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hiiiiii........ using integration by parts....... here I = e-t f(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-t f'(x-t)dt I=-f(x-t)e-t -[ -f(x-t)e-t +e-t f(x-t)dt] I =-f(x-t)e-t +f(x-t)e-t -e-t f(x-t)dt 2I =0 I =0 so therfore f(x) =x2 +I f(x) =x2 as I=0 answer f(x) =x2 is da method correct.....plzz correct if wrong...... thank u......
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Moderation Team
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