let f(x)=2x−(√(x−1)) find ∫ f^(−1) (x)dx .


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Question Number 41136 by math khazana by abdo last updated on 02/Aug/18

$l e t f (x) = 2 x - \sqrt{x - 1}$ $f i n d \int f^{- 1} (x) d x .$
	Answered by MJS last updated on 02/Aug/18
	$f(x) defined for x∈[1; +∞] f′(x)=2−(1/(2(√(x−1))))=0 ⇒ x=((17)/(16)) f′′(x)=(1/(4(x−1)^(3/2) )); f′′(((17)/(16)))=16>0 ⇒ min at ((((17)/(16))),(((15)/8)) ) range(f(x))=[((15)/8); +∞[ x=2y−(√(y−1)) (√(y−1))=2y−x y−1=4y^2 −4xy+x^2 y^2 −(x+(1/4))y+((x^2 +1)/4)=0 y_1 =(x/2)+(1/8)−((√(8x−15))/8) defined for x∈[((15)/8); +∞] y_1 ′=(1/2)−(1/(2(√(8x−15))))=0 ⇒ x=2 y_1 ′′=(2/((8x−15)^(3/2) )); x=2 ⇒ y_2 ′′=2>0 ⇒ min at ((2),(1) ) range=[1; +∞] y_2 =(x/2)+(1/8)+((√(8x−15))/8) defined for x∈[((15)/8); +∞] y_2 ′=(1/2)+(1/(2(√(8x−15))))=0 ⇒ no solution range=[((17)/(16)); +∞] f^(−1) (x)= { (((x/2)+(1/8)−((√(8x−15))/8); x∈[((15)/8); 2])),(((x/2)+(1/8)+((√(8x−15))/8); x∈]((15)/8); +∞[)) :} ∫f^(−1) (x)dx= { (((x^2 /4)+(x/8)−(((8x−15)^(3/2) )/(96)); x∈[((15)/8); 2])),(((x^2 /4)+(x/8)+(((8x−15)^(3/2) )/(96)); x∈]((15)/8); +∞[)) :} but of course if you want the area between this function and the x−axis you must take ∫f^(−1) (x)dx= { (((x^2 /4)+(x/8)−(((8x−15)^(3/2) )/(96)); x∈[((15)/8); 2])),(((x^2 /4)+(x/8)+(((8x−15)^(3/2) )/(96)); x∈]2; +∞[)) :}$
	$f (x) defined for x \in [1; + \infty]$ $f^{'} (x) = 2 - \frac{1}{2 \sqrt{x - 1}} = 0 \Rightarrow x = \frac{17}{16}$ $f^{″} (x) = \frac{1}{4 {(x - 1)}^{\frac{3}{2}}}; f^{″} (\frac{17}{16}) = 16 > 0 \Rightarrow \min at (\begin{matrix} \frac{17}{16} \\ \frac{15}{8} \end{matrix})$ $range (f (x)) = [\frac{15}{8}; + \infty [$ $x = 2 y - \sqrt{y - 1}$ $\sqrt{y - 1} = 2 y - x$ $y - 1 = 4 y^{2} - 4 x y + x^{2}$ $y^{2} - (x + \frac{1}{4}) y + \frac{x^{2} + 1}{4} = 0$ $y_{1} = \frac{x}{2} + \frac{1}{8} - \frac{\sqrt{8 x - 15}}{8}$ $defined for x \in [\frac{15}{8}; + \infty]$ $y_{1}^{'} = \frac{1}{2} - \frac{1}{2 \sqrt{8 x - 15}} = 0 \Rightarrow x = 2$ $y_{1}^{″} = \frac{2}{{(8 x - 15)}^{\frac{3}{2}}}; x = 2 \Rightarrow y_{2}^{″} = 2 > 0 \Rightarrow \min at (\begin{matrix} 2 \\ 1 \end{matrix})$ $range = [1; + \infty]$ $y_{2} = \frac{x}{2} + \frac{1}{8} + \frac{\sqrt{8 x - 15}}{8}$ $defined for x \in [\frac{15}{8}; + \infty]$ $y_{2}^{'} = \frac{1}{2} + \frac{1}{2 \sqrt{8 x - 15}} = 0 \Rightarrow no solution$ $range = [\frac{17}{16}; + \infty]$ $f^{- 1} (x) = {\begin{cases} \frac{x}{2} + \frac{1}{8} - \frac{\sqrt{8 x - 15}}{8}; x \in [\frac{15}{8}; 2] \\ \frac{x}{2} + \frac{1}{8} + \frac{\sqrt{8 x - 15}}{8}; x \in] \frac{15}{8}; + \infty [ \end{cases}$ $\int f^{- 1} (x) d x = {\begin{cases} \frac{x^{2}}{4} + \frac{x}{8} - \frac{{(8 x - 15)}^{\frac{3}{2}}}{96}; x \in [\frac{15}{8}; 2] \\ \frac{x^{2}}{4} + \frac{x}{8} + \frac{{(8 x - 15)}^{\frac{3}{2}}}{96}; x \in] \frac{15}{8}; + \infty [ \end{cases}$ $but of course if you want the area between$ $this function and the x - axis you must take$ $\int f^{- 1} (x) d x = {\begin{cases} \frac{x^{2}}{4} + \frac{x}{8} - \frac{{(8 x - 15)}^{\frac{3}{2}}}{96}; x \in [\frac{15}{8}; 2] \\ \frac{x^{2}}{4} + \frac{x}{8} + \frac{{(8 x - 15)}^{\frac{3}{2}}}{96}; x \in] 2; + \infty [ \end{cases}$
	Commented by math khazana by abdo last updated on 03/Aug/18

	$t h a n k y o u s i r m j s f o r t h i s h a r d w o r k .$
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