4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

Good job, you are almost there.

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.