# Intergration :(Watch

This discussion is closed.
#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

is this anywhere near what i should be doing?

thanks
0
13 years ago
#2
(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

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
0
13 years ago
#3
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 !!!
0
#4
ok thanks for your help. I will give it a go tommorow and report back any problems
0
#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 ?
0
13 years ago
#6
(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

(hee hee - like playing with that!)
0
13 years ago
#7
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.
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Sheffield Hallam University
Thu, 13 Dec '18
• University of Buckingham
Thu, 13 Dec '18
• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18

### Poll

Join the discussion

Yes (18)
31.03%
No (40)
68.97%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.