>>112343412 + sin(x) is a wavy line bouncing between 3 and 1.
take the square root, it's a wavy line bouncing between and 1. So about 1.7 and 1.
Taking the integral of that, I know it will be greater than the integral of 1, but less than the integral of 1.7.
Denoting the value of the Integral as I(x), we know that but not necessarily .
Ok, so where does ? Some simple algebra should tell you at about 1.27 (or more exactly: . Therefore for all numbers past 1.27, it's proven.
Notice that , so this intersection occurs within the first upswing of the sine function. We really just need to prove it in this quadrant. How to do that? Well the derivative of the sine function is maximized at x=0, and this should be true for the square root of a sign function. Taking the derivative, we find that it's at x=0. The derivative of is . At x=0 this is , and it only increases with "x". Since we know this function is always going to be greater than the sine function (at least within that first quadrant of it's motion). So we can definitively say that it's always greater. Although I don't know why this is useful to do.
Note that at x=0, sin(x) has a slope of 1. A simple taylor expansion tells us the slope
