let-f-x-0-1-dt-x-2-t-with-x-gt-0-1-determine-a-explicit-form-for-f-x-2-determine-also-g-x-0-1-dt-x-2-x-2-3-give-f-n-x-at-form-of-integral-4-calculate-0-1-dt

Question Number 64160 by mathmax by abdo last updated on 14/Jul/19

l e t f (x) = \int_{0}^{1} \frac{d t}{x + 2^{t}} w i t h x > 0

1) d e t e r m i n e a e x p l i c i t f o r m f o r f (x)

2) d e t e r m i n e a l s o g (x) = \int_{0}^{1} \frac{d t}{{(x + 2^{x})}^{2}}

3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l

4) c a l c u l a t e \int_{0}^{1} \frac{d t}{1 + 2^{t}} a n d \int_{0}^{1} \frac{d t}{{(1 + 2^{t})}^{2}}

Commented by mathmax by abdo last updated on 14/Jul/19

1) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{x + 2^{t}} (x > 0) c h a n g e m e n t 2^{t} = u g i v e

e^{t l n 2} = u \Rightarrow t l n (2) = l n u \Rightarrow t = \frac{l n (u)}{l n (2)} \Rightarrow f (x) = \int_{1}^{2} \frac{d u}{l n (2) u (x + u)}

= \frac{1}{l n (2)} \frac{1}{x} \int_{1}^{2} {\frac{1}{u} - \frac{1}{x + u}} d u = \frac{1}{x l n (2)} {[l n (\frac{u}{x + u})]}_{1}^{2}

= \frac{1}{x l n (2)} {l n (\frac{2}{x + 2}) - l n (\frac{1}{x + 1})} = \frac{1}{x l n (2)} {l n 2 - l n (x + 2) + l n (x + 1)}

f (x) = \frac{1}{x} + \frac{l n (\frac{x + 1}{x + 2})}{x l n (2)}

Commented by mathmax by abdo last updated on 14/Jul/19

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{d t}{{(x + 2^{t})}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)

w e h a v e f (x) = \frac{1}{x} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} \Rightarrow

f^{^{'}} (x) = - \frac{1}{x^{2}} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} + \frac{1}{x} {\frac{1}{(x + 1) l n 2} - \frac{1}{(x + 2) l n (2)}}

\Rightarrow g (x) = \frac{1}{x^{2}} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} - \frac{1}{x} {\frac{1}{(x + 1) l n 2} - \frac{1}{(x + 2) l n 2}}

Commented by mathmax by abdo last updated on 14/Jul/19

3) w e h a v e f^{(n)} (x) = \int_{0}^{1} \frac{\partial^{n}}{\partial x^{n}} (\frac{1}{x + 2^{t}}) d t

= \int_{0}^{1} \frac{{(- 1)}^{n} n!}{{(x + 2^{t})}^{n + 1}} d t = {(- 1)}^{n} n! \int_{0}^{1} \frac{d t}{{(x + 2^{t})}^{n + 1}}

Commented by mathmax by abdo last updated on 14/Jul/19

4) \int_{0}^{1} \frac{d t}{1 + 2^{t}} = f (1) = 1 + \frac{l n (\frac{2}{3})}{l n (2)} = 1 + \frac{l n (2) - l n (3)}{l n 2} = 2 - \frac{l n (3)}{l n 2}

Commented by mathmax by abdo last updated on 14/Jul/19

\int_{0}^{1} \frac{d t}{{(1 + 2^{t})}^{2}} = g (1) = 1 + \frac{l n (2) - l n (3)}{l n 2} - (\frac{1}{2 l n 2} - \frac{1}{3 l n 2})

= 2 - \frac{l n (3)}{l n 2} + (\frac{1}{3} - \frac{1}{2}) \frac{1}{l n (2)} = 2 - \frac{l n 3}{l n 2} - \frac{1}{6 l n 2}

Answered by Eminem last updated on 14/Jul/19

1) l e t u = 2^{t} d t = \frac{d u}{u l n (2)}

f (x) = \int_{1}^{2} \frac{d u}{u l n (2) (x + u)} = \int_{1}^{2} \frac{d u}{x u l n (2)} - \int_{1}^{2} \frac{d u}{x l n (2) (x + u)} = \frac{1}{x} - \frac{1}{x l n (2)} [l n (x + 2) - l n (x + 1)]

Answered by Eminem last updated on 14/Jul/19

2) S a m e i d e a i t s g (x) = \int \frac{d t}{{(x + 2^{t})}^{2}} ?

f^{(n)} l i b n e i z f o r m u l a e x p l i c i t e

w i t h e i n t e g r a l j u s t s w i t h e d e r i v a t i o n w i t h e i n t e g r a l

f^{n} (x) = \int \frac{d}{{d x}^{n}} \frac{d t}{(x + 2^{t})} = \int \frac{{(- 1)}^{n} (n)!}{{(x + 2^{t})}^{n + 1}} d t

4 . p u t x = 1 i n f (x) 2 n d x = 1 i n f^{^{'}} (x)

Facebook Tweet Pin