let-f-x-x-x-2-x-arctan-e-x-t-dt-calculate-f-x-and-f-0-

Question Number 68466 by mathmax by abdo last updated on 11/Sep/19

l e t f (x) = \int_{x^{}}^{x^{2} - x} a r c t a n (e^{- x - t}) d t

c a l c u l a t e f^{^{'}} (x) a n d f^{^{'}} (0) .

Commented by mathmax by abdo last updated on 11/Sep/19

w e h a v e g (x, t) = a r c t a n (e^{- x - t}), u (x) = x a n d v (x) = x^{2} - x \Rightarrow

f^{^{'}} (x) = \int_{u (x)}^{v (x)} \frac{\partial g}{\partial x} (x, t) d t + v^{^{'}} g (x, v) - u^{^{'}} g (x, u)

= \int_{x}^{x^{2} - x} \frac{- e^{- x - t}}{1 + e^{- 2 x - 2 t}} d t + (2 x - 1) g (x, x^{2} - x) - g (x, x)

= - e^{- x} \int_{x}^{x^{2} - x} \frac{e^{- t}}{1 + e^{- 2 t - 2 x}} d t + (2 x - 1) a r c t a n (e^{- x^{2}})

- a r c t a n (e^{- 2 x}) l e t f i n d I = \int_{x}^{x^{2} - x} \frac{e^{- t}}{1 + e^{- 2 t - 2 x}} d t \Rightarrow

I = \int_{x}^{x^{2} - 1} \frac{e^{t}}{e^{2 t} + e^{- 2 x}} d t =_{e^{t} = u} \int_{e^{x}}^{e^{x^{2} - x}} \frac{u}{u^{2} + e^{- 2 x}} \frac{d u}{u}

= \int_{e^{x}}^{e^{x^{2} - x}} \frac{d u}{u^{2} + e^{- 2 x}} =_{u = e^{- x} z} \int_{e^{2 x}}^{e^{x^{2}}} \frac{e^{- x} d z}{e^{- 2 x} (z^{2} + 1)} = e^{x} {[a r c t a n z]}_{e^{2 x}}^{e^{x^{2}}}

= e^{x} {a r c t a n (e^{x^{2}}) - a r c t a n (e^{2 x})} \Rightarrow

f^{^{'}} (x) = - a r c t a n (e^{x^{2}}) + e^{- x} a r c t a n (e^{2 x}) + (2 x - 1) a r c t a n (e^{- x^{2}})

- a r c t a n (e^{- 2 x}) .

Commented by mathmax by abdo last updated on 11/Sep/19

f^{^{'}} (0) = - \frac{π}{4} + \frac{π}{4} - \frac{π}{4} - \frac{π}{4} \Rightarrow f^{^{'}} (0) = - \frac{π}{2}

Answered by meme last updated on 11/Sep/19

f^{^{'}} (x) = a r c t a n (e^{- x - x^{2} + x}) - a r c t a n (e^{- x - x})

= a r c t a n (e^{- x^{2}}) - a r c t a n (e^{- 2 x})

f^{^{'}} (0) = a r c t a n (1) - a r c t a n (1)

= 0

Commented by turbo msup by abdo last updated on 11/Sep/19

y o u r a n s w e r i s n o t c o r r e c t s i r .