soit-o-x-f-t-x-0-x-t-2-f-t-dt-determiner-f-1-

Question Number 156475 by SANOGO last updated on 11/Oct/21

s o i t : \int_{o}^{x} f (t) = x + \int_{0}^{x} t^{2} f (t) d t

d e t e r m i n e r f (1)

Commented by SANOGO last updated on 11/Oct/21

o k t h a n k y o u

Commented by mr W last updated on 11/Oct/21

f (x) = 1 + x^{2} f (x)

f (x) = \frac{1}{1 - x^{2}}

f (1) {d o e s n}^{'} t e x i s t .

Commented by SANOGO last updated on 11/Oct/21

c a n y o u d e t a i l f o r m e

s o i u n d e r s t a n d ?

Commented by mr W last updated on 11/Oct/21

\frac{d}{d x} [\int_{a (x)}^{b (x)} f (t) d t] = f [b (x)] \frac{d b (x)}{d x} - f [a (x)] \frac{d a (x)}{d x}

Answered by Mathspace last updated on 11/Oct/21

b y d e r i v a t i o n w e g e t

f (x) = 1 + x^{2} f (x) \Rightarrow (1 - x^{2}) f (x) = 1 \Rightarrow

f (x) = \frac{1}{1 - x^{2}} \Rightarrow f (1) d o n t e x i s t!

Commented by SANOGO last updated on 11/Oct/21

t h a n k y o u

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