Answer
Error $(S_2)=0$, so $S_2$ gives the exact value of the integral.
Work Step by Step
We are given that $f(x)=x-x^3$
$f\prime (x)=1-3x^2; f\prime \prime (x)=-6x; f\prime \prime \prime (x)=-6 $ and $f \prime\prime\prime \prime=0$
Error $(S_N) \leq \dfrac{k_4(b-a)^3}{180N^4} $
For $N=2$, we have:
Error $(S_N) \leq \dfrac{k_4(b-a)^3}{180N^4} \\ \leq 0$
We see that Error $(S_2)=0$, so $S_2$ gives the exact value of the integral.
