left[ int_{0}^{2} sqrt{x + sqrt{x + sqrt{x + | vedclass

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(A) Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.Squaring both sides,we get $y^2 = x + y$,which implies $y^2 - y - x = 0$.Solving for $y$ u

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$3$

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(A) Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.Squaring both sides,we get $y^2 = x + y$,which implies $y^2 - y - x = 0$.Solving for $y$ using the quadratic formula,$y = \frac{1 + \sqrt{1 + 4x}}{2}$ (since $y > 0$).Now,we need to evaluate $I = \int_{0}^{2} \frac{1 + \sqrt{1 + 4x}}{2} \, dx$.$I = \frac{1}{2} \int_{0}^{2} 1 \, dx + \frac{1}{2} \int_{0}^{2} (1 + 4x)^{1/2} \, dx$.$I = \frac{1}{2} [x]_{0}^{2} + \frac{1}{2} \left[ \frac{(1 + 4x)^{3/2}}{3/2 \cdot 4} \right]_{0}^{2}$.$I = \frac{1}{2} (2) + \frac{1}{12} [(1 + 8)^{3/2} - (1 + 0)^{3/2}]$.$I = 1 + \frac{1}{12} [27 - 1] = 1 + \frac{26}{12} = 1 + \frac{13}{6} = \frac{19}{6} \approx 3.166$.The Greatest Integer Function $[I] = [3.166] = 3$.

$\left[ \int_{0}^{2} \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}} \, dx \right]$ is equal to (where $[\cdot]$ is the $G.I.F.$)

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