Minimum value of an integral with least square?

Question

I have a problem. The question is: given a parabola $$p(t) = a + bt + ct^2$$ I need to evaluate the least squares straight line $$A + Bt$$ for which this integral is minimum:

$$ \int_{t_1}^{t_2} \left(p(t) - A - Bt\right)^2 dt $$

How could I solve this? I don't understand well.. :( Thanks to everybody!!!

Answer 1

I can calculate the integral but it's quite a mess.. I mean it should be:

$$\int_{t_1}^{t_2} \left(a + bt + ct^2 - A - Bt\right)^2 dt$$

I can make che shifts $\alpha = a - A$ and $\beta = b - B$ in order to get a simpler form:

$$\int_{t_1}^{t_2}\left(\alpha + \beta t + ct^2\right)^2 dt$$

in that way I can (Should I?) expand the bracket:

$$ \int_{t_1}^{t_2} \left(\alpha^2 + \beta^2 t^2 + c^2 t^4 + 2\alpha\beta t + 2\alpha ct^2 + 2\beta ct^3 \right) dt $$

and now?

Answer 2

HINT:

$$ E(A,B)=\int_{t_1}^{t_2} \left(p(t) - A - Bt\right)^2 dt $$

You want to minimize $E(A,B)$ thus

Find : $$\frac{\partial}{\partial A} (E(A,B))= 0$$ $$ \frac{\partial}{\partial B} (E(A,B))= 0$$ And then solve $A,B$

$$ \frac{\partial}{\partial A} (E(A,B))=\int_{t_1}^{t_2} \frac{\partial}{\partial A}( \left(p(t) - A - Bt\right)^2) dt$$

$$ \frac{\partial}{\partial B} (E(A,B))=\int_{t_1}^{t_2} \frac{\partial}{\partial B}( \left(p(t) - A - Bt\right)^2) dt$$

Hope it helps