Hard integration Watch

Civ-217
Badges: 1
Rep:
?
#1
Report Thread starter 10 years ago
#1
How do you integrate \int \ (1-x^7)^{1/3}  dx?

Let  u = (1-x^7)^{1/3}  ,  \frac{dv}{dx} = 1
So  \frac{du}{dx} = \frac{-7x^6}{3}(1-x^7)^{-2/3} ,  v=x

\int \ (1-x^7)^{1/3}  dx
=  x(1-x^7)^{1/3}  - \int \frac{-7x^7}{3}(1-x^7)^{-2/3}} dx
=  x(1-x^7)^{1/3}  + \frac{7}{3} \int x^7(1-x^7)^{-2/3}} dx

Now let  w = x and  \frac{ds}{dx}= (x^6)(1-x^7)^{-2/3}
So  \frac{dw}{dx}=1 and  s = \frac{-3}{7}(1-x^7)^{1/3}

=  x(1-x^7)^{1/3}  + \frac{7}{3} ( \frac{-3x}{7}(1-x^7)^{1/3} + \frac{3}{7} \int (1-x^7)^{1/3} dx )
=  x(1-x^7)^{1/3} - x(1-x^7)^{1/3} +  \int (1-x^7)^{1/3} dx )
At the end, I just get
\int \ (1-x^7)^{1/3}  dx = \int \ (1-x^7)^{1/3}  dx

This is actually not the full question, the actual question is

\int^1_0 \ (1-x^7)^{1/3} - (1-x^3)^{1/7}  dx

Anyone has clue how to do it? Thank you.
0
reply
DFranklin
Badges: 18
Rep:
?
#2
Report 10 years ago
#2
Nothing obvious. Where does it come from?
0
reply
Civ-217
Badges: 1
Rep:
?
#3
Report Thread starter 10 years ago
#3
(Original post by DFranklin)
Nothing obvious. Where does it come from?
It's from 'problem plus' of the book Calculus by James Stewart. They are the challenging exercises in the book.

http://www.amazon.com/Calculus-James...5437323&sr=8-2
0
reply
Dystopia
Badges: 8
Rep:
?
#4
Report 10 years ago
#4
Well, the integral can't be found in terms of elementary functions.

You could use the beta function, but I don't know if that will give a nice answer...
0
reply
bballer4life
Badges: 1
Rep:
?
#5
Report 10 years ago
#5
when you intergrate by parts twice, you have to use the same terms for u and (dv/dx) if you don't you basically undo the intergration you done, and go back to the origianl eqation
0
reply
Notnek
  • Forum Helper
Badges: 20
Rep:
?
#6
Report 10 years ago
#6
(Original post by Civ-217)
It's from 'problem plus' of the book Calculus by James Stewart. They are the challenging exercises in the book.

http://www.amazon.com/Calculus-James...5437323&sr=8-2
Can you tell me which chapter/question it is? I know someone who has this book and you (and I) will be interested to know how they did it.
0
reply
pyrolol
Badges: 0
Rep:
?
#7
Report 10 years ago
#7
Maple can't do it (it wants to use Hypergeometrics), I haven't got Axiom installed so I can't try with that unfortunately.
0
reply
DFranklin
Badges: 18
Rep:
?
#8
Report 10 years ago
#8
If there's a nice answer, my gut tells me it's going to be zero.
0
reply
ycntv
Badges: 0
#9
Report 10 years ago
#9
Perhaps this is of help? \displaystyle \int^{b}_{a}f(x)\mathrm{d}x=(b-a)\int^{1}_{0}f[a+(b-a)x]\mathrm{d}x

Thus, \displaystyle\int^{1}_{0} (1-x^7)^{1/3} \mathrm{d}x=-\int^{0}_{1} (x^7)^{1/3} \mathrm{d}x...

I probably have made a very obvious mistake somewhere, though.
0
reply
Dystopia
Badges: 8
Rep:
?
#10
Report 10 years ago
#10
(Original post by DFranklin)
If there's a nice answer, my gut tells me it's going to be zero.
That's what I got from using the beta function.
0
reply
pyrolol
Badges: 0
Rep:
?
#11
Report 10 years ago
#11
Hey those limits weren't there before >_<. That makes the question so much easier lol...
0
reply
infernal_rustage
Badges: 0
Rep:
?
#12
Report 10 years ago
#12
Nope, that formula only gives
b-a = 1-0=1 and
f (a+(b-a)x) = f(0+(1-0)x) = f(x)
hence it just gives you the original integrand

Maybe trying u=x^7?
0
reply
bballer4life
Badges: 1
Rep:
?
#13
Report 10 years ago
#13
i do think the answer is 0, i mean, when you look at the (1-x^7)^1/3

whenever you put 1 into it, it becomes 0
and if its being multiplied by a power of x, whne you put the 0 in, it also becomes 0.

so my guess is that it equals 0
0
reply
DFranklin
Badges: 18
Rep:
?
#14
Report 10 years ago
#14
I = \int_0^1 (1-x^7)^{1/3} \,dx Let y = (1-x^7), dy/dx = -7x^6 = -7(1-y)^{6/7}. and so the integral becomes
I=\frac{1}{7} \int_0^1 y^{1/3}(1-y)^{-6/7}\,dy. Integrate by parts with du = (1-y)^(-6/7):
I = \frac{1}{7}[-7(1-y)^{1/7}y^{1/3}]_0^1 + \frac{1}{3}\int_0^1 y^{-2/3}(1-y)^{1/7}\,dy Do a t = 1-y sub and get
I = \frac{1}{3} \int_0^1 t^{1/7} (1-t)^{-2/3} \,dt

But this equals \int_0^1 (1-x^3)^{1/7}\,dx by a similar argument to the first two lines, but subbing t=(1-x^3)
0
reply
qgujxj39
Badges: 15
#15
Report 10 years ago
#15
I plugged it into my calculator, and it crashed :confused:
reply
Dystopia
Badges: 8
Rep:
?
#16
Report 10 years ago
#16
(Original post by DFranklin)
I = \int_0^1 (1-x^7)^{1/3} \,dx Let y = (1-x^7), dy/dx = -7x^6 = -7(1-y)^{6/7}. and so the integral becomes
I=\frac{1}{7} \int_0^1 y^{1/3}(1-y)^{-6/7}\,dy. Integrate by parts with du = (1-y)^(-6/7):
I = \frac{1}{7}[-7(1-y)^{1/7}y^{1/3}] + \int_0^1 y^{-2/3}(1-y)^{1/7}\,dy Do a t = 1-y sub and get
I = \int_0^1 t^{1/7} (1-t)^{-2/3} \,dt

But this equals \int_0^1 (1-x^3)^{1/7}\,dx by a similar argument to the first two lines, but subbing t=(1-x^3)
Nice.
0
reply
DFranklin
Badges: 18
Rep:
?
#17
Report 10 years ago
#17
Well, it's really just an explicit version of doing it with beta functions... Oh, and I differentiated y^1/3 wrong (now fixed).
0
reply
Dystopia
Badges: 8
Rep:
?
#18
Report 10 years ago
#18
(Original post by DFranklin)
Well, it's really just an explicit version of doing it with beta functions... Oh, and I've differentiated y^1/3 wrong - need to fix that...
Yeah, I realised that after I posted.

Quite a cool result, I think.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (189)
39.54%
No - but I will (28)
5.86%
No - I don't want to (33)
6.9%
No - I can't vote (<18, not in UK, etc) (228)
47.7%

Watched Threads

View All