if F(x)=∫_(u(x)) ^(v(x)) g(x,t)dt determine a expression for F^′ (x).


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 67972 by mathmax by abdo last updated on 02/Sep/19

$i f F (x) = \int_{u (x)}^{v (x)} g (x, t) d t d e t e r m i n e a e x p r e s s i o n f o r F^{^{'}} (x) .$
	Answered by Tanmay chaudhury last updated on 03/Sep/19

	$\frac{d F}{d x} = \int_{u (x)}^{v (x)} \frac{\partial}{\partial x} (g (x, t) d x) + \frac{d v (x)}{d x} g (x, v (x)) - \frac{d u (x)}{d x} g (x, u (x))$
	Commented by mathmax by abdo last updated on 03/Sep/19

	$c a n y o u s i n d t h e s o u r s e o f t h i s f o r m u l a s i r t a n m a y o r$ $s e n d t h e p r o o f . . .$
	Commented by Tanmay chaudhury last updated on 03/Sep/19

	$o k s i r . . .$
	Commented by mind is power last updated on 03/Sep/19

	$i n m y c o m m e n t F (x) = \int_{v}^{u} g (x, t) d t$
	Commented by mathmax by abdo last updated on 03/Sep/19
	$let verify this formulae by F(x) =∫_x ^x^2 e^(−xt) dt we have g(x,t) =e^(−xt) , u(x)=x and v(x)=x^2 gormulae give F^′ (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v)−u^′ g(x,u) =∫_x ^x^2 −t e^(−xt) dt +(2x)e^(−x^3 ) −e^(−x^2 ) by partswe get ∫_x ^x^2 t e^(−xt) dt =[−(t/x)e^(−xt) ]_x ^x^2 −∫_x ^x^2 (−(1/x))e^(−xt) dt =e^(−x^2 ) −x e^(−x^3 ) +(1/x) ∫_x ^x^2 e^(−xt) dt =e^(−x^2 ) −x e^(−x^3 ) +(1/x)[−(1/x)e^(−xt) ]_x ^x^2 =e^(−x^2 ) −xe^(−x^3 ) −(1/x^2 ){ e^(−x^3 ) −e^(−x^2 ) } =(1+(1/x^2 ))e^(−x^2 ) −(x+(1/x^2 ))e^(−x^3 ) ⇒F^′ (x) =−(1+(1/x^2 ))e^(−x^2 ) +(x+(1/x^2 ))e^(−x^3 ) +2x e^(−x^3 ) −e^(−x^2 ) =−(2+(1/x^2 ) )e^(−x^2 ) (3x+(1/x^2 ))e^(−x^3 ) now let find F(x) directly F(x) =[−(1/x)e^(−xt) ]_x ^x^2 =−(1/x){ e^(−x^3 ) −e^(−x^2 ) } =(1/x){e^(−x^2 ) −e^(−x^3 ) } ⇒ F^′ (x) =−(1/x^2 ){ e^(−x^2 ) −e^(−x^3 ) } +(1/x){−2x e^(−x^2 ) +3x^2 e^(−x^3 ) } =−(1/x^2 )e^(−x^2 ) +(1/x^2 )e^(−x^3 ) −2e^(−x^2 ) +3x e^(−x^3 ) =−((1/x^2 ) +2)e^(−x^2 ) +(3x+(1/x^2 ))e^(−x^2 ) we get the same result so the formulae is correct .$
	$l e t v e r i f y t h i s f o r m u l a e b y F (x) = \int_{x}^{x^{2}} e^{- x t} d t w e h a v e$ $g (x, t) = e^{- x t}, u (x) = x a n d v (x) = x^{2} g o r m u l a e g i v e$ $F^{^{'}} (x) = \int_{u (x)}^{v (x)} \frac{\partial g}{\partial x} (x, t) d t + v^{^{'}} g (x, v) - u^{^{'}} g (x, u)$ $= \int_{x}^{x^{2}} - t e^{- x t} d t + (2 x) e^{- x^{3}} - e^{- x^{2}} b y p a r t s w e g e t$ $\int_{x}^{x^{2}} t e^{- x t} d t = {[- \frac{t}{x} e^{- x t}]}_{x}^{x^{2}} - \int_{x}^{x^{2}} (- \frac{1}{x}) e^{- x t} d t$ $= e^{- x^{2}} - x e^{- x^{3}} + \frac{1}{x} \int_{x}^{x^{2}} e^{- x t} d t = e^{- x^{2}} - x e^{- x^{3}} + \frac{1}{x} {[- \frac{1}{x} e^{- x t}]}_{x}^{x^{2}}$ $= e^{- x^{2}} - {x e}^{- x^{3}} - \frac{1}{x^{2}} {e^{- x^{3}} - e^{- x^{2}}}$ $= (1 + \frac{1}{x^{2}}) e^{- x^{2}} - (x + \frac{1}{x^{2}}) e^{- x^{3}} \Rightarrow F^{^{'}} (x) = - (1 + \frac{1}{x^{2}}) e^{- x^{2}} + (x + \frac{1}{x^{2}}) e^{- x^{3}}$ $+ 2 x e^{- x^{3}} - e^{- x^{2}} = - (2 + \frac{1}{x^{2}}) e^{- x^{2}} (3 x + \frac{1}{x^{2}}) e^{- x^{3}} n o w l e t f i n d F (x) d i r e c t l y$ $F (x) = {[- \frac{1}{x} e^{- x t}]}_{x}^{x^{2}} = - \frac{1}{x} {e^{- x^{3}} - e^{- x^{2}}} = \frac{1}{x} {e^{- x^{2}} - e^{- x^{3}}} \Rightarrow$ $F^{^{'}} (x) = - \frac{1}{x^{2}} {e^{- x^{2}} - e^{- x^{3}}} + \frac{1}{x} {- 2 x e^{- x^{2}} + 3 x^{2} e^{- x^{3}}}$ $= - \frac{1}{x^{2}} e^{- x^{2}} + \frac{1}{x^{2}} e^{- x^{3}} - 2 e^{- x^{2}} + 3 x e^{- x^{3}}$ $= - (\frac{1}{x^{2}} + 2) e^{- x^{2}} + (3 x + \frac{1}{x^{2}}) e^{- x^{2}} w e g e t t h e s a m e r e s u l t$ $s o t h e f o r m u l a e i s c o r r e c t .$
	Answered by mind is power last updated on 03/Sep/19
	$let G(x,t)=∫g(x,t)dt F(x)=G(x,v(x))−G(x,u(x)) F′(x)=(∂/∂x){G(x,v(x))−G(x,u(x))}..1 (∂/∂x)(f(u,v))=(∂u/∂x).((∂f(u,v))/∂x)+(∂v/∂x).((∂f(u,v))/∂v) we apply this in 1 F′(x)= (∂/∂x)G(x,u(x))+(∂u/∂x) ((∂ G(x,u(x)))/∂t)−((∂G(x,v(x))/∂x)−(∂v/∂x).((∂G(x,v(x))/∂t) withe ((∂G(x,t))/∂t)=g(x,t) ((∂G(x,t))/∂x)=(∂/∂x)∫g(x,t)dt=∫((∂g(x,t))/∂x)dt we get ∫_0 ^u ((∂g(x,t)dt)/∂x)+u^′ g(x,u)−∫_0 ^v ((∂g(x,t))/∂x)dt−v^, g(x,v) =∫_v ^u ((∂g(x,t))/∂x)dt+u^′ g(x,u)−v^′ g(x,v)$
	$l e t G (x, t) = \int g (x, t) d t$ $F (x) = G (x, v (x)) - G (x, u (x))$ $F^{'} (x) = \frac{\partial}{\partial x} {G (x, v (x)) - G (x, u (x))} . . 1$ $\frac{\partial}{\partial x} (f (u, v)) = \frac{\partial u}{\partial x} . \frac{\partial f (u, v)}{\partial x} + \frac{\partial v}{\partial x} . \frac{\partial f (u, v)}{\partial v}$ $w e a p p l y t h i s i n 1$ $F^{'} (x) = \frac{\partial}{\partial x} G (x, u (x)) + \frac{\partial u}{\partial x} \frac{\partial G (x, u (x))}{\partial t} - \frac{\partial G (x, v (x)}{\partial x} - \frac{\partial v}{\partial x} . \frac{\partial G (x, v (x)}{\partial t}$ $w i t h e \frac{\partial G (x, t)}{\partial t} = g (x, t)$ $\frac{\partial G (x, t)}{\partial x} = \frac{\partial}{\partial x} \int g (x, t) d t = \int \frac{\partial g (x, t)}{\partial x} d t$ $w e g e t$ $\int_{0}^{u} \frac{\partial g (x, t) d t}{\partial x} + u^{^{'}} g (x, u) - \int_{0}^{v} \frac{\partial g (x, t)}{\partial x} d t - v^{,} g (x, v)$ $= \int_{v}^{u} \frac{\partial g (x, t)}{\partial x} d t + u^{^{'}} g (x, u) - v^{^{'}} g (x, v)$
	Commented by mathmax by abdo last updated on 10/Sep/19

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

Question and Answers Forum