Calculate the value

AREAS 
$4$ The area bounded by $y^{2}=4x$ with the lines 
$x=2$ and $x=5$ is 
$\dfrac {8} {3}\left(5\sqrt{5} -2\sqrt{2} \right)$ $2\right)$ $\dfrac {4} {3}\left(5\sqrt{5} -2\sqrt{2} \right)$ 
$1\right)$ $\dfrac {8} {3}\left(5\sqrt{5} $ 
$3\right)$ $5\sqrt{5} -2\sqrt{2} $ $4\right)$ $\dfrac {16} {3}$ 

What is the inverse of $g\left(x\right)=\sqrt{x-1} $ 
$Og\left(x\right)=\sqrt{x+1} $ 
$+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ 
$Og^{'}\left(x\right)=x^{2}+1$ 
$○g\left(x\right)=x^{2}-1$ 
« Previous Ne 

Retiep 



$1\right)$ 

$2\sqrt{6} \dfrac {-.\sqrt{5} } {\sqrt{5} -2\sqrt{6} }$ $3$ 

355-256 



The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these 

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