN'T JUST MAKE UR OWN RULES UP!!!

I didn't see that at the beginning.
Plus you didn't finish your integration solution and i'm still lolling at your solution.

 

[Timothy.Siu]

YOU CANT DO THAT IF IT ASKS U TO USE THE SUBSTITUTUION U CAN'T JUST MAKE UR OWN RULES UP!!!

at least he got it right

 

[Continuum]

but the textbook is telling me to use that substitution.

If you have to use that then:

 

[3unitz]

integrate the following using the given substitution

x. [root(x+1)] when x = (u^2)-1

Int x. [root(x+1)] dx

dx = 2u du
u = root(x+1)

Int [(u^2)-1]. u 2u du

Int 2u^3 - 2u^2 du

u^4/2 - 2u^3/3 + C

(x+1)^2/2 - 2(x+1)^(3/2)/3 + C

 

[shaon0]

at least he got it right

Yeah.

 

[jellybelly59]

ty all.

 

[shaon0]

Int x. [root(x+1)] dx

dx = 2u du
u = root(x+1)

Int [(u^2)-1]. u 2u du

Int 2u^3 - 2u^2 du

u^4/2 - 2u^3/3 + C

(x+1)^2/2 - 2(x+1)^(3/2)/3 + C

How come we get different solutions?