Recovering a quadratic polynomial from three values using calculus I'm asked to solve this using calculus:

Let $$ f(x) = ax^2 + bx +c .$$ If $ f(1) = 3 $,  $f(2) = 7$, $f(3) = 13$, then find $a$, $b$, and $f(0)$.

I know I can solve this using solving three equations simultaneously. And I can also solve this using Gauss Jordan or Gaussian elimination method by writing the augmented matrix.  But I'm wondering is there any other method to solve this. 
Solving by any method it turns out that $a = b = c = 1$.

Use calculus to solve this...  Well, I can wedge some calculus in here...
The derivative of $f(x) = a x^2 + b x + c$ with respect to $x$ is $2ax+b$.  By the Fundamental Theorem of Calculus, we have that \begin{align*}
    f(x) &= f(0) + \int_0^x 2 a t + b \,\mathrm{d}t  \\
         &= f(0) + (a t^2 + bt)_{t=x} - (a t^2 + bt)_{t=0}  \\
         &= f(0) + a x^2 + bx \text{.}
\end{align*}  (This is not a shocking revelation: $c = f(0)$.)
Consequently, \begin{align*}
    f(1) = 3 &= f(0) + \int_0^1 2 a t + b \,\mathrm{d}t = f(0) + a + b \text{,}  \\
    f(2) = 7 &= f(0) + \int_0^2 2 a t + b \,\mathrm{d}t = f(0) + 4a + 2b  \text{, and}  \\
    f(3) = 13 &= f(0) + \int_0^3 2 a t + b \,\mathrm{d}t = f(0) + 9a + 3b  \text{.}
\end{align*}
This has the convenient property that it exactly contains the expressions, $a$, $b$, and $f(0)$, that are sought in the Question.  Now we look at how the function changes between the given points.  \begin{align*}
    f(2) - f(1) = 4 &= \int_1^2 2 a t + b \,\mathrm{d}t = 3a+b \text{ and}  \\
    f(3) - f(2) = 6 &= \int_2^3 2 a t + b \,\mathrm{d}t = 5a + b\text{.}
\end{align*}
These tell us $2a=2$, or $a=1$.  Then the first tells us that $b=1$.  From $3 = f(1) = f(0) + a + b$, we find $f(0) = 1$.
