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# 数学代写|微积分代写Calculus代考|Approximating area with midpoint sums

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## 数学代写|微积分代写Calculus代考|Approximating area with midpoint sums

A third way to approximate areas with rectangles is to make each rectangle cross the curve at the midpoint of its top side. A midpoint sum is a much better estimate of area than either a left or a right sum. Figure 8-7 shows why.

You can see in Figure 8-7 that the part of each rectangle that’s above the curve looks about the same size as the gap between the rectangle and the curve. A midpoint sum produces such a good estimate because these two errors roughly cancel out each other.
For the three rectangles in Figure 8-8, the widths are 1 and the heights are $f(0.5)=1.25 f(1.5)=3.25$, and $f(2.5)=7.25$. The total area comes to 11.75. Table $8-3$ lists the midpoint sums for the same number of rectangles used in Tables 8-1 and 8-2.

Estimates of the Area under $\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^2+1$ Given by Increasing Numbers of “Midpoint” Rectangles
\begin{tabular}{l|l|}
\hline Number of Rectangles & Area Estimate \
\hline 3 & 11.75 \
\hline 6 & 11.9375 \
12 & $\sim 11.9844$ \
24 & $\sim 11.9961$ \
48 & $\sim 11.9990$ \
96 & $\sim 11.9998$ \
192 & $\sim 11.9999$ \
384 & $\sim 11.99998$ \
\hline
\end{tabular}

Sorry to give away the ending, but you’ll soon see that the exact area is 12. And to see how much faster the midpoint approximations approach the exact answer of 12 than the left or right approximations, compare the three tables. The error with 6 midpoint rectangles is about the same as the error with 192 left or right rectangles!

## 数学代写|微积分代写Calculus代考|The Midpoint Rule

The Midpoint Rule: You can approximate the exact area under a curve between $a$ and $b, \int_a^b f(x) d x$, with a sum of midpoint rectangles given by the following formula. In general, the more rectangles, the better the estimate.
$$\begin{gathered} M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+\ldots \ldots \ldots . . .\right. \ \left.+f\left(\frac{x_{n-1}+x_n}{2}\right)\right] \end{gathered}$$
where $n$ is the number of rectangles, $\frac{b-a}{n}$ is the width of each, and the function values are the heights of the rectangles.

The left, right, and midpoint sums are called Riemann sums after the great German mathematician G. F. B. Riemann (1826-66).

## 数学代写|微积分代写Calculus代考|Approximating area with midpoint sums

\begin{tabular}{l|l|}
\hline Number of Rectangles & Area Estimate \hline 3 & 11.75 \hline 6 & 11.9375 \12 &$\sim 11.9844$ \24 &$\sim 11.9961$ \48 &$\sim 11.9990$ \96 &$\sim 11.9998$ \192 &$\sim 11.9999$ \384 &$\sim 11.99998$ \hline
\end{tabular}

## 数学代写|微积分代写Calculus代考|The Midpoint Rule

$$\begin{gathered} M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+\ldots \ldots \ldots . . .\right. \ \left.+f\left(\frac{x_{n-1}+x_n}{2}\right)\right] \end{gathered}$$

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