calculate ∫_0 ^1 (dx/(3+2^x ))


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Question Number 64159 by mathmax by abdo last updated on 14/Jul/19

$c a l c u l a t e \int_{0}^{1} \frac{d x}{3 + 2^{x}}$
Commented by turbo msup by abdo last updated on 15/Jul/19
$chsngement 2^x =t give e^(xln(2)) =t ⇒ xln(2)=ln(t)⇒x=((ln(t))/(ln2)) ∫_0 ^1 (dx/(3+2^x )) =∫_1 ^2 (dt/(tln2(3+t))) =(1/(3ln2)) ∫_1 ^2 ((1/t)−(1/(3+t)))dt =(1/(3ln2))[ln((t/(t+3)))]_1 ^2 =(1/(3ln2)){ln((2/5))−ln((1/4))} =(1/(3ln2)){ln2−ln5+2ln2} =(1/(3ln2)){3ln2−ln5}$
$c h s n g e m e n t 2^{x} = t g i v e e^{x l n (2)} = t \Rightarrow$ $x l n (2) = l n (t) \Rightarrow x = \frac{l n (t)}{l n 2}$ $\int_{0}^{1} \frac{d x}{3 + 2^{x}} = \int_{1}^{2} \frac{d t}{t l n 2 (3 + t)}$ $= \frac{1}{3 l n 2} \int_{1}^{2} (\frac{1}{t} - \frac{1}{3 + t}) d t$ $= \frac{1}{3 l n 2} {[l n (\frac{t}{t + 3})]}_{1}^{2}$ $= \frac{1}{3 l n 2} {l n (\frac{2}{5}) - l n (\frac{1}{4})}$ $= \frac{1}{3 l n 2} {l n 2 - l n 5 + 2 l n 2}$ $= \frac{1}{3 l n 2} {3 l n 2 - l n 5}$
	Answered by Hope last updated on 15/Jul/19

	$f (x) = \frac{1}{3 + 2^{x}}$ $f (0) = \frac{1}{4} = 0.25$ $f (1) = \frac{1}{5} = 0.2$ $\int_{0}^{1} f (x) d x = \frac{1}{2} (0.25 + 0.20) \times 1 = 0.225 (a p p r o x)$ $T a n m a y i s h o p e . . .$
	Commented by Hope last updated on 15/Jul/19

	$s t i l l h o p e p r e s e n t . . i h a v e c h a n g e d m y p r o f i l e n a m e$ $b e c a u s e m y a n s w e r d o n o t g e t a c c e p t e d b y a [f e w$ $s o i t r i e d t o m i n i m i s e m y a c t i v i t y i n t h i s$ $p l a t f o r m . . 0 k g o o d b y e$
	Commented by mathmax by abdo last updated on 15/Jul/19

	$s i r t a n m a y y o u a r e w r o n g i f y o u s a y t h i s w h a t s y o u r p r o o f$ $t h a t y o u r a n s w e r i s n o t a c c e p t e d y o u a r e a l w a y s a c t i f i n t h a t$ $f o r u m a n d c o n s i d e r e d . . . .$
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