Find a function f(x) such that f'(x)=8x^2+5 and f (0) =1
You are looking for the antiderivative of f'(x)
The antiderivative of x2 is x3/3, that of x0 is x1/1 (Note that x0 is just 1 and x1/1 is just x).
So f(x) = 8x3/3 + 5x + k where k is a constant to be found by using the fact that f(0) = 1.