Infinite Radicals!

$x + 2 = \sqrt{(x + 2)^2} = \sqrt{1 + (x^2 + 4x + 3)} = \sqrt{1 + (x + 1)(x + 3)} =$

$\sqrt{1 + (x + 1)\sqrt{(x + 3)^2}} = \sqrt{1 + (x + 1)\sqrt{1 + x^2 + 6x + 8}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)(x + 4)}} = \sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{(x + 4)^2}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + x^2 + 8x + 15}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)(x + 5)}}} =$

$\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{(x + 5)^2}}}} =$

To show that this pattern continues indefinitely we just need to show:

$\sqrt{ 1 + (x + n)\sqrt{(x + n + 2)^2}} =$ $\sqrt{1 + (x + n)\sqrt{1 + (x + n + 1)(x + n + 3)}}$

but it's fairly obvious that $(x + n + 2)^2 = (x + n)^2 + 4(x + n) + 4$

and

$1 + (x + n + 1)(x + n + 3) = 1 + (x + n)^2 + 4(x + n) + 3 = (x + n)^2 + 4(x + n) + 4$

$\therefore f(x) = x + 2 = \sqrt{x + \sqrt{(x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + ...}}}}}$

$\implies \displaystyle\int_{1}^{a} \sqrt{x + \sqrt{(x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + ...}}}}} \:\ dx =$

$\displaystyle\int_{1}^{a} x + 2 \:\ dx = \dfrac{x^2}{2} + 2x|_{1}^{a} = \dfrac{a^2 + 4a - 5}{2} = \dfrac{7}{2} \implies a^2 + 4a - 12 = 0 \implies (a + 6)(a - 2) = 0 \:\ a > 1 \implies a = \boxed{2}$ .

$$

Alternative approach to finding the area:

The graph above shows the desired area of the trapezoid(Area under the curve

$f(x) = x + 2$ on $[1,a]$ . The graph also shows the function(red line)

$\sqrt{x + \sqrt{(x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + (x + 4)\sqrt{1 + (x + 5)\sqrt{1 + (x + 6)\sqrt{1 + (x + 7)}}}}}}}}$ .

$$

The Area of the trapezoid $A = \dfrac{1}{2}(a + 5)(a - 1) = \dfrac{7}{2} \implies a^2 + 4a - 12 = 0 \implies (a + 6)(a - 2) = 0 \:\ a > 1 \implies a = \boxed{2}$