IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

hsbhatt

Find₀

¹ ³

(1-x⁷) - ⁷

1-x³dx

hsbhatt

Is this problem so easy that it is not even worth a dekko?

konichiwa2x

working on it now...

elastiboysai

Hi sir,

I got a method, but final ans involves somethin in terms of

_{[0 ]}

^{[1 ]} 1/7*t^1/3*(1-t)^-6/7dt and _{[0 ]}^{[1 ] t^1/7(1-t)^-2/3dt}

^{I was able to simplify it in terms of beta function as}

^{1/7B(4/3,1/7)-1/3B(8/7,1/3)}

^{but i don know how beta fn works for fractions}

^{tell me if rt, ill post my method}

hsbhatt

No, you needn't go as far as that. It's meant to be for IIT aspirants not cosmologists!

rockey

i think its zero. if my ans is correct i give my solution

hsbhatt

go ahead! right or wrong

priyesh

refer http://integrals.wolfram.com/index.jsp
u might get some help
i am unable to do this one

hsbhatt

you are a funny if resourceful guy priyesh

priyesh

wat did u find funny

elastiboysai

Re:Instructive

elastiboysai

dis is wen i got stuc

n i thought of beta fn

ill try a new approach

hsbhatt

First a confession. I had no clue how to go about this one. I have filched the solution. The reason I've put it in this forum is that I thought this is a typical JEE problem which tries to lam you under the chin with a very simple concept.

We have to find ₀

¹f(x) - g(x) dx where f(x) = ³

(1-x⁷) and g(x) = ⁷

1-x³

The solution becomes very evident when you recognise that g = f^-1.

I mean if y = ³

(1-x⁷) then y³ = 1-x⁷ and x = ⁷

1-y³.

Also, we have when x=0, y=1 and when x=1, y=0 which also means when y=0,x =1 and when y=1,x=0.

The "graph"ologists among you will recognise that the area under the curves for both f and g between the limits 0 and 1 are same.

Hence the answer is 0.

hsbhatt

Why are you guys rating me for a problem I didnt solve?

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