The two-point open Newton-Cotes formula provides an estimate of the integral

$$\int_{-1}^{2} (x^3 - x^2 + x)dx$$

The error term is expressed in terms of $f''(z)$. The problem asks to find $z$.

I know that the two-point open Newton-Cotes formula is

$$\int_{x_{-1}}^{x_{2}} f(x)dx = \frac{3h}{2}[f(x_{0}) + f(x_{1})] + \frac{3h^3}{4}f''(z)$$ where $$x_{-1} < z < x_{2}$$

We know that $h=\frac{2-(-1)}{1+2}=1$, so $x_{-1}=-1, x_{0}=0, x_{1}=1,$ and $x_{2}=2$. With the provided information, I have found that

$$\int_{-1}^{2} (x^3 - x^2 + x)dx = \frac{3}{2} + \frac{3}{4}f''(z)$$

At this point, I can find bounds for the error, but I'm not sure what it would mean to find $z$. Using integration gives 2.25 for the above definite integral. According to my homework, the answer is $\frac{1}{2}$. Perhaps there is a typo in regards to what to find?


Good job, you are almost there.

We have $f''(z) = 6z-2$, so we solve

$$\dfrac{9}{4} = \dfrac{3}{2} + \dfrac{3}{4} (6 z - 2) \implies z = \dfrac{1}{2}$$


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