let f(x) =∫_x^ ^(x^2 −x) arctan(e^(−x−t) )dt calculate f^′ (x) and f^′ (0).


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
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
$we have g(x,t)=arctan(e^(−x−t) ) ,u(x)=x and v(x)=x^2 −x ⇒ f^′ (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v)−u^′ g(x,u) =∫_x ^(x^2 −x) ((−e^(−x−t) )/(1+e^(−2x−2t) ))dt +(2x−1)g(x,x^2 −x)−g(x,x) =−e^(−x) ∫_x ^(x^2 −x) (e^(−t) /(1+e^(−2t−2x) ))dt +(2x−1)arctan(e^(−x^2 ) ) −arctan(e^(−2x) ) let find I =∫_x ^(x^2 −x) (e^(−t) /(1+e^(−2t−2x) ))dt ⇒ I =∫_x ^(x^2 −1) (e^t /(e^(2t) +e^(−2x) )) dt =_(e^t =u) ∫_e^x ^e^(x^2 −x) (u/(u^2 +e^(−2x) )) (du/u) =∫_e^x ^e^(x^2 −x) (du/(u^2 +e^(−2x) )) =_(u=e^(−x) z) ∫_e^(2x) ^e^x^2 ((e^(−x) dz)/(e^(−2x) (z^2 +1))) =e^x [arctanz]_e^(2x) ^e^x^2 =e^x { arctan(e^x^2 )−arctan(e^(2x) )} ⇒ f^′ (x) =−arctan(e^x^2 )+e^(−x) arctan(e^(2x) )+(2x−1)arctan(e^(−x^2 ) ) −arctan(e^(−2x) ) .$
$w e h a v e g (x, t) = a r c t a n (e^{- x - t}), u (x) = x a n d v (x) = x^{2} - x \Rightarrow$ $f^{^{'}} (x) = \int_{u (x)}^{v (x)} \frac{\partial g}{\partial x} (x, t) d t + v^{^{'}} g (x, v) - u^{^{'}} g (x, u)$ $= \int_{x}^{x^{2} - x} \frac{- e^{- x - t}}{1 + e^{- 2 x - 2 t}} d t + (2 x - 1) g (x, x^{2} - x) - g (x, x)$ $= - e^{- x} \int_{x}^{x^{2} - x} \frac{e^{- t}}{1 + e^{- 2 t - 2 x}} d t + (2 x - 1) a r c t a n (e^{- x^{2}})$ $- a r c t a n (e^{- 2 x}) l e t f i n d I = \int_{x}^{x^{2} - x} \frac{e^{- t}}{1 + e^{- 2 t - 2 x}} d t \Rightarrow$ $I = \int_{x}^{x^{2} - 1} \frac{e^{t}}{e^{2 t} + e^{- 2 x}} d t =_{e^{t} = u} \int_{e^{x}}^{e^{x^{2} - x}} \frac{u}{u^{2} + e^{- 2 x}} \frac{d u}{u}$ $= \int_{e^{x}}^{e^{x^{2} - x}} \frac{d u}{u^{2} + e^{- 2 x}} =_{u = e^{- x} z} \int_{e^{2 x}}^{e^{x^{2}}} \frac{e^{- x} d z}{e^{- 2 x} (z^{2} + 1)} = e^{x} {[a r c t a n z]}_{e^{2 x}}^{e^{x^{2}}}$ $= e^{x} {a r c t a n (e^{x^{2}}) - a r c t a n (e^{2 x})} \Rightarrow$ $f^{^{'}} (x) = - a r c t a n (e^{x^{2}}) + e^{- x} a r c t a n (e^{2 x}) + (2 x - 1) a r c t a n (e^{- x^{2}})$ $- a r c t a n (e^{- 2 x}) .$
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

	$y o u r a n s w e r i s n o t c o r r e c t s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum