# 微积分网课代修|积分学代写Integral Calculus代考|MA1030C Riemann sums

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## 微积分作业代写calclulus代考|Riemann sums

The identity
$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$
that we have just seen might be improved by subdividing the interval $[a, b]$ by intermediate points:
$$a=a_{1}<b_{1}=a_{2}<b_{2}=a_{3}<\cdots<a_{n}<b_{n}=b$$
This expresses the interval $[a, b]$ as the union of a collection of nonoverlapping, compact subintervals:
$$\left[a_{1}, b_{1}\right],\left[a_{2}, b_{2}\right],\left[a_{3}, b_{3}\right], \ldots,\left[a_{n}, b_{n}\right]$$
The mean-value theorem of the differential calculus, as before, asserts that we can select a point $\xi_{i}$ inside each interval $\left[a_{i}, b_{i}\right]$ so that
$$F\left(b_{i}\right)-F\left(a_{i}\right)=f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right) .$$
This leads to a new covering relation
$$\pi=\left{\left(\left[a_{1}, b_{1}\right], \xi_{1}\right),\left(\left[a_{2}, b_{2}\right], \xi_{2}\right),\left(\left[a_{3}, b_{3}\right], \xi_{3}\right), \ldots,\left(\left[a_{n}, b_{n}\right], \xi_{n}\right}\right.$$
which is called a partition. The partition is denoted as $\pi$ (the letter is chosen so as to use the Greek letter corresponding to “P,” not to have anything to do with areas of circles).
Using this partition $\pi$ we have
$$F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(b_{i}\right)-F\left(a_{i}\right)\right]=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$
and consequently
$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$

## 微积分作业代写calclulus代考|Riemann sums constructed from the derivative

We have seen that if $f$ is Newton integrable on an interval $[a, b]$ then for some partition of that interval
$$\pi=\left{\left(\left[a_{i}, b_{i}\right], \xi_{i}\right): i=1,2, \ldots, n\right}$$
we have the identity
$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$

## 微积分作业代写calclulus代考|Riemann sums

$$\int_{a}^{b} f(x) d x=f(\xi)(b-a)$$

$$a=a_{1}<b_{1}=a_{2}<b_{2}=a_{3}<\cdots<a_{n}<b_{n}=b$$

$$\left[a_{1}, b_{1}\right],\left[a_{2}, b_{2}\right],\left[a_{3}, b_{3}\right], \ldots,\left[a_{n}, b_{n}\right]$$

$$F\left(b_{i}\right)-F\left(a_{i}\right)=f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right) .$$

$$F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(b_{i}\right)-F\left(a_{i}\right)\right]=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$

$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$

## 微积分作业代写calclulus代考|Riemann sums constructed from the derivative

|pi=|left{\left(|left[a_{i}, b_{i}}right], |xi_{i}\right): i $=1,2$, |ldots, n|right $}$

$$\int_{a}^{b} f(x) d x=\sum_{i=1}^{n} f\left(\xi_{i}\right)\left(b_{i}-a_{i}\right)$$