Consider the ellipse
$$\frac{x^2}{4} + \frac{y^2}{2} =1$$
The line integral $I$ of the ellipse in the first quadrant is
$$I=\int^2_0 \Big[ 1+(y'(x))^2 \Big]^{1/2} dx$$
Find the cubic polynomial $P_3(x)$ that interpolates y(x) in the first quadrant at $x=0, 1, \sqrt{2}, 2.$ We would then have
$$I \approx \int^2_0 \Big[ 1+(P'(x))^2 \Big]^{1/2}$$
Find the cubic polynomial $Q_3 (x)$ that interpolates $\sqrt{1+(P_3' (x)')^2}$ at these points. Hence estimate $I.$
Attempt
Rearranging $y=\sqrt{2 \left( 1- \frac{x^2}{4} \right)}$ we get
We will have the following Lagrange polynomial:
$$p_3(x) = y(x_0) \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0 - x_1)(x_0-x_2) (x_0-x_3)}+y(x_1) \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1 - x_0)(x_1-x_2) (x_1-x_3)} +y(x_2) \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2 - x_0)(x_2-x_1) (x_2-x_3)} + y(x_3) \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3 - x_0)(x_3-x_1) (x_3-x_2)}$$
But when I substitute the numbers in, the resulting polynomial looks extremely messy so I am not sure if I am on the right track. And when we have this polynomial how do we exactly need to proceed to find $Q_3 (x)$?
Any help is appreciated.