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.

You are correct.  Now do the integral, plug in those values, and see what you get.  The last unknown cancels out and you get a numerical answer.  
