let f(x)=∫_(2x) ^(4x) (dt/(t^2 −2t +3)) 1)find f(x) 2) calculate lim_(x→0) f(x) and lim


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Question Number 57416 by Abdo msup. last updated on 03/Apr/19

$l e t f (x) = \int_{2 x}^{4 x} \frac{d t}{t^{2} - 2 t + 3}$ $1) f i n d f (x)$ $2) c a l c u l a t e {l i m}_{x \to 0} f (x) a n d {l i m}_{x \to + \infty} f (x)$
Commented by maxmathsup by imad last updated on 04/Apr/19
$1)we have f(x)=∫_(2x) ^(4x) (dt/(t^2 −2t+1+2)) =∫_(2x) ^(4x) (dt/((t−1)^2 +2)) =_(t−1=(√2)u) ∫_((2x−1)/(√2)) ^((4x−1)/(√2)) (((√2)du)/(2(1+u^2 ))) =(1/(√2)) [arctanu]_((2x−1)/(√2)) ^((4x−1)/(√2)) =(1/(√2)){arctan(((4x−1)/(√2)))−arctan(((2x−1)/(√2)))} 2) we have lim_(x→0) f(x)=(1/(√2)){ −arctan((1/(√2))) +arctan((1/(√2)))}=0 also lim_(x→+∞) f(x) =(1/(√2)){(π/2) −(π/2)} =0 .$
$1) w e h a v e f (x) = \int_{2 x}^{4 x} \frac{d t}{t^{2} - 2 t + 1 + 2} = \int_{2 x}^{4 x} \frac{d t}{{(t - 1)}^{2} + 2} =_{t - 1 = \sqrt{2} u} \int_{\frac{2 x - 1}{\sqrt{2}}}^{\frac{4 x - 1}{\sqrt{2}}} \frac{\sqrt{2} d u}{2 (1 + u^{2})}$ $= \frac{1}{\sqrt{2}} {[a r c t a n u]}_{\frac{2 x - 1}{\sqrt{2}}}^{\frac{4 x - 1}{\sqrt{2}}} = \frac{1}{\sqrt{2}} {a r c t a n (\frac{4 x - 1}{\sqrt{2}}) - a r c t a n (\frac{2 x - 1}{\sqrt{2}})}$ $2) w e h a v e {l i m}_{x \to 0} f (x) = \frac{1}{\sqrt{2}} {- a r c t a n (\frac{1}{\sqrt{2}}) + a r c t a n (\frac{1}{\sqrt{2}})} = 0 a l s o$ ${l i m}_{x \to + \infty} f (x) = \frac{1}{\sqrt{2}} {\frac{π}{2} - \frac{π}{2}} = 0 .$
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