Let-x-f-x-e-f-x-0-e-f-x-dx-

Question Number 72296 by naka3546 last updated on 27/Oct/19

L e t x = f (x) e^{f (x)}

\int_{0} \overset{e}{} f (x) d x = ?

Commented by naka3546 last updated on 27/Oct/19

e^{\ln x} = x

Commented by mr W last updated on 27/Oct/19

Commented by mind is power last updated on 27/Oct/19

Sorry sir Mr W i dealt my communt in same time yours

Commented by mr W last updated on 27/Oct/19

{t h a t}^{'} s o k . i p u t s o m e m a t e r i a l i f o u n d

i n w e k i p e d i a .

Answered by mind is power last updated on 27/Oct/19

\int f (x) dx = [xf (x)] - \int {xf}^{'} (x) dx

x = f (x) e^{f (x)} \Rightarrow 1 = (f^{'} (x) + f^{'} (x) . f (x)) e^{f (x)}

\Rightarrow f^{'} (x) = \frac{1}{(1 + f (x)) e^{f (x)}} = \frac{f (x)}{x (1 + f (x))} \Rightarrow {xf}^{'} (x) = \frac{f (x)}{1 + f (x)}

\Rightarrow \int f (x) dx = xf (x) - \int \frac{f (x)}{1 + f (x)} dx

let u = f (x)

\Rightarrow du = f^{'} (x) dx = \frac{f (x)}{x (1 + f (x))} dx

x = f (x) e^{f (x)} = {ue}^{u} \Rightarrow du = \frac{u}{{ue}^{u} (1 + u)} dx \Rightarrow dx = \frac{{ue}^{u} du}{u} . \frac{1 + u}{}

\int \frac{f (x)}{1 + f (x)} dx = \int \frac{u}{1 + u} . \frac{{ue}^{u} du}{u} . (1 + u) = \int {ue}^{u} = (u - 1) e^{u} + c

= (f (x) - 1) e^{f (x)} + c = x - e^{f (x)} + c = x (1 - \frac{1}{f (x)}) + c

\Rightarrow \int f (x) dx = xf (x) - x (1 - \frac{1}{f (x)}) + c = x (f (x) - 1 + \frac{1}{f (x)}) + c

\int_{0}^{e} f (x) dx = e (f (e) - 1 + \frac{1}{f (e)})

f (e) e^{f (e)} = e = 1 . e^{1} . . z

f (t) = {te}^{t} \Rightarrow f^{'} (t) = (t + 1) e^{t} ⩾ 0 \forall t \in {IR}^{+}

f (1) = e \Rightarrow {xe}^{x} = e \Leftrightarrow x \in {1}

z \Rightarrow f (e) = 1

\int_{0}^{e} f (x) dx = e (1 - 1 + \frac{1}{1}) - \lim_{x \to 0} x (f (x) - 1 + \frac{1}{f (x)})

\frac{1}{f (x)} = \frac{e^{f (x)}}{x}

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

\lim_{x \to 0} x (f (x) - 1) + e^{f (x)} = e^{f (0)}

f (0) e^{f (0)} = 0 \Rightarrow f (0) = 0 \Rightarrow e^{f (0)} = 1

\int_{0}^{e} f (x) dx = e - 1

Commented by mr W last updated on 27/Oct/19

p l e a s e c h e c k a g a i n s i r . t h e r e s u l t

s h o u l d b e e - 1 .

Commented by mr W last updated on 27/Oct/19

Commented by mind is power last updated on 27/Oct/19

yeah i got it x (f (x) - 1 + \frac{1}{f (x)}) in 0 i did zero

but its not [defind cause f (0) = 0

we should tack \lim