let f(x) =(√(x^2 +3))−2x −1 1) calculate ∫_0 ^1 f(x)dx 2)determine f^(−1) (x) and calculate ∫ f^(−1) (x)dx .


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Question Number 56831 by maxmathsup by imad last updated on 24/Mar/19

$l e t f (x) = \sqrt{x^{2} + 3} - 2 x - 1$ $1) c a l c u l a t e \int_{0}^{1} f (x) d x$ $2) d e t e r m i n e f^{- 1} (x) a n d c a l c u l a t e \int f^{- 1} (x) d x .$
Commented by kaivan.ahmadi last updated on 25/Mar/19
$I=∫_0 ^1 (√(x^2 +3))dx x=(√3)tgα⇒dx=(√3)sec^2 αdα (√(x^2 +3))=(√3)secα { ((x=0⇒α=0)),((x=1⇒α=(π/6))) :}⇒ I=3∫_0 ^(π/6) sec^3 αdα= (1/2)secαtgα+(1/2)ln∣secα+tgα∣]_0 ^(π/6) = ((1/2)×((√3)/2)×((√3)/3)+(1/2)ln(((√3)/2)+((√3)/3)))−0= (1/4)+(1/2)ln(((5(√3))/6)) J=∫_0 ^1 (−2x−1)dx=−x^2 −x]_0 ^1 = (−1−1)−0=−2 ∫_0 ^1 f(x)dx=I+J=((−7)/4)+(1/2)ln(((5(√3))/6))$
$I = \int_{0}^{1} \sqrt{x^{2} + 3} d x$ $x = \sqrt{3} t g α \Rightarrow d x = \sqrt{3} {s e c}^{2} α d α$ $\sqrt{x^{2} + 3} = \sqrt{3} s e c α$ ${\begin{cases} x = 0 \Rightarrow α = 0 \\ x = 1 \Rightarrow α = \frac{π}{6} \end{cases} \Rightarrow$ $I = 3 \int_{0}^{\frac{π}{6}} {s e c}^{3} α d α =$ ${\frac{1}{2} s e c α t g α + \frac{1}{2} l n ∣ s e c α + t g α ∣]}_{0}^{\frac{π}{6}} =$ $(\frac{1}{2} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{3} + \frac{1}{2} l n (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{3})) - 0 =$ $\frac{1}{4} + \frac{1}{2} l n (\frac{5 \sqrt{3}}{6})$ ${J = \int_{0}^{1} (- 2 x - 1) d x = - x^{2} - x]}_{0}^{1} =$ $(- 1 - 1) - 0 = - 2$ $\int_{0}^{1} f (x) d x = I + J = \frac{- 7}{4} + \frac{1}{2} l n (\frac{5 \sqrt{3}}{6})$
Commented by kaivan.ahmadi last updated on 24/Mar/19

$y = \sqrt{x^{2} + 3} - 2 x - 1 \Rightarrow y + 1 = \sqrt{x^{2} + 3} - 2 x$ $\Rightarrow {(y + 1)}^{2} = x^{2} + 3 + 4 x^{2} - 4 x \sqrt{x^{2} + 3} \Rightarrow$ ${(y + 1)}^{2} = 5 x^{2} - 4 x \sqrt{x^{2} + 3} + 3 =$ $5 x^{2} - 4 x (y + 2 x + 1) + 3 =$ $- 3 x^{2} - 4 x y - 4 x + 3 \Rightarrow$ $- 3 x^{2} - (4 y + 4) x + 3 - {(y + 1)}^{2} = 0 \Rightarrow$ $Δ = {(4 y + 4)}^{2} + 36 - 12 {(y + 1)}^{2} =$ $4 {(y + 1)}^{2} + 36 = 4 ({(y + 1)}^{2} + 9) =$ $x = \frac{(4 y + 4) \pm \sqrt{Δ}}{- 6} = \frac{4 (y + 1) \pm 2 \sqrt{{(y + 1)}^{2} + 9}}{- 6} =$ $\Rightarrow f^{- 1} (x) = \frac{- 2}{3} (x + 1) \pm \frac{- 1}{3} \sqrt{{(x + 1)}^{2} + 9}$
Commented by Abdo msup. last updated on 25/Mar/19

$s i r A h m a d i w h e r a r e y o u f r o m . . . .$
Commented by kaivan.ahmadi last updated on 25/Mar/19

$H i d e a r A b d o, i^{'} m I r a n i a n$
Commented by maxmathsup by imad last updated on 25/Mar/19

$w i c h y o u g o o d l u c k s i r i h o p e t o v i s i t I r a n s o m e d a y s . . . .$
Commented by maxmathsup by imad last updated on 26/Mar/19
$1) ∫_0 ^1 f(x)dx =∫_0 ^1 (√(x^2 +3))dx −∫_0 ^1 (2x+1)dx ∫_0 ^1 (2x+1)dx =[x^2 +x]_0 ^1 =2 ∫_0 ^1 (√(x^2 +3))dx =_(x=(√3)sh(t)) ∫_0 ^(arsh((1/(√3)))) (√3)ch(t)(√3)ch(t)dt =3 ∫_0 ^(argsh((1/(√3)))) ((1+ch(2t))/2)dt =(3/2) argsh((1/(√3))) +(3/4) [sh(2t)]_0 ^(argsh((1/(√3)))) =(3/2)ln((1/(√3)) +(√(4/3))) +(3/8)[e^(2t) −e^(−2t) ]_0 ^(argsh((1/(√3)))) =(3/2)ln((√3)) +(3/8){ 3−(1/3)} =(3/4)ln(3) +1 ⇒∫_0 ^1 f(x)dx =(3/4)ln(3)−1 .$
$1) \int_{0}^{1} f (x) d x = \int_{0}^{1} \sqrt{x^{2} + 3} d x - \int_{0}^{1} (2 x + 1) d x$ $\int_{0}^{1} (2 x + 1) d x = {[x^{2} + x]}_{0}^{1} = 2$ $\int_{0}^{1} \sqrt{x^{2} + 3} d x =_{x = \sqrt{3} s h (t)} \int_{0}^{a r s h (\frac{1}{\sqrt{3}})} \sqrt{3} c h (t) \sqrt{3} c h (t) d t = 3 \int_{0}^{a r g s h (\frac{1}{\sqrt{3}})} \frac{1 + c h (2 t)}{2} d t$ $= \frac{3}{2} a r g s h (\frac{1}{\sqrt{3}}) + \frac{3}{4} {[s h (2 t)]}_{0}^{a r g s h (\frac{1}{\sqrt{3}})} = \frac{3}{2} l n (\frac{1}{\sqrt{3}} + \sqrt{\frac{4}{3}}) + \frac{3}{8} {[e^{2 t} - e^{- 2 t}]}_{0}^{a r g s h (\frac{1}{\sqrt{3}})}$ $= \frac{3}{2} l n (\sqrt{3}) + \frac{3}{8} {3 - \frac{1}{3}} = \frac{3}{4} l n (3) + 1 \Rightarrow \int_{0}^{1} f (x) d x = \frac{3}{4} l n (3) - 1 .$
Commented by kaivan.ahmadi last updated on 26/Mar/19

$t h a n k s i r, w h e r e a r e y o u f r o m ?$
Commented by maxmathsup by imad last updated on 26/Mar/19

$i a m f r o m m o r o c c o s i r .$
Commented by maxmathsup by imad last updated on 30/Mar/19

$2) f (x) = y \Leftrightarrow x = f^{- 1} (y) a n d f (x) = y \Rightarrow \sqrt{x^{2} + 3} - 2 x - 1 = y \Rightarrow$ $(2 x + 1 + y) = \sqrt{x^{2} + 3} \Rightarrow {(2 x + 1 + y)}^{2} = x^{2} + 3 \Rightarrow$ ${(2 x + 1)}^{2} + 2 (2 x + 1) y + y^{2} = x^{2} + 3 \Rightarrow$ $4 x^{2} + 4 x + 1 + 4 y x + 2 y + y^{2} - x^{2} - 3 = 0 \Rightarrow$ $3 x^{2} + (4 + 4 y) x + y^{2} + 2 y - 2 = 0 l e t f i n d x i n t e r m s o f y$ $Δ^{,} = {(2 + 2 y)}^{2} - 3 (y^{2} + 2 y - 2) = 4 + 8 y + 4 y^{2} - 3 y^{2} - 6 y + 6$ $= y^{2} + 2 y + 10 = {(y + 1)}^{2} + 9 \Rightarrow x_{1} = \frac{- 2 (y + 1) + \sqrt{{(y + 1)}^{2} + 9}}{3}$ $x_{2} = \frac{- 2 (y + 1) - \sqrt{{(y + 1)}^{2} + 9}}{3} w e m u s t h a v e 2 x + 1 + y ⩾ 0 l e t v e r i f y t h i s$ $c o n d i t i o n$ $2 x_{1} + 1 + y = \frac{- 4 y - 4 + 2 \sqrt{{(y + 1)}^{2} + 9} + 3 + 3 y}{3} = \frac{2 \sqrt{y^{2} + 2 y + 10} - y - 1}{3}$ $w e h a v e 4 (y^{2} + 2 y + 10) - {(y + 1)}^{2} = 4 y^{2} + 6 y + 40 - y^{2} - 2 y - 1$ $= 3 y^{2} + 4 y + 39 > \Rightarrow x_{1} i s t h e s o l u t i o n \Rightarrow f^{- 1} (x) = \frac{\sqrt{{(x + 1)}^{2} + 9} - 2 (x + 1)}{3}$
Commented by maxmathsup by imad last updated on 30/Mar/19

$\int f^{- 1} (x) d x = \frac{1}{3} \int \sqrt{{(x + 1)}^{2} + 9} d x - \frac{2}{3} \int (x + 1) d x$ $\int (x + 1) d x = \frac{x^{2}}{2} + x + c_{1}$ $\int \sqrt{{(x + 1)}^{2} + 9} d x =_{x + 1 = 3 s h (t)} \int 3 c h (t) 3 c h (t) = 9 \int {c h}^{2} t d t$ $= 9 \int \frac{1 + c h (2 t)}{2} d t = \frac{9}{2} t + \frac{9}{4} s h (2 t) = \frac{9}{2} t + \frac{9}{2} c h (t) s h (t) + c_{2}$ $= \frac{9}{2} a r g s h (\frac{x + 1}{3}) + \frac{9}{2} \frac{x + 1}{3} \sqrt{1 + {(\frac{x + 1}{3})}^{2}}$ $= \frac{9}{2} l n (\frac{x + 1}{3} + \sqrt{1 + {(\frac{x + 1}{3})}^{2}}) + \frac{3}{2} (x + 1) \sqrt{1 + {(\frac{x + 1}{3})}^{2}} + c_{2} \Rightarrow$ $\int f^{- 1} (x) d x = \frac{3}{z} l n (\frac{x + 1}{3} + \sqrt{1 + {(\frac{x + 1}{3})}^{2}}) + \frac{x + 1}{2} \sqrt{1 + {(\frac{x + 1}{3})}^{2}} - \frac{1}{3} x^{2} - \frac{2}{3} x + C$
	Answered by MJS last updated on 25/Mar/19

	$y = \sqrt{x^{2} + 3} - 2 x - 1$ $\Rightarrow x = - \frac{1}{3} (2 y + 2 \pm \sqrt{y^{2} + 2 y + 10})$ $but we have to square the equation to get$ $this \Rightarrow one solution could be wrong$ $with y = x we get$ $x = \sqrt{x^{2} + 3} - 2 x - 1 \Rightarrow x = \frac{1}{4}$ $x = - \frac{1}{3} (2 (x + 1) - \sqrt{x^{2} + 2 x + 10}) \Rightarrow x = \frac{1}{4}$ $x = - \frac{1}{3} (2 (x + 1) + \sqrt{x^{2} + 2 x + 10}) \Rightarrow x = - 1$ $\Rightarrow f^{- 1} (x) = - \frac{1}{3} (2 (x + 1) - \sqrt{x^{2} + 2 x + 10})$ $I get$ $\int f (x) d x = - x (x + 1) + \frac{1}{2} x \sqrt{x^{2} + 3} + \frac{3}{2} \ln (x + \sqrt{x^{2} + 3}) + C$ $\int f^{- 1} (x) d x = - \frac{1}{3} x (x + 2) + \frac{1}{6} (x + 1) \sqrt{x^{2} + 2 x + 10} + \frac{3}{2} \ln (x + 1 + \sqrt{x^{2} + 2 x + 10}) + C$
	Commented by kaivan.ahmadi last updated on 25/Mar/19

	$i f F (x) = x (x + 1) + \frac{1}{2} x \sqrt{x^{2} + 3} + \frac{3}{2} l n (x + \sqrt{x^{2} + 3}) \Rightarrow$ $t h e n f (x) = F^{'} (x)$ $M r M J S c a n y o u p r o v e i t ?$
	Commented by maxmathsup by imad last updated on 25/Mar/19

	$t h a n k s s i r .$
	Commented by MJS last updated on 25/Mar/19

	$F (x) = - x (x + 1) + \frac{1}{2} x \sqrt{x^{2} + 3} + \frac{3}{2} \ln (x + \sqrt{x^{2} + 3})$ $F^{'} (x) = \frac{d}{d x} [- x (x + 1)] + \frac{d}{d x} [\frac{1}{2} x \sqrt{x^{2} + 3}] + \frac{d}{d x} [\frac{3}{2} \ln (x + \sqrt{x^{2} + 3})]$ $\frac{d}{d x} [- x (x + 1)] = - 2 x - 1$ $\frac{d}{d x} [\frac{1}{2} x \sqrt{x^{2} + 3}] = \frac{1}{2} (\sqrt{x^{2} + 3} + \frac{x^{2}}{\sqrt{x^{2} + 3}}) = \frac{2 x^{2} + 3}{2 \sqrt{x^{2} + 3}}$ $\frac{d}{d x} [\frac{3}{2} \ln (x + \sqrt{x^{2} + 3})] = \frac{3}{2} (\frac{1}{x + \sqrt{x^{2} + 3}} (1 + \frac{x}{\sqrt{x^{2} + 3}})) =$ $= \frac{3}{2} (\frac{x - \sqrt{x^{2} + 3}}{- 3} \times \frac{x + \sqrt{x^{2} + 3}}{\sqrt{x^{2} + 3}}) = \frac{3}{2 \sqrt{x^{2} + 3}}$ $- 2 x - 1 + \frac{2 x^{2} + 3}{2 \sqrt{x^{2} + 3}} + \frac{3}{2 \sqrt{x^{2} + 3}} =$ $- 2 x - 1 + \frac{2 x^{2} + 6}{2 \sqrt{x^{2} + 3}} = - 2 x - 1 + \frac{x^{2} + 3}{\sqrt{x^{2} + 3}} =$ $= \sqrt{x^{2} + 3} - 2 x - 1$
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