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Question Number 87538 by Power last updated on 04/Apr/20

Commented by mathmax by abdo last updated on 04/Apr/20
$I =∫_(−1) ^1 f^(−1) (x)dx changement f^(−1) (x)=t give x =f(t)⇒ dx =f^′ (t)dt ⇒ I =∫_(f^(−1) (−1)) ^(f^(−1) (1)) t f^′ (t)dt =_(by parts) =[(t^2 /2)f(t)]_(f^(−1) (−1)) ^(f^(−1) (1)) −∫_(f^(−1) (−1)) ^(f^(−1) (1)) (t^2 /2)f(t)dt =(1/2){ (f^(−1) (1))^2 + (f^(−1) (−1))^2 }−(1/2)∫_(f^(−1) (−1)) ^(f^(−1) (1)) t^2 (t^3 +t−1)dt we hsve ∫_(f^(−1) (−1)) ^(f^(−1) (1)) (t^5 +t^3 −t^2 )dt [(t^6 /6)+(t^4 /4)−(t^3 /3)]_(f^(−1) (−1)) ^(f^(−1) (1)) rest calculus of f^(−1) (1) and f^(−1) (−1) ...be continued...$
$I = \int_{- 1}^{1} f^{- 1} (x) d x c h a n g e m e n t f^{- 1} (x) = t g i v e x = f (t) \Rightarrow$ $d x = f^{^{'}} (t) d t \Rightarrow I = \int_{f^{- 1} (- 1)}^{f^{- 1} (1)} t f^{^{'}} (t) d t =_{b y p a r t s}$ $= {[\frac{t^{2}}{2} f (t)]}_{f^{- 1} (- 1)}^{f^{- 1} (1)} - \int_{f^{- 1} (- 1)}^{f^{- 1} (1)} \frac{t^{2}}{2} f (t) d t$ $= \frac{1}{2} {{(f^{- 1} (1))}^{2} + {(f^{- 1} (- 1))}^{2}} - \frac{1}{2} \int_{f^{- 1} (- 1)}^{f^{- 1} (1)} t^{2} (t^{3} + t - 1) d t$ $w e h s v e$ $\int_{f^{- 1} (- 1)}^{f^{- 1} (1)} (t^{5} + t^{3} - t^{2}) d t$ ${[\frac{t^{6}}{6} + \frac{t^{4}}{4} - \frac{t^{3}}{3}]}_{f^{- 1} (- 1)}^{f^{- 1} (1)} r e s t c a l c u l u s o f f^{- 1} (1) a n d f^{- 1} (- 1)$ $. . . b e c o n t i n u e d . . .$
Commented by Power last updated on 04/Apr/20

$sir step by step solution please$
Commented by MJS last updated on 04/Apr/20

$\frac{5}{4}$
Commented by mathmax by abdo last updated on 04/Apr/20
$let solve f(x)=y ⇒x^3 +x−1 =y let x =u+v ⇒ (u+v)^3 +u+v−1−y =0 ⇒u^3 +v^3 +3uv(u+v) +u+v −1−y =0 ⇒ u^3 +v^3 −1−y +(u+v)(3uv +1) =0 ⇒ { ((u^3 +y^3 =1+y)),((uv =−(1/3))) :} ⇒ { (( u^3 +v^3 =1+y)),((u^3 v^3 =−(1/(27)))) :} ⇒ u^3 and v^3 are solution of X^2 −(1+y)X −(1/(27)) =0 Δ =(1+y)^2 +(4/(27)) ⇒X_1 =((1+y +(√((1+y)^2 +(4/(27)))))/2) X_2 =((1+y−(√((1+y)^2 +(4/(27)))))/2) ⇒ u =^3 (√((1+y+(√((1+y)^2 +(4/(27)))))/2)) and v =^3 (√((1+y−(√((1+y)^2 +(4/(27)))))/2)) or u =−^3 (√(...)) and v =−^3 (√(....)) ( uv<0) ⇒ f^(−1) (x) =^3 (√((1+x+(√((1+x)^2 +(4/(27)))))/2)) +^3 (√((1+x−(√((1+x)^2 +(4/(27)))))/2)) ⇒f^(−1) (−1) =^3 (√(1/(27)))−^3 (√(1/(27))) =0 f^(−1) (1) =^3 (√((2+(√(4+(4/(27)))))/2)) +^3 (√((2−(√(4+(4/(27)))))/2))$
$l e t s o l v e f (x) = y \Rightarrow x^{3} + x - 1 = y l e t x = u + v \Rightarrow$ ${(u + v)}^{3} + u + v - 1 - y = 0 \Rightarrow u^{3} + v^{3} + 3 u v (u + v) + u + v - 1 - y = 0 \Rightarrow$ $u^{3} + v^{3} - 1 - y + (u + v) (3 u v + 1) = 0 \Rightarrow {\begin{cases} u^{3} + y^{3} = 1 + y \\ u v = - \frac{1}{3} \end{cases}$ $\Rightarrow {\begin{cases} u^{3} + v^{3} = 1 + y \\ u^{3} v^{3} = - \frac{1}{27} \end{cases}$ $\Rightarrow u^{3} a n d v^{3} a r e s o l u t i o n o f X^{2} - (1 + y) X - \frac{1}{27} = 0$ $Δ = {(1 + y)}^{2} + \frac{4}{27} \Rightarrow X_{1} = \frac{1 + y + \sqrt{{(1 + y)}^{2} + \frac{4}{27}}}{2}$ $X_{2} = \frac{1 + y - \sqrt{{(1 + y)}^{2} + \frac{4}{27}}}{2} \Rightarrow$ $u =^{3} \sqrt{\frac{1 + y + \sqrt{{(1 + y)}^{2} + \frac{4}{27}}}{2}} a n d v =^{3} \sqrt{\frac{1 + y - \sqrt{{(1 + y)}^{2} + \frac{4}{27}}}{2}}$ $o r u = -^{3} \sqrt{. . .} a n d v = -^{3} \sqrt{. . . .} (u v < 0) \Rightarrow$ $f^{- 1} (x) =^{3} \sqrt{\frac{1 + x + \sqrt{{(1 + x)}^{2} + \frac{4}{27}}}{2}} +^{3} \sqrt{\frac{1 + x - \sqrt{{(1 + x)}^{2} + \frac{4}{27}}}{2}}$ $\Rightarrow f^{- 1} (- 1) =^{3} \sqrt{\frac{1}{27}} -^{3} \sqrt{\frac{1}{27}} = 0$ $f^{- 1} (1) =^{3} \sqrt{\frac{2 + \sqrt{4 + \frac{4}{27}}}{2}} +^{3} \sqrt{\frac{2 - \sqrt{4 + \frac{4}{27}}}{2}}$
Commented by Power last updated on 05/Apr/20

$sir pls solution$
Commented by Power last updated on 05/Apr/20

$Mjs sir 5 / 4 correct answer$
Commented by mathmax by abdo last updated on 05/Apr/20

$t r y t o f i n i s h t h e a n s w e r s i r . . .$
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