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.
 A: 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!
A: 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.  
A: It could be helpful to do the integration step as well:
\begin{align}
\int_{-1}^1p(x)dx&=\int_{-1}^1\left(ax^3+bx^2+cx+d\right)\,\text{d}x\\
&=\color{blue}{\frac{a}{4}x^4\big|_{x=-1}^{x=1}}+\color{red}{\frac b3x^3 \big|_{x=-1}^{x=1}}+\color{green}{\frac c2 x^2\big|_{x=-1}^{x=1}}+\color{black}{dx\big|_{x=-1}^{x=1}}\\
&=\color{blue}{0}+\color{red}{\frac b3 \times 2} + \color{green}{0} + \color{black}{d \times 2}
\end{align}
Note that with $d=2$ you have $a+b+c=1$ and $-a+b-c=2$. If you sum the latter euqations on both sides you obtain $2b=3$ or $b=\frac32$. Hence your integral becomes $$\frac{\frac32}{3}\times 2+2 \times 2 = 5.$$
A: As the integration range is symmetric, you can only care about the even part of the polynomial,
$$e(x):=\frac{p(x)+p(-x)}2=bx^2+d,$$ giving the integral
$$I=2\int_0^1(bx^2+d)\,dx.$$
Clearly $e(0)=d=2$ and $e(1)=b+d=\dfrac{3+4}2$, so that $I=2\left(\dfrac32\dfrac13+2\right)=5$.
