Explanation needed: $g \,\colon [0,2] \to \Bbb R \,$ be a twice continuously differentiable function

Question

I am stuck on the following problem:

Let $\,\,g \, \colon [0,2] \to \Bbb R \,$ be a twice continuously differentiable function. If $\displaystyle \int_{0}^{2}g(x)dx \approx 2 g(1),$ then the error in the approximation is :

$\frac{g'(\xi)}{12}$ for some $\,\,\xi \in (0,2)$

$\frac{g'(\xi)}{2}$ for some $\,\,\xi \in (0,2)$

$\frac{g''(\xi)}{3}$ for some $\,\,\xi \in (0,2)$

$\frac{g''(\xi)}{6}$ for some $\,\,\xi \in (0,2)$

Can someone help me out ?

If you don't have to prove the result, you can pick the right answer by considering the function $g(x)=(x-1)^2$. Calculate the error in the approximation, and the second derivative, and compare them. — TonyK, Jun 02 '13 at 08:55
@TonyK By taking $g(x)=(x-1)^2,$ I get $g(1) \approx \frac {1}{12}$. Now what? — learner, Jun 02 '13 at 09:07
I just noticed that those are first derivatives in your question. Shouldn't they be second derivatives? — TonyK, Jun 02 '13 at 09:34
This smells like "use Taylor's theorem with remainder", doesn't it? — kahen, Jun 02 '13 at 18:17

score 0 · Accepted Answer · answered Jun 03 '13 at 04:52

The error of midpoint rule is $\dfrac{g''(\xi)(b-a)^3}{24}$ (if the interval is not divided into subintervals). In your example $a=0$ and $b=2$.

The answers with the first derivative could be ruled out without knowing the precise error term. The midpoint rule is exact for linear functions; the error comes from convexity/concavity which is expressed by the second derivative.

Explanation needed: $g \,\colon [0,2] \to \Bbb R \,$ be a twice continuously differentiable function

1 Answers1