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Question Number 1985 by 123456 last updated on 27/Oct/15

f : [a, b] \to R

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

does

π \underset{a}{\int^{b}} f^{2} d x = 2 π \underset{a}{\int^{b}} f \sqrt{1 + {(\frac{d f}{d x})}^{2}} d x ?

Commented by Yozzi last updated on 28/Oct/15

\int_{a}^{b} f d x = \int_{a}^{b} \sqrt{1 + {(\frac{d f}{d x})}^{2}} d x

(a r e a = a r c l e n g t h f o r x \in [a, b])

\Rightarrow \underset{a}{\int^{b}} (f - \sqrt{1 + {(\frac{d f}{d x})}^{2}}) d x = 0 \dots . . (1)

I f b \neq a, (1) i s t r u e i f f

f - \sqrt{1 + {(\frac{d f}{d x})}^{2}} = 0 .

\Rightarrow f^{2} = 1 + {(\frac{d f}{d x})}^{2}

\Rightarrow f^{2} - 1 = {(\frac{d f}{d x})}^{2}

\Rightarrow \frac{d f}{d x} = \pm \sqrt{f^{2} - 1} (∣ f ∣> 1)

\Rightarrow \int \frac{d f}{\sqrt{f^{2} - 1}} = \pm \int d x

L e t u = {c o s h}^{- 1} (f / a) a \neq 0

c o s h u = f / a

D i f f e r e n t i a t i n g w r t f

\Rightarrow s i n h u \times u^{^{'}} = 1 / a

u^{^{'}} = \frac{1}{a s i n h u} = \frac{1}{a \sqrt{{c o s h}^{2} u - 1}}

u^{^{'}} = \frac{1}{a \sqrt{\frac{f^{2}}{a^{2}} - 1}} = \frac{1}{a \times \frac{1}{a} \sqrt{f^{2} - a^{2}}} = \frac{1}{\sqrt{f^{2} - a^{2}}}

L e t a = 1, \frac{d u}{d f} = \frac{1}{\sqrt{f^{2} - 1}} \Rightarrow u = \int \frac{d f}{\sqrt{f^{2} - 1}}

∴ {c o s h}^{- 1} f = \pm x + C

f = c o s h (C \pm x) C, x \in R \Rightarrow f : [a, b] \to R

L e t f = c o s h (C + x) . ∴ \frac{d f}{d x} = s i n h (C + x)

Missing \left or extra \right

\Rightarrow \sqrt{1 + {(\frac{d f}{d x})}^{2}} =∣ c o s h (x + C) ∣

N o w, c o s h (x + C) ⩾ 1 > 0 \forall x, C \in R .

∴ ∣ c o s h (x + C) ∣= c o s h (x + C)

\Rightarrow \sqrt{1 + {(\frac{d f}{d x})}^{2}} = c o s h (x + C)

L e t I = 2 π \int_{a}^{b} f \sqrt{1 + {(\frac{d f}{d x})}^{2}} d x = 2 π \int_{a}^{b} c o s h (x + C) \times c o s h (x + C) d x

I = 2 π \underset{a}{\int^{b}} {c o s h}^{2} (x + C) d x

N o w, l e t Q = π \int_{a}^{b} f^{2} d x = π \int_{a}^{b} {c o s h}^{2} (x + C) d x

∴ I = 2 Q \Rightarrow Q \neq I i f f = c o s h (x + C) .

W e d e d u c e I \neq Q a l s o i f f = c o s h (C - x) .

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