Lost on this problem:
Find y = f(x) given f^ii(x) = 12x^2 and f^i (1) = 10 and f(2) = 26
How long did you try to figure this out?
$\displaystyle f''(x) = 12x^2$
$\displaystyle \Rightarrow f'(x) = \int f''(x)~dx = 12 \int x^2 ~dx = 4x^3 + C$
Now $\displaystyle f'(1) = 10$
$\displaystyle \Rightarrow 10 = 4(1)^3 + C \implies C = 6$
So, $\displaystyle f'(x) = 4x^3 + 6$
$\displaystyle \Rightarrow f(x) = \int f'(x)~dx = \int \left( 4x^3 + 6 \right)~dx = x^4 + 6x + C$
Now $\displaystyle f(2) = 26$
$\displaystyle \Rightarrow 26 = (2)^4 + 6(2) + C \implies C = -2$
So, finally, $\displaystyle \boxed { f(x) = x^4 + 6x - 2 }$
$\displaystyle \displaystyle f''(x)=12x^2\Rightarrow f'(x)=12\cdot \frac{x^3}{3}+c_1=4x^3+c_1$
$\displaystyle f'(1)=10\Rightarrow 4+c_1=10\Rightarrow c_1=6\Rightarrow f'(x)=4x^3+6$
$\displaystyle \displaystyle \Rightarrow f(x)=4\cdot \frac{x^4}{4}+6x+c_2=x^4+6x+c_2$
$\displaystyle f(2)=26\Rightarrow c_2=-2\Rightarrow f(x)=x^4+6x-2$