## Calculus (3rd Edition)

Error $(S_2)=0$, so $S_2$ gives the exact value of the integral.
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.