Home » Ask & Discuss » Mathematics. » Integral Calculus « Back to Discussion

Integral Calculus

Blazing goIITian

Joined:	25 Aug 2007
Post:	964

25 Aug 2007 21:32:25 IST

0 People liked this

417

integration problem

None

INTEGRATION PROBLEM for x>0

_{[ ]}

^{[ ]} (x^3m+x^2m+x^m)(2x^2m+3x^m+6)^1/mdx

Share this article on:

Comments (2)

punterjack

Hot goIITian

Joined: 18 May 2007

Posts: 144

26 Aug 2007 15:43:52 IST

Like 2 people liked this

a very good Question.......(an iit problem of 2001(or02) i suppose)

the exp. can be written as (x^3m-1+x^2m-1+x^m-1)(2x^3m+3x^2m+6x^m)^1/m

U V

here if V is =t (say)

then Udx= 1/6mdt

thus integral= t ^(1+m)/m/6(1+m)......

this can be finalised by substuting for t......

rate me if you found my expln. satisfactory or do correct me if i were wrong.....................

Priyesh

Blazing goIITian

Joined: 18 Feb 2007

Posts: 1042

26 Aug 2007 15:58:50 IST

Like 2 people liked this

Hi gonik

see

_{[ ]}^{[ ]} (x^3m+x^2m+x^m)(2x^2m+3x^m+6)^1/mdx

= _{[ ]}^{[ ]} (x^3m-1+x^2m-1+x^m-1)(2x^3m+3x^2m+6x^m)^1/mdx (by taking x out from first bracket expression & putting it inside second bracket expression)

now put (2x^3m+3x^2m+6x^m) = t^m

=> 6m(x^3m-1+x^2m-1+x^m-1)= mt^m-1dt

=> therefore the first bracket expression in the integral along with dx becomes

1/6 t^m-1dt

therefore the integral becomes

1/6_{[ ]}^{[ ]} tt^m-1dt = 1/6_{[ ]}^{[ ]} t^mdt

therefore the ans(in terms of t) is 1/6 t^m+1/(m+1)

now substitute back value of t to get the ans. in terms of x

this question first came in a russian olympiad & then came in IIT-JEE.

hope you understood

cheers!!!!!!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team