If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$, then what is $\int_{-1}^{1} p(x)dx$? Q.If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$,Then $\int_{-1}^{1} p(x)dx$=__?
My attempt:

Let $p(x)$ be $ax^3+bx^2+cx+d$
$p(0)=d=2$
$p(1)=a+b+c+d=3$
$p(-1)=-a+b-c+d=4$

From them,we get $b=3.5$,$d=2$ and $a+c=-0.5$
I could not progress any further.

This method will work.
However easier is to apply Simpson's rule for approximating an integral (https://en.wikipedia.org/wiki/Simpson%27s_rule). You have been given all the required information from the polynomial and fortunately Simpson's rule is exact for polynomials up to degree 3!
