Alternating signs

$\begin{aligned} y & = \sqrt{x-\color{#3D99F6}\sqrt{x + \sqrt{x-\sqrt{x+\sqrt{\cdots}}}}} & \small \color{#3D99F6} \text{Let }z = \sqrt{x + \sqrt{x-\sqrt{x+\sqrt{\cdots}}}} \\ & = \sqrt{x-z} \\ y^2 & = x - z \quad ...(1) \end{aligned}$

$\begin{aligned} z & = \sqrt{x+\sqrt{x - \sqrt{x+\sqrt{x-\sqrt{\cdots}}}}} \\ z & = \sqrt{x+y} \\ z^2 & = x + y \quad ... (2) \end{aligned}$

Therefore,

$\begin{aligned} (2)-(1): \quad z^2 - y^2 & = y+z \\ (z+y)(z-y) & = y+z & \small \color{#3D99F6} \text{For } y \ne z \\ z-y & = 1 \\ z & = y+1 \end{aligned}$

$\begin{aligned} (1): \quad y^2 & = x - z \\ & = x - 1 - y \end{aligned}$

$\begin{aligned} y^2 + y + 1 - x & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }y \\ \implies y & = \frac {-1+\sqrt{1-4+4x}}2 & \small \color{#3D99F6} y > 0 \\ & = - \frac 12 + \sqrt{x - \frac 34} \end{aligned}$

Then we have:

$\begin{aligned} \int_{1.75}^{4.75} y \ dx & = \int_\frac 74 ^{\frac {19}4} - \frac 12 + \sqrt{x - \frac 34} \ dx \\ & = - \frac x2 + \frac 23 \left( x - \frac 34 \right)^\frac 32 \bigg|_\frac 74 ^{\frac {19}4} \\ & = - \frac 32 + \frac 23 (8-1) \\ & = \frac {19}6 \end{aligned}$

$\implies a + b = 19+6 = \boxed{25}$

Let $y= (x-z)^ {1/2}$ Hence z is the similar nested root alternating with a positive sign first.

Hence $z= (x+y) ^ {1/2}$

Now square both these expressions and subtract them. Since is y not equal to negative of z we get

$y=z-1$

Substitute this in the first expression and solve for z. Now obtain y and integrate

Good question Rohith

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