Given-f-x-0-x-dt-f-t-2-and-0-2-dt-f-t-2-6-1-3-Then-then-the-value-of-f-9-is-

Question Number 118663 by benjo_mathlover last updated on 19/Oct/20

G i v e n f (x) = \int \underset{0}{\overset{x}{}} \frac{d t}{{[f (t)]}^{2}} a n d \int \underset{0}{\overset{2}{}} \frac{d t}{{[f (t)]}^{2}} = \sqrt[3]{6}

T h e n t h e n t h e v a l u e o f f (9) i s__

Commented by mr W last updated on 19/Oct/20

f^{'} (x) = \frac{1}{{(f (x))}^{2}}

y^{2} d y = d x

\frac{y^{3}}{3} = x + \frac{C}{3}

y^{3} = 3 x + C

{(\sqrt[3]{6})}^{3} = 3 \times 2 + C

\Rightarrow C = 0

\Rightarrow y^{3} = 3 x

\Rightarrow y = f (x) = \sqrt[3]{3 x}

f (9) = \sqrt[3]{3 \times 9} = 3

Commented by 1549442205PVT last updated on 19/Oct/20

Great! Sir .

Answered by benjo_mathlover last updated on 19/Oct/20

b y T h e o r e m F u n d a m e n t a l C a l c u l u s - 1

f^{'} (x) = \underset{0}{\int^{x}} f (t) d t . G i v e n e q u a t i o n

f (x) = \underset{0}{\int^{x}} \frac{d t}{{[f (t)]}^{2}} \Rightarrow f^{'} (x) = \frac{1}{{[f (x)]}^{2}}

\Rightarrow {[f (x)]}^{2} d (f (x)) = d x, i n t e g r a t i n g

b o t h s i d e s \int {[f (x)]}^{2} d x = \int d x

\frac{1}{3} {[f (x)]}^{3} = x + C

\Rightarrow s o f (x) = \sqrt[3]{3 x + λ}, λ = 3 C

t h u s \int_{0}^{2} \frac{d t}{{[f (t)]}^{2}} = f (2) = \sqrt[3]{6}

w e g e t f (2) = \sqrt[3]{6 + λ} = \sqrt[3]{6}, g i v e λ = 0

T h u s f (x) = \sqrt[3]{3 x} a n d f (9) = \sqrt[3]{27} = 3