f " (x) = X^2 + X -1, f ' (0)=5, f (0)=10

First we need to determine f'(x) by calculating the integral of f''(x)

We know that:

f'(x) = intg f''(x)

==> f'(x) = intg (x^2 + x -1) dx

             = x^3/3 + x^2/2 - x + C

==> f'(x) = (1/3)x^2 + (1/2)x^2 - x + C

Given that:

f'(0) = 5

==> f'(x) = (1/3)x^3 + (1/2)x^2 - x + 5

Now let us calculate f(x)"

We know that:

f(x) = intg f'(x)

        = intg ((1/3)x^3 + (1/2)x^2 - x + 5) dx

         = (1/3)x^4/4 + (1/2)x^3/3 - x^2/2 + 5x + C

        = (1/12)x^4 + (1/6)x^3 - (1/2)x^2 + 5x + C

Given that:

f(0) = 10 ==> c = 10

==> f(x) =  (1/12)x^4 + (1/6)x^3 - (1/2)x^2 + 5x + 10