<s>[INST] Find $f$ such that $f'(x)=ax^2+bx$, given the values of $f'(1)$, $f''(1)$, and $\int_0^2 f(x)\,dx$ The question is : Find the solution $f(x)$ if $f'(x)=ax^2+bx$, and 
(i) $f'(1)=6$, 
  (ii)  $f''(1)=18$,
  (iii) $\int_0^2 f(x)dx=18$.
My solution is:
According to $(i)$, we know $6=a+b$, and $a=6-b$.
Since $f''(x)=2ax+b$, and according to $(ii)$, we know $18=2a+b$.
Therefore, $b=-6, a=12$. And, $f'(x)=12x^2-6x$.
Now, we have
$$
f(x)=\int f'(x)dx=\int (12x^2-6x)dx=4x^3-3x^2+C
$$
According to $(iii)$,
$$
\bigl |_0^2 f(x) = 4*8-12-0=20
$$
It means $(\bigr|_0^2 f(x) )+ C=18$, and $C=-2$. The solution of $f(x)$ is $4x^3-3x^2-2$.
But the correct answer should be $4x^3-3x^2+5$. Where is my mistake?
Thank you!!
 [/INST] You used $f'$ instead of $f$ in (iii), which should read $$18=\int_0^2f(x)\,\mathrm dx=\int_0^2(4x^3-3x^2+C)\,\mathrm dx=\left.x^4-x^3+Cx\,\right|_0^2=8+2C.$$
</s>