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-

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 = \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 \dots .

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 \dots .

Commented by maxmathsup by imad last updated on 26/Mar/19

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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