Midpoint integral approx.

  • Thread starter DivGradCurl
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In summary, your calculations for both parts are correct, but the answer given in the textbook may have used a different method or rounding.
  • #1
DivGradCurl
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(a) Use the Midpoint Rule and the given data to estimate the value of the integral [tex]\int _0 ^{3.5} f(x)\: dx[/tex]

[tex]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[/tex]

[tex]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[/tex]

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


The answers given in my textbook are:

(a) [tex]23.44[/tex]
(b) [tex]0.341\overline{3}[/tex]

Anyhow, this is what I have:

[tex]\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[/tex]

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

[tex]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[/tex]

which makes it possible to obtain

[tex]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[/tex]

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

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

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

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

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. :smile:
 
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  • #2


Your calculations for part (a) are correct. The answer given in your textbook might be using the trapezoidal rule instead of the midpoint rule, which would result in a slightly different value for the integral.

For part (b), your calculation of the error bound is also correct. However, the answer given in your textbook seems to have rounded the value to 0.341\overline{3}. It is possible that they used a different method to calculate the error bound, resulting in a slightly different value.

Overall, your calculations and approach are correct. It is possible that there is a slight difference in the rounding or method used in the textbook answer.
 
  • #3


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.
 

1. What is a midpoint integral approximation?

A midpoint integral approximation is a method used in calculus to estimate the area under a curve by dividing it into smaller rectangles and finding the sum of their areas. The midpoint of each rectangle is used as the height to calculate the area, hence the name "midpoint" approximation.

2. How is the midpoint approximation formula derived?

The formula for midpoint approximation is derived from the Riemann sum, which is a way of approximating the area under a curve using rectangles. In midpoint approximation, the width of each rectangle is the same, and the midpoint of each rectangle is used to calculate the area.

3. What is the advantage of using midpoint approximation over other methods of approximation?

Midpoint approximation can provide a more accurate estimate of the area under a curve compared to other methods, such as left endpoint or right endpoint approximation, because it uses the midpoint of each rectangle which is closer to the actual curve.

4. How is midpoint approximation used in real life?

Midpoint approximation is used in many real-life applications, such as in physics, engineering, and economics, to estimate the area under a curve. It is also used in computer algorithms to calculate integrals and in data analysis to estimate the total value of a dataset.

5. Are there any limitations to using midpoint approximation?

One limitation of midpoint approximation is that it can only provide an estimate of the area under a curve, and the accuracy of the estimate depends on the number of rectangles used. It may also not be suitable for curves with irregular shapes or when the interval is large.

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