let f(t) =∫_0 ^3 (√(t +x +x^2 ))dx with t ≥(1/4) 1) find a explicit form of f(t) 2) find also g(t) = ∫_0 ^3 (dx/(√(t+x +x^2 ))) 3) calculate ∫_0 ^3 (√(1+x+x^2 ))dx , ∫_0 ^3 (√(2 +x+x^2 ))dx ∫


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Question Number 60498 by abdo mathsup 649 cc last updated on 21/May/19

$l e t f (t) = \int_{0}^{3} \sqrt{t + x + x^{2}} d x w i t h t ⩾ \frac{1}{4}$ $1) f i n d a e x p l i c i t f o r m o f f (t)$ $2) f i n d a l s o g (t) = \int_{0}^{3} \frac{d x}{\sqrt{t + x + x^{2}}}$ $3) c a l c u l a t e \int_{0}^{3} \sqrt{1 + x + x^{2}} d x, \int_{0}^{3} \sqrt{2 + x + x^{2}} d x$ $\int_{0}^{3} \frac{d x}{\sqrt{2 + x + x^{2}}} .$
Commented by maxmathsup by imad last updated on 22/May/19
$let x^2 +x +t →Δ=1−4t but 4t≥1 ⇒4t−1≥0 ⇒1−4t ≤0 case 1 1−4t <0 ⇒x^2 +x +t =x^2 +2(x/2) +(1/(4 )) +t−(1/4) =(x+(1/2))^2 +((4t−1)/4) we do the changement (x+(1/2)) =((√(4t−1))/2)u ⇒u=((2x+1)/(√(4t−1))) f(t) =∫_(1/(√(4t−1))) ^(7/(√(4t−1))) ((√(4t−1))/2)(√(1+u^2 ))((√(4t−1))/2) du =((4t−1)/4) ∫_(1/(√(4t−1))) ^(7/(√(4t−1))) (√(1+u^2 ))du =_(u =sh(α)) ((4t−1)/4) ∫_(argsh((1/(√(4t−1))))) ^(argsh((7/(√(4t−1))))) ch(α)ch(α)dα =((4t−1)/8) ∫_(ln((1/(√(4t−1))) +(√(1+(1/(4t−1)))))) ^(ln((7/(√(4t−1))) +(√(1+((49)/(4t−1)))))) (1+ch(2t))dt =((4t−1)/8){ln((7/(√(4t−1)))+(√((4t+48)/(4t−1))))−ln((1/(√(4t−1)))+((2(√t))/(√(4t−1))))} +((4t−1)/(16))[ sh(2t)]_(ln((1/((√(4t))−1)) +(√((4t)/(4t−1))))) ^(ln((7/((√(4t))−1)) +(√((4t+48)/(4t−1))))) =((4t−1)/8){ ln((7/(√(4t−1))) +(√((4t+48)/(4t−1))))−ln((1/(√(4t−1))) +(√((4t)/(4t−1))))} ((4t−1)/(32))[ e^(2t) −e^(−2t) ]_(...) ^(...) ⇒ f(t)=((4t−1)/8){ ln((7/(√(4t−1))) +(√((4t+48)/(4t−1)))) −ln((1/(√(4t−1))) +(√((4t)/(4g−1))))} +((4t−1)/(32)){ ((7/(√(4t−1))) +(√((4t+48)/(4t−1))))^2 −(1/(((7/(√(4t−1)))+(√((4t +48)/(4t−1))))^2 )) −((1/(√(4t−1))) +(√((4t)/(4t−1))))^2 +(1/(((1/(√(4t−1)))+(√((4t)/(4t−1))))^2 ))}$
$l e t x^{2} + x + t \to Δ = 1 - 4 t b u t 4 t ⩾ 1 \Rightarrow 4 t - 1 ⩾ 0 \Rightarrow 1 - 4 t ⩽ 0$ $c a s e 1 1 - 4 t < 0 \Rightarrow x^{2} + x + t = x^{2} + 2 \frac{x}{2} + \frac{1}{4} + t - \frac{1}{4} = {(x + \frac{1}{2})}^{2} + \frac{4 t - 1}{4}$ $w e d o t h e c h a n g e m e n t (x + \frac{1}{2}) = \frac{\sqrt{4 t - 1}}{2} u \Rightarrow u = \frac{2 x + 1}{\sqrt{4 t - 1}}$ $f (t) = \int_{\frac{1}{\sqrt{4 t - 1}}}^{\frac{7}{\sqrt{4 t - 1}}} \frac{\sqrt{4 t - 1}}{2} \sqrt{1 + u^{2}} \frac{\sqrt{4 t - 1}}{2} d u = \frac{4 t - 1}{4} \int_{\frac{1}{\sqrt{4 t - 1}}}^{\frac{7}{\sqrt{4 t - 1}}} \sqrt{1 + u^{2}} d u$ $=_{u = s h (α)} \frac{4 t - 1}{4} \int_{a r g s h (\frac{1}{\sqrt{4 t - 1}})}^{a r g s h (\frac{7}{\sqrt{4 t - 1}})} c h (α) c h (α) d α$ $= \frac{4 t - 1}{8} \int_{l n (\frac{1}{\sqrt{4 t - 1}} + \sqrt{1 + \frac{1}{4 t - 1}})}^{l n (\frac{7}{\sqrt{4 t - 1}} + \sqrt{1 + \frac{49}{4 t - 1}})} (1 + c h (2 t)) d t$ $= \frac{4 t - 1}{8} {l n (\frac{7}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t + 48}{4 t - 1}}) - l n (\frac{1}{\sqrt{4 t - 1}} + \frac{2 \sqrt{t}}{\sqrt{4 t - 1}})}$ $+ \frac{4 t - 1}{16} {[s h (2 t)]}_{l n (\frac{1}{\sqrt{4 t} - 1} + \sqrt{\frac{4 t}{4 t - 1}})}^{l n (\frac{7}{\sqrt{4 t} - 1} + \sqrt{\frac{4 t + 48}{4 t - 1}})}$ $= \frac{4 t - 1}{8} {l n (\frac{7}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t + 48}{4 t - 1}}) - l n (\frac{1}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t}{4 t - 1}})}$ $\frac{4 t - 1}{32} {[e^{2 t} - e^{- 2 t}]}_{. . .}^{. . .} \Rightarrow$ $f (t) = \frac{4 t - 1}{8} {l n (\frac{7}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t + 48}{4 t - 1}}) - l n (\frac{1}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t}{4 g - 1}})}$ $+ \frac{4 t - 1}{32} {{(\frac{7}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t + 48}{4 t - 1}})}^{2} - \frac{1}{{(\frac{7}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t + 48}{4 t - 1}})}^{2}}$ $- {(\frac{1}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t}{4 t - 1}})}^{2} + \frac{1}{{(\frac{1}{\sqrt{4 t - 1}} + \sqrt{\frac{4 t}{4 t - 1}})}^{2}}}$
Commented by maxmathsup by imad last updated on 22/May/19

$c a s e 2 t = \frac{1}{4} \Rightarrow f (t) = \int_{0}^{3} \sqrt{x^{2} + x + \frac{1}{4}} d x = \int_{0}^{3} \sqrt{{(x + \frac{1}{2})}^{2}} d x$ $= \int_{0}^{3} (x + \frac{1}{2}) d x = {[\frac{x^{2}}{2} + \frac{1}{2} x]}_{0}^{3} = \frac{9}{2} + \frac{3}{2} = 6$ $2) w e h a v e f^{^{'}} (t) = \int_{0}^{3} \frac{d x}{2 \sqrt{t + x + x^{2}}} = \frac{1}{2} g (t) \Rightarrow g (x) = 2 f^{^{'}} (t) r e s t t o {c a l c u l a t e f}^{^{'}} (t)$
Commented by maxmathsup by imad last updated on 22/May/19
$3) ∫_0 ^3 (√(1+x+x^2 ))dx =f(1) =(3/8){ln((7/(√3)) +((√(52))/(√3)))−ln((1/(√3)) +(2/(√3)))} +(3/(32)){ ((7/(√3)) +((√(52))/(√3)))^2 −(1/(((7/(√3))+((√(52))/(√3)))^2 )) −((1/(√3)) +(2/(√3)))^2 +(1/(((1/(√3)) +(2/(√3)))^2 ))} . ∫_0 ^3 (√(2+x+x^2 ))dx =f(2)=....$
$3) \int_{0}^{3} \sqrt{1 + x + x^{2}} d x = f (1) = \frac{3}{8} {l n (\frac{7}{\sqrt{3}} + \frac{\sqrt{52}}{\sqrt{3}}) - l n (\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}$ $+ \frac{3}{32} {{(\frac{7}{\sqrt{3}} + \frac{\sqrt{52}}{\sqrt{3}})}^{2} - \frac{1}{{(\frac{7}{\sqrt{3}} + \frac{\sqrt{52}}{\sqrt{3}})}^{2}} - {(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}^{2} + \frac{1}{{(\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}^{2}}} .$ $\int_{0}^{3} \sqrt{2 + x + x^{2}} d x = f (2) = . . . .$
Commented by maxmathsup by imad last updated on 22/May/19

$l e t c a l c u l a t e I = \int_{0}^{3} \frac{d x}{\sqrt{x^{2} + x + 2}} w e h a v e x^{2} + x + 2 = x^{2} + x + \frac{1}{4} + 2 - \frac{1}{4}$ $= {(x + \frac{1}{2})}^{2} + \frac{7}{4} l e t u s e t h e c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{7}}{2} t a n θ \Rightarrow 2 x + 1 = \sqrt{7} t a n θ$ $I = \int_{a r c t a n (\frac{1}{\sqrt{7}})}^{a r c t a n (\sqrt{7})} \frac{1}{\sqrt{\frac{7}{4} (1 + {t a n}^{2} θ})} \frac{\sqrt{7}}{2} d θ = \int_{a r c t a n (\frac{1}{\sqrt{7}})}^{a r c t a n (\sqrt{7})} \frac{d θ}{\sqrt{1 + {t a n}^{2} θ}}$ $= \int_{a r c t a n (\frac{1}{\sqrt{7}})}^{a r c t a n (\sqrt{7})} c o s θ d θ = {[s i n θ]}_{a r c t a n (\frac{1}{\sqrt{7}})}^{a r c t a n (\sqrt{7})} = s i n (a r c t a n (\sqrt{7})) - s i n (a r c t a n (\frac{1}{\sqrt{7}}))$ $s i n (a r c t a n x) = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow s i n (a r c t a n (\sqrt{7})) = \frac{\sqrt{7}}{\sqrt{8}} = \frac{\sqrt{7}}{2 \sqrt{2}}$ $s i n (a r c t a n (\frac{1}{\sqrt{7}})) = \frac{1}{\sqrt{7} \sqrt{1 + \frac{1}{7}}} = \frac{1}{\sqrt{7} \frac{\sqrt{8}}{\sqrt{7}}} = \frac{1}{2 \sqrt{2}} \Rightarrow$ $I = \frac{\sqrt{7} - 1}{2 \sqrt{2}} .$
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