y-x-1-x-a-a-x-t-t-t-t-t-dt-x-a-a-gt-0-y-x-gt-0-Find-y-2a-

Question Number 1625 by 112358 last updated on 27/Aug/15

y (x) = \frac{1}{x - a} \int_{a}^{x} \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \dots}}}}} d t

x \neq a, a > 0, y (x) > 0 .

F i n d y (2 a) .

Answered by Rasheed Soomro last updated on 28/Aug/15

y (x) = \frac{1}{x - a} \int_{a}^{x} (\sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \dots}}}}}) d t

L e t \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \dots}}}}} = u

S q u a r i n g b o t h s i d e s :

t + \sqrt{t + \sqrt{t + \sqrt{t + \sqrt{t + \dots}}}} = u^{2}

t + u = u^{2} \Rightarrow u^{2} - u - t = 0

u = \frac{1 \pm \sqrt{1 + 4 t}}{2}

y (x) = \frac{1}{x - a} \int_{a}^{x} (\frac{1 \pm \sqrt{1 + 4 t}}{2}) d t

= \frac{1}{2 (x - a)} \int_{a}^{x} (1 \pm \sqrt{1 + 4 t}) d t

= \frac{1}{2 (x - a)} {[t \pm \frac{1}{6} {(1 + 4 t)}^{\frac{3}{2}}]}_{a}^{x}

= \frac{1}{2 (x - a)} ([x \pm \frac{\sqrt{{(1 + 4 x)}^{3}}}{6}] - [a \pm \frac{\sqrt{{(1 + 4 a)}^{3}}}{6}])

= \frac{1}{2 (x - a)} \times \frac{6 (x - a) \pm \sqrt{{(1 + 4 x)}^{3}} \mp \sqrt{{(1 + 4 a)}^{3}}}{6}

= \frac{6 (x - a) \pm (\sqrt{{(1 + 4 x)}^{3}} - \sqrt{{(1 + 4 a)}^{3}})}{12 (x - a)}

y (x) = \frac{1}{2} \pm \frac{\sqrt{{(1 + 4 x)}^{3}} - \sqrt{{(1 + 4 a)}^{3}}}{12 (x - a)}

y (2 a) = \frac{1}{2} \pm \frac{\sqrt{{(1 + 4 (2 a))}^{3}} - \sqrt{{(1 + 4 a)}^{3}}}{12 ((2 a) - a)}

= \frac{1}{2} \pm \frac{\sqrt{{(1 + 8 a)}^{3}} - \sqrt{{(1 + 4 a)}^{3}}}{12 a}

Commented by 123456 last updated on 28/Aug/15

\underset{a}{\int^{x}} 1 \pm \sqrt{1 + 4 t} d t = \underset{a}{\int^{x}} d t \pm \underset{a}{\int^{x}} \sqrt{1 + 4 t} d t

\underset{a}{\int^{x}} d t = x - a

\underset{a}{\int^{x}} \sqrt{1 + 4 t} d t

u = 1 + 4 t \Rightarrow d u = 4 d t

t = a \Rightarrow u = 1 + 4 a

t = x \Rightarrow u = 1 + 4 x

\underset{a}{\int^{x}} \sqrt{1 + 4 t} d t = \frac{1}{4} \underset{1 + 4 a}{\int^{1 + 4 x}} \sqrt{u} d u = {[\frac{1 / 4}{3 / 2} \sqrt{u^{3}}]}_{1 + 4 a}^{1 + 4 x}

= \frac{\sqrt{{(1 + 4 x)}^{3}} - \sqrt{{(1 + 4 a)}^{3}}}{6}

\underset{a}{\int^{x}} 1 \pm \sqrt{1 + 4 t} d t = x - a \pm \frac{\sqrt{{(1 + 4 x)}^{3}} - \sqrt{{(1 + 4 a)}^{3}}}{6}

Commented by 123456 last updated on 28/Aug/15

y (x) = \frac{1}{2 (x - a)} \times \frac{6 (x - a) \pm \sqrt{{(1 + 4 x)}^{3}} \mp \sqrt{{(1 + 4 a)}^{3}}}{6}

Commented by Rasheed Soomro last updated on 28/Aug/15

Thanks to mention my calculation mistake . I am going to

correct it .

Commented by 123456 last updated on 28/Aug/15

\lim_{x \to a} y (x) is like {[\frac{d}{d x} \int f (x) d x]}_{x = a} :)

\lim_{x \to a} \frac{\underset{a}{\int^{x}} f (x) d x}{x - a} = \lim_{x \to a} \frac{F (x) - F (a)}{x - a} = f (a)

Commented by 112358 last updated on 29/Aug/15

I c r e a t e d t h i s q u e s t i o n h a v i n g t h e

f o r m u l a f o r t h e a v e r a g e v a l u e

o f a f u n c t i o n i n m i n d :

y_{a v g} = \frac{1}{b - a} \int_{a}^{b} f (x) d x .

y_{a v g} = \frac{F (b) - F (a)}{b - a} (*)

(*) r e s e m b l e s t h e r e l a t i o n g i v e n

b y t h e m e a n v a l u e t h e o r e m w h i c h

s t a t e s t h a t \exists c \in [a, b] s u c h t h a t

\frac{d}{d x} f (x) ∣_{x = c} = \frac{f (a) - f (b)}{b - a}

g i v e n t h a t f (x) i s c o n t i n u o u s

\forall x \in [a, b] a n d d i f f e r e n t i a b l e

\forall x \in (a, b) .

Commented by 123456 last updated on 29/Aug/15

or mean value theorem for integrails

if f is continuous and integrable into [a, b]

them \exists ξ \in [a, b]

\underset{a}{\int^{b}} f (x) d x = f (ξ) (b - a)

so for some ξ \in [a, b]

y_{a v g} = \frac{1}{b - a} \underset{a}{\int^{b}} f (x) d x = f (ξ)

Commented by 112358 last updated on 29/Aug/15

N i c e . T h a n k s!

N i c e . T h a n k s!

Facebook Tweet Pin