let f(x) =∫_0 ^1 (dt/(x +2^t )) with x>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/(1+2^t )) and ∫


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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 bymathmax by abdo last updated on 14/Jul/19
$1)we have f(x)=∫_0 ^1 (dt/(x+2^t )) (x>0) changement 2^t =u give e^(tln2) =u ⇒tln(2)=lnu ⇒t=((ln(u))/(ln(2))) ⇒f(x) =∫_1 ^2 (du/(ln(2)u(x+u))) =(1/(ln(2))) (1/x)∫_1 ^2 {(1/u)−(1/(x+u))}du =(1/(xln(2)))[ln((u/(x+u)))]_1 ^2 =(1/(xln(2))){ln((2/(x+2)))−ln((1/(x+1)))}=(1/(xln(2))){ln2−ln(x+2)+ln(x+1)} f(x)=(1/x) +((ln(((x+1)/(x+2))))/(xln(2)))$
$1) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{x + 2^{t}} (x > 0) c h a n g e m e n t 2^{t} = u g i v e$ $e^{t l n 2} = u \Rightarrow t l n (2) = l n u \Rightarrow t = \frac{l n (u)}{l n (2)} \Rightarrow f (x) = \int_{1}^{2} \frac{d u}{l n (2) u (x + u)}$ $= \frac{1}{l n (2)} \frac{1}{x} \int_{1}^{2} {\frac{1}{u} - \frac{1}{x + u}} d u = \frac{1}{x l n (2)} {[l n (\frac{u}{x + u})]}_{1}^{2}$ $= \frac{1}{x l n (2)} {l n (\frac{2}{x + 2}) - l n (\frac{1}{x + 1})} = \frac{1}{x l n (2)} {l n 2 - l n (x + 2) + l n (x + 1)}$ $f (x) = \frac{1}{x} + \frac{l n (\frac{x + 1}{x + 2})}{x l n (2)}$
Commented bymathmax by abdo last updated on 14/Jul/19
$2) we have f^′ (x) =−∫_0 ^1 (dt/((x+2^t )^2 )) =−g(x) ⇒g(x)=−f^′ (x) we have f(x)=(1/x){1 +((ln(x+1)−ln(x+2))/(ln2))} ⇒ f^′ (x) =−(1/x^2 ){1+((ln(x+1)−ln(x+2))/(ln2))}+(1/x){(1/((x+1)ln2))−(1/((x+2)ln(2)))} ⇒g(x)=(1/x^2 ){1+((ln(x+1)−ln(x+2))/(ln2))}−(1/x){(1/((x+1)ln2))−(1/((x+2)ln2))}$
$2) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{d t}{{(x + 2^{t})}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $w e h a v e f (x) = \frac{1}{x} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} \Rightarrow$ $f^{^{'}} (x) = - \frac{1}{x^{2}} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} + \frac{1}{x} {\frac{1}{(x + 1) l n 2} - \frac{1}{(x + 2) l n (2)}}$ $\Rightarrow g (x) = \frac{1}{x^{2}} {1 + \frac{l n (x + 1) - l n (x + 2)}{l n 2}} - \frac{1}{x} {\frac{1}{(x + 1) l n 2} - \frac{1}{(x + 2) l n 2}}$
Commented bymathmax 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 bymathmax 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 bymathmax 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

	$2) S a m e i d e a i t s g (x) = \int \frac{d t}{{(x + 2^{t})}^{2}} ?$ $f^{(n)} l i b n e i z f o r m u l a e x p l i c i t e$ $w i t h e i n t e g r a l j u s t s w i t h e d e r i v a t i o n w i t h e i n t e g r a l$ $f^{n} (x) = \int \frac{d}{{d x}^{n}} \frac{d t}{(x + 2^{t})} = \int \frac{{(- 1)}^{n} (n)!}{{(x + 2^{t})}^{n + 1}} d t$ $4 . p u t x = 1 i n f (x) 2 n d x = 1 i n f^{^{'}} (x)$
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