# Homework Help: Midpoint integral approx.

1. Feb 10, 2005

(a) Use the Midpoint Rule and the given data to estimate the value of the integral $$\int _0 ^{3.5} f(x)\: dx$$

$$x=0.0\quad 0.4\quad 0.8\quad 1.2\quad 1.6\quad 2.0\quad 2.4\quad 2.8\quad 3.2$$

$$f(x)=6.8\quad 6.5\quad 6.3\quad 6.4\quad 6.9\quad 7.6\quad 8.4\quad 8.8\quad 9.0$$

(b) If it is known that $$-4 \leq f^{\prime \prime} (x) \leq 1$$ for all $$x$$, estimate the error involved in the approximation in part (a).

The answers given in my textbook are:

(a) $$23.44$$
(b) $$0.341\overline{3}$$

Anyhow, this is what I have:

$$\bar{x}_i=0.2\quad 0.6\quad 1.0\quad 1.4\quad 1.8\quad 2.2\quad 2.6\quad 3.0$$

Then, I simply figured out the arithmetic mean of subsequent values of $$f(x)$$ (taken two at a time)

$$f(\bar{x}_i)=6.65\quad 6.4\quad 6.35\quad 6.65\quad 7.25\quad 8.0\quad 8.6\quad 8.9$$

which makes it possible to obtain

$$M_8 = \Delta x \sum _{i=1} ^{8} f(\bar{x}_i) = \frac{3.2-0}{8}(6.65+ 6.4+ 6.35+ 6.65+ 7.25+ 8.0+ 8.6+ 8.9) = 23.52$$

For part (b), I just figured out the error bound

$$\left| E_M \right| \leq \frac{K(b-a)^3}{24n^2}$$

We are given $$-4 \leq f^{\prime \prime} (x) \leq 1$$, then $$\left| f^{\prime \prime} (x) \right| \leq 4 = K$$. So, it follows

$$\left| E_M \right| \leq \frac{4(3.2-0)^3}{24(8)^2} = 0.085\overline{3}$$

As you can see, my results are a bit different. In part (a), the author of my textbook could have applied some weighted average (I really don't know!!!). In part (b), there is nothing different I could do (so I wonder what is wrong with my calculations).

Any help is highly appreciated.

Last edited: Feb 10, 2005