We'll recall that the critical points of a function are the zeroes of the first derivative of the function.
Therefore, to determine the critical points of h(x), we'll have to find out the first derivative of h(x).
But h(x) = Int f(x)dx => h'(x) = f(x)
Since h(x) is the definite integral of the function f(x), we'll apply Leibniz Newton formula, to determine h'(x).
h'(x) = f(x) - f(0)
But h'(x) = 0 => f(x) - f(0) = 0 => f(x) = f(0)
We'll calculate f(0) = (e^0)*(0) = 0
f(x) = 0
(e^x)*(x^4) = 0
Since e^x cannot be zero for any value of x => x^4 = 0
x = 0
Since the derivative of the function h(x) has only one root, therefore the function has only one critical value, x = 0.