Okay, I am moving my comment as an answer:
Use derivative test to see if the function is increasing or decreasing. A strictly increasing/ decreasing function must be injective, so it can have at most one zero. Note that the function $e^x + x - 2$ takes both positive and negative values by evaluating at 0 and 1. What can you conclude by the intermediate value theorem?
Also, you should not try to find extremum points to solve this question, for the zeroes of $e^x + x -2$ need not be extremum points.
You mention distance and velocity. Yes, in this case finding the area under a curve is used. However, this situation can be generalized. The relationship between velocity of distance is that velocity is the first derivative of distance (displacement) with respect to time. Therefore, integrating (finding the area under the curve) of velocity with respect to time gives you change in displacement.
In general, when you have an quantity changing with respect to time - a rate - (or with respect to anything, technically), you would integrate that (area under curve) to find how much the quantity changed.
For example, I could say that that the amount of water in a tank increases at a certain rate (a number of gallons per minute). I could then integrate (find the area under) that curve to find how much the water in the tank increased over the given time.
I could take it further and say that the rate by which the amount of water in the tank is changing, is changing. This is the second derivative (akin to acceleration when talking about displacement) of the amount of water in the tank. Find the area under this curve would tell you how much the rate of water into the tank changed over the given time.
And so on. This is calculus.
Best Answer
Let $f(x) = x^2 - x\sin x - \cos x$, for $x \ge 0$. Then $f$ is continuous with $f(0) = -1 < 0$ and $f(\pi) = \pi^2 + 1 > 0$. So by the intermediate value theorem, $f$ has a zero in $(0,\pi)$. Since
$$f'(x) = 2x - x\cos x = x(2 - \cos x) \ge x(2 - 1) = x > 0$$
on $(0,\infty)$, $f$ is strictly increasing on $(0,\infty)$. Therefore, $f$ has a unique root in $(0,\infty)$.