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数学代写|微积分note A Simple Procedure for Finding Integrals

9.6 A Simple Procedure for Finding Integrals
Suppose f is a continuous function and F is an increasing integrator function. How do you find baf(x)dF? Is there some sort of easy way to do it which will handle lots of simple cases? It turns out there is a way. It is based on Lemma 9.3.22. First of all
$$
F(x+) \equiv \lim {y \rightarrow x+} F(y), F(x-) \equiv \lim {y \rightarrow x-} F(y)
Foranincreasingfunction$F$,thejumpofthefunctionat$x$equals$F(x+)F(x)$.Procedure9.6.1Suppose$f$iscontinuouson$[a,b]$and$F$isanincreasingfunctiondefinedon$[a,b]$suchthattherearefinitelymanyintervalsdeterminedbythepartition$a=x0<x1<<xn=b$whichhavethepropertythaton$[xi,xi+1]$,thefollowingfunctionisdifferentiableandhasacontinuousderivative.
G_{i}(x) \equiv\left{F(x) on (xi,xi+1) F(xi+) when x=xi F(xi+1) when x=xi+1\right.
Alsoassume$F(a)=F(a+),F(b)=F(b)$.Then \int_{a}^{b} f(x) d F=\sum_{j=0}^{n-1} \int_{x_{j}}^{x_{j+1}} f(x) G_{j}^{\prime}(x) d x+\sum_{i=1}^{n-1} f\left(x_{i}\right)\left(F\left(x_{i}+\right)-F\left(x_{i}-\right)\right) 202CHAPTER9.INTEGRATIONHereiswhythisprocedureworks.Let$δ$beverysmallandconsiderthepartition
a=x0<x1δ<x1<x1+δ <x2δ<x2<x2+δ< xn1δ<xn1<xn1+δ<xnδ<xn=b
where$δ$isalsosmallenoughthatwhenever$|xy|<δ$,itfollows$|f(x)f(y)|<ε$.Thenfromthepropertiesoftheintegralpresentedabove,
x1δafdF+x2δx1+δfdF++bxn1+δfdF+n1i=1(f(xi)ε)(F(xi+δ)F(xiδ)) bafdFx1δafdF+x2δx1+δfdF++bxn1+δfdF +n1i=1(f(xi)+ε)(F(xi+δ)F(xiδ))
ByLemma9.3.22thisimplies
x1δafG0dx+x2δx1+δfG1dx++bxn1+δfGn1dx +n1i=1(f(xi)ε)(F(xi+δ)F(xiδ))bafdF
$$

9.6 求积分的简单程序
假设f 是一个连续函数,F 是一个递增积分函数。您如何找到 baf(x)dF 是否有某种简单的方法可以处理许多简单的情况?事实证明有一种方法。它基于引理 9.3.22。首先
$$
F(x+) \equiv \lim {y \rightarrow x+} F(y), F(x-) \equiv \lim {y \rightarrow x-} F(y)
$F$$x$$F(x+)F(x)$9.6.1$f$$[a,b]$$F$$[a,b]$$a=x0<x1<<xn=b$$[xi,xi+1]$
G_{i}(x) \equiv\left{\begin{array}{l}
F(x) \text { on }\left(x_{i}, x_{i+1}\right) \
F\left(x_{i}+\right) \text { 当 } x=x_{i} \
F\left(x_{i+1}-\right) \text { 当 } x=x_{i+1}
\end{数组}\对。
$F(a)=F(a+),F(b)=F(b)$ \int_{a}^{b} f(x) d F=\sum_{j=0}^{n-1} \int_{x_{j}}^{x_{j+1}} f (x) G_{j}^{\prime}(x) d x+\sum_{i=1}^{n-1} f\left(x_{i}\right)\left(F\left(x_{ i}+\right)-F\left(x_{i}-\right)\right) 202|9$δ$
\开始{对齐}
一个 &=x_{0}<x_{1}-\delta<x_{1}<x_{1}+\delta \
&<x_{2}-\delta<x_{2}<x_{2}+\delta<\
\cdots x_{n-1}-\delta &<x_{n-1}<x_{n-1}+\delta<x_{n}-\delta<x_{n}=b
\end{对齐}
$δ$$|xy|<δ$$|f(x)f(y)|<ε$
\开始{聚集}
\int_{a}^{x_{1}-\delta} fd F+\int_{x_{1}+\delta}^{x_{2}-\delta} fd F+\cdots+\int_{x_{n-1 }+\delta}^{b} fd F+\sum_{i=1}^{n-1}\left(f\left(x_{i}\right)-\varepsilon\right)\left(F\left (x_{i}+\delta\right)-F\left(x_{i}-\delta\right)\right) \
\leq \int_{a}^{b} fd F \leq \int_{a}^{x_{1}-\delta} fd F+\int_{x_{1}+\delta}^{x_{2}- \delta} fd F+\cdots+\int_{x_{n-1}+\delta}^{b} fd F \
+\sum_{i=1}^{n-1}\left(f\left(x_{i}\right)+\varepsilon\right)\left(F\left(x_{i}+\delta\right )-F\left(x_{i}-\delta\right)\right)
\结束{聚集}
9.3.22
\开始{对齐}
&\int_{a}^{x_{1}-\delta} f G_{0}^{\prime} d x+\int_{x_{1}+\delta}^{x_{2}-\delta} f G_{1}^{\prime} d x+\cdots+\int_{x_{n-1}+\delta}^{b} f G_{n-1}^{\prime} dx \
&+\sum_{i=1}^{n-1}\left(f\left(x_{i}\right)-\varepsilon\right)\left(F\left(x_{i}+\delta\右)-F\left(x_{i}-\delta\right)\right) \leq \int_{a}^{b} fd F \leq
\end{对齐}
$$
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