Find equation of quadratic function

Question

If I know the vertex (assume $(0,0)$), as well as the definite integral of a quadratic function from $0$ to $t$, is there a way to derive the actual equation of the function?

Answer 1

By the fundamental theorem of calculus $\frac d{dt} \int_a^t f(x)\, dx = f(t)$. So if you know the definite integral from $0$ to $t$ as a function of $t$ all you have to do is take the derivative.

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Edit: I misunderstood the question slightly. Suppose that the vertex of the polynomial is $(x_0,y_0)$ and for some $a$, $\int_{a}^b f(x)\,dx=b$. Then we know that for some $k$, $f(x)=k(x-x_0)^2+y_0$, so $\int_{a}^b f(x)\, dx = \int_{a}^b k(x-x_0)^2+y_0\,dx=\left[\frac{k}{3}(x-x_0)^3+y_0x\right]_a^b=\frac{k}{3}(a-x_0)^3+y_0a-\frac{k}{3}(b-x_0)^3+y_0b=c$. However, are simply numbers $a,b,c,x_0,y_0$,so you can substitute them in and solve for $k$. After doing that you the quadratic is $k(x-x_0)+y_0$.

For example, let's use the numbers in your comment. The vertex is $(0,0)$, $\int_0^{10} f(x)\,dx= 500$. Then, $f(x)=kx^2$. That means that \begin{align} 500 &= \int_0^{10} kx^2 \\ &= \left[\frac k 3 x^3\right]^{10}_0\\ &= \frac k 3(10)^3-\frac k 3 (0)^3 \\ &= \frac{1000k}3$. \end{align} Basic algebra givs that $k=\frac{15}{10}$ and so $f(x)=\frac{15}{10}x^2$.