if-1-2-f-x-dx-2-then-1-4-1-x-f-x-dx-is-please-help-me-Sir-I-ve-been-trying-this-for-2-days-and-getting-stuck-

Question Number 56147 by afachri last updated on 11/Mar/19

if \underset{1}{\int^{2}} f (x) d x = \sqrt{2}, then \underset{1}{\int^{4}} \frac{1}{\sqrt{x}} f (x) d x

is ? ?

please help me Sir . I^{'} ve been trying

this for 2 days and getting stuck .

Commented by 121194 last updated on 11/Mar/19

f (x) = \sqrt{2}

\underset{1}{\int^{2}} f (x) d x = \sqrt{2}

\underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x = 2 \sqrt{2}

\underset{1}{\int^{2}} \sqrt{x} d x = \frac{2}{3} (2 \sqrt{2} - 1)

f (x) = \frac{3 \sqrt{2 x}}{2 (2 \sqrt{2} - 1)}

\underset{1}{\int^{2}} f (x) d x = \sqrt{2}

\underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x = \frac{9 \sqrt{2}}{2 (2 \sqrt{2} - 1)} = \frac{9 (4 + \sqrt{2})}{14}

\underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x

u^{2} = x \Rightarrow u = \sqrt{x}; d u = \frac{d x}{2 \sqrt{x}} \Leftrightarrow 2 d u = \frac{d x}{\sqrt{x}}

x = 1 \Rightarrow u = 1

x = 4 \Rightarrow u = 2

\underset{1}{\int^{4}} f (x) \frac{d x}{\sqrt{x}} = 2 \underset{1}{\int^{2}} f (u^{2}) d u

Commented by 121194 last updated on 11/Mar/19

are you sure this question is right ?

Commented by afachri last updated on 11/Mar/19

yes Sir, {that}^{'} s the question is S i r

Commented by afachri last updated on 11/Mar/19

p a r d o n m e S i r . I {d o n}^{'} t u n d e r s r a n d

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

c l e a r l y S i r .

Commented by 121194 last updated on 11/Mar/19

wierd, i {don}^{'} t think that condition is enought to

determinate the other .

because unless i made a typo, there 2 diferet function

such \underset{1}{\int^{2}} f (x) d x = \sqrt{2}, but \underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x gives diferent values

also given g : [1, 4] \to R integable on that domain,

you can always build a solution to it with

f (x) = \frac{\sqrt{2} g (x)}{\underset{1}{\int^{2}} g (x) d x}; \underset{1}{\int^{2}} g (x) d x \neq 0

Commented by prakash jain last updated on 12/Mar/19

If we modify the question \int_{1}^{4} \frac{f (\sqrt{} x)}{\sqrt{x}} d x

x = u^{2}

d x = 2 u d u

\int_{1}^{4} \frac{f (\sqrt{} x)}{\sqrt{x}} d x = \int_{1}^{2} 2 f (u) d u

Answered by MJS last updated on 11/Mar/19

if we {don}^{'} t know the nature of f (x) we {can}^{'} t

solve . the number of functions satisfying the

1^{st} condition is \infty

Commented by MJS last updated on 11/Mar/19

even if f (x) is linear {it}^{'} s not determined :

f (x) = a x + b

\underset{1}{\int^{2}} f (x) d x = \frac{3 a}{2} + b = \sqrt{2} \Rightarrow b = \sqrt{2} - \frac{3 a}{2}

f (x) = a x + \sqrt{2} - \frac{3 a}{2}; a \in R

\underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x = \frac{5 a}{3} + 2 \sqrt{2}; a \in R

if b = 0 \Rightarrow a = \frac{2 \sqrt{2}}{3} \Rightarrow \underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x = \frac{28 \sqrt{2}}{9}

if a = 0 \Rightarrow b = \sqrt{2} \Rightarrow \underset{1}{\int^{4}} \frac{f (x)}{\sqrt{x}} d x = 2 \sqrt{2}

\dots

Commented by afachri last updated on 11/Mar/19

thank you for all of Y o u S i r .

that was i thinking before . how

to or is that any possible to do to

find the real f (x)

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