Suppose we know that f"(x) = 5x^4 + 2x^3 + x + 3 and f'(0) = 4 and f(1) = 9.883333.  What is f(x)?

We have to integrate f''(x) first to get f'(x).

f'(x) = x^5 + (1/4)x^4 + (1/2)x^2 + 3x + c, since f'(0) = 4, we get

4 = 0^5 + (1/4)(0)^4 +(1/2)(0)^2 + 3(0) = c, therefore c = 4

f'(x) = x^5 + (1/4)x^4 + (1/2)x^2 + 3x + 4.  We get f(x) by integrating f'(x).

Therefore f(x) = (1/6)x^6 + (1/20)x^5 + (1/6)x^3 + (3/2)x^2 + 3x + c. Since we know f(1) = 9.883333, we get

9.883333 = 1/6 + 1/20 + 1/6 + 3/2 + 3  + c.   Therefore, c =5 and f(x) is

f(x) = (1/6)x^6 + (1/20)x^5 + (1/6)x^3 + (3/2)x^2 + 3x + 5.