Let $$\int_0^x \sqrt{1 - (y'(t))^2} \, dt = \int_0^x y(t) \, dt, \, 0 \le x \le 3, \, y \ge 0, \, y(0) = 0$$. Then at $$x = 2$$, $$y'' + y + 1$$ is equal to

A

1

B

2

C

$$\sqrt{2}$$

D

$$1/2$$