Intergration :( Watch

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Cortez
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#1
Report Thread starter 13 years ago
#1
I got an intergration assessment at college and i havent got a clue where to start with substitution

First one is this
2x(1-x)^7 suggested substitution [u = 1-x]

this is what ive done
u = 1 - x
du/dx = -1
dx = du/-1

integral of 2x(u)^7 du/-1

now what :confused: :confused:

is this anywhere near what i should be doing?

thanks
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Bezza
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#2
Report 13 years ago
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(Original post by chrisgill369)
I got an intergration assessment at college and i havent got a clue where to start with substitution

First one is this
2x(1-x)^7 suggested substitution [u = 1-x]

this is what ive done
u = 1 - x
du/dx = -1
dx = du/-1

integral of 2x(u)^7 du/-1

now what :confused: :confused:

is this anywhere near what i should be doing?

thanks
You're almost there, just use x = 1-u so you're integrating -2(1-u)*u^7 du
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VCVT17
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#3
Report 13 years ago
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Yeah, you are doing in right way:

Just need more work to finish it:

2x(1-x)^7 can be rewritten as: -2( 1 - u )*u^7 du = ( 2*u^8 - 2*u^7 ) du
Integrating with respect to u:
= 2/9*u^9 - 2/8*u^8 + K ( with K is constant)
= 2/9*u^9 - 1/4*u^8 + K

Rewrite it again for x: = 2/9*( 1 - x)^9 - 1/4*(1-x)^8

Hopefully it right !!!
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Cortez
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#4
Report Thread starter 13 years ago
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ok thanks for your help. I will give it a go tommorow and report back any problems
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Cortez
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#5
Report Thread starter 13 years ago
#5
well i come up with this

((-2u^8)/8) + ((2u^9)/9) + C


i got another small question now .....
if you intergrate 2e^u do you just get 2e^u again ?
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Bezza
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#6
Report 13 years ago
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(Original post by chrisgill369)
well i come up with this

((-2u^8)/8) + ((2u^9)/9) + C


i got another small question now .....
if you intergrate 2e^u do you just get 2e^u again ?
Thats the same as what VCVT17 got so sounds about right. you might have to sub back in u = 1-x, but doesn't really matter. And yes


\fontsize{5}

\Bigint 2e^u\,du = 2e^u + c
(hee hee - like playing with that!)
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Newton
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#7
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f(x)=INT[2x((1-x)^7)]dx

Let u=1-x=>x=1-u=>(dx/du)=-1=>dx=-du

f(x)=INT[-2(1-u)(u^7)]du=-2INT[(1-u)(u^7)]du

f(x)=-2INT[(u^7)-(u^8)]du

f(x)=-2[(1/8)(u^8)-(1/9)(u^9)]+C

f(x)=-2(u^8)[(1/8)-(1/9)u]+C

f(x)=-(1/36)(u^8)[9-8u]+C

* u=1-x

f(x)=-(1/36)((1-x)^8)[9-8(1-x)]+C

f(x)=-(1/36)((1-x)^8)[1+8x]+C

f(x)=C-(1/36)((1-x)^8)[1+8x)]

Newton.
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