Let-f-x-2-x-dt-1-t-6-Prove-that-1-730-lt-f-3-lt-1-65-

Question Number 50186 by rahul 19 last updated on 14/Dec/18

L e t f (x) = \int_{2}^{x} \frac{d t}{1 + t^{6}} .

P r o v e t h a t : \frac{1}{730} < f (3) < \frac{1}{65} .

Answered by mr W last updated on 16/Dec/18

F (t) = \frac{1}{1 + t^{6}} i s a d e c r e a s i n g f u n c t i o n

f (3) = \int_{2}^{3} \frac{d t}{1 + t^{6}}

f (3) < F (2) \times (3 - 2) = F (2) = \frac{1}{1 + 2^{6}} = \frac{1}{1 + 64} = \frac{1}{65}

f (3) > F (3) \times (3 - 2) = F (3) = \frac{1}{1 + 3^{6}} = \frac{1}{1 + 9 \times 9 \times 9} = \frac{1}{1 + 729} = \frac{1}{730}

\Rightarrow \frac{1}{730} < f (3) < \frac{1}{65}

Commented by mr W last updated on 14/Dec/18

Commented by mr W last updated on 15/Dec/18

i t i s e x a c t a s y o u c a n s e e i n t h e d i a g r a m .

t h e r e e x i s t s s u c h a p o i n t ξ . w e k n o w

i t e x i s t s, b u t w e {d o n}^{'} t k n o w w h e r e i t

i s .

\int_{a}^{b} F (t) d t = a r e a u n d e r t h e c u r v e f r o m a t o b .

i f F (t) i s d e c r e a s i n g, {i t}^{'} s o b v i o u s t h a t

\int_{a}^{b} F (t) d t < (b - a) F (a) = r e c t a n g l e w i t h m a x . h e i g h t

\int_{a}^{b} F (t) d t > (b - a) F (b) = r e c t a n g l e w i t h m i n . h e i g h t

i f F (t) i s i n c r e a s i n g, {i t}^{'} s o b v i o u s t h a t

\int_{a}^{b} F (t) d t > (b - a) F (a)

\int_{a}^{b} F (t) d t < (b - a) F (b)

Commented by rahul 19 last updated on 15/Dec/18

S i r, f (3) = F (ξ) (3 - 2) i s j u s t a n

a p p r o x i m a t i o n, n o t e x a c t l y e q u a l, r i g h t ?

T h a n k y o u S i r!

Commented by rahul 19 last updated on 15/Dec/18

Perfect explaination!��