The value of f( ln 5 ) is 0.
[tex] \tt \: \: \implies \: f(x) = \int\limits_{0}^{x} f(t)\, dt \\ \\ [/tex]
Applying Newton Leibniz Theorem.
[tex] \tt \: \: \implies \: \frac{d}{dx} \bigg(f(x) \bigg)= \frac{d}{dx} \bigg( \int\limits_{0}^{x} f(t)\, dt \bigg)\\ \\ [/tex]
[tex] \tt \: \: \implies \: f'(x) = f(x) - 0 \\ \\ [/tex]
[tex] \tt \: \: \implies \: \frac{ f'(x) }{f(x)} = 1.\\ \\ [/tex]
Integrating both sides.
[tex] \tt \: \: \implies \: \int\frac{ f'(x) }{f(x)} dx=\int dx.\\ \\ [/tex]
[tex] \tt \: \: \implies \: ln | f(x) | =x+ c.\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) ={e}^{x + c} \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) ={e}^{x}{e}^{c}\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) = P{e}^{x} \: \: \: \: \: --(1)\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(0) = \int\limits_{0}^{0} f(t)\, dt = 0. \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(0) = 0. \: \: \: \: \: - - (2)\\ \\ [/tex]
From equation ( 2 )
[tex] \tt \: \: \implies \: P =0 . \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) = P{e}^{x}[/tex]
put p = 0.
[tex] \tt \: \: \implies \: f(x) = 0. \\ \\ [/tex]
Therefore,
[tex] \tt \: \: \implies \: f(ln5) = 0. \\ \\ [/tex]
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Answer :
The value of f( ln 5 ) is 0.
Solution :
[tex] \tt \: \: \implies \: f(x) = \int\limits_{0}^{x} f(t)\, dt \\ \\ [/tex]
Applying Newton Leibniz Theorem.
[tex] \tt \: \: \implies \: \frac{d}{dx} \bigg(f(x) \bigg)= \frac{d}{dx} \bigg( \int\limits_{0}^{x} f(t)\, dt \bigg)\\ \\ [/tex]
[tex] \tt \: \: \implies \: f'(x) = f(x) - 0 \\ \\ [/tex]
[tex] \tt \: \: \implies \: \frac{ f'(x) }{f(x)} = 1.\\ \\ [/tex]
Integrating both sides.
[tex] \tt \: \: \implies \: \int\frac{ f'(x) }{f(x)} dx=\int dx.\\ \\ [/tex]
[tex] \tt \: \: \implies \: ln | f(x) | =x+ c.\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) ={e}^{x + c} \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) ={e}^{x}{e}^{c}\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) = P{e}^{x} \: \: \: \: \: --(1)\\ \\ [/tex]
[tex] \tt \: \: \implies \: f(0) = \int\limits_{0}^{0} f(t)\, dt = 0. \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(0) = 0. \: \: \: \: \: - - (2)\\ \\ [/tex]
From equation ( 2 )
[tex] \tt \: \: \implies \: P =0 . \\ \\ [/tex]
[tex] \tt \: \: \implies \: f(x) = P{e}^{x}[/tex]
put p = 0.
[tex] \tt \: \: \implies \: f(x) = 0. \\ \\ [/tex]
Therefore,
[tex] \tt \: \: \implies \: f(ln5) = 0. \\ \\ [/tex]