IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shoreline

if f(x)= x²+ ₀

^x e^-tf(x-t) dt

then fin f(x) ie make it free from the variable 't'

Arjun

i hav done it in da supr post...check it out.....

magiclko

e^-tf(x-t)dt

I=f(x-t)e^-t.dt+f'(x-t).(-e^-t)dt

I=-f(x-t)e^-t -

f'(x-t).-

e^-tdt

I =-f(x-t)e^-t -[ e^-t.

f'(x-t)dt +

e^-tf'(x-t)dt

I=-f(x-t)e^-t -[ -f(x-t)e^-t +

e^-tf(x-t)dt]

I =-f(x-t)e^-t+f(x-t)e^-t -

e^-tf(x-t)dt

2I =0

I =0

so therfore

f(x) =x² +I

f(x) =x²

shoreline

if u wanna do this q.........
let me give u a hint
u will need some differentiation........

shakirshafi12

hey shoreline even then magciklo,s answer is right

f(x)= x²+ ₀

^x e^-tf(x-t) dt

here f(0)=0+ ₀

⁰ e^-tf(-t) dt =0

we differenciate

f`(x)=2x+e^-x*f(0)

f`(x)=2x

so answer will still be same

11.nakul

every1 have done it wrong....

f(x) can b rewritten as

f(x)=x^2₊₀

^xe^-(x-t)f(t) dt

f(x)=x^2₊e^-x ₀

^x e^tf(t) dt...............(1)

now on differentiating

f ' (x)= 2x + e^-x e^xf(x) - e^-x ₀^x e^tf(t) dt

f ' (x)= 2x + f(x) - e^-x ₀^x e^tf(t) dt............(2)

on adding (1) and (2)...

f(x)+ f ' (x) = x^2₊2x + f(x)

f ' (x) = x^2₊2x

now on integrating...

f (x)= x³/ 3 +x²+ c

but f(0)=0 so, c=0

f (x)= x³/ 3 +x²

shoreline

ya 11.nakul its perfect

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