How Accurate Is the Midpoint Rule for Estimating Integrals?

[DivGradCurl]

(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.

 

[wais]

Your calculations seem to be correct, but there are a few things to note.

Firstly, in part (a), the textbook answer of 23.44 is most likely due to rounding. Your answer of 23.52 is more accurate.

Secondly, in part (b), the textbook answer of 0.341\overline{3} is also due to rounding. Your answer of 0.085\overline{3} is more accurate.

Lastly, in part (b), the error bound formula should use the maximum value of \left| f^{\prime \prime} (x) \right|, which in this case is 4. So the correct error bound would be:

\left| E_M \right| \leq \frac{4(3.2-0)^3}{24(8)^2} = 0.341\overline{3}

So your calculations are correct, it's just a matter of rounding and using the maximum value in the error bound formula.