作业帮The limitis the limit of a right Riemann sum for a certain definite integralusing a partition of the interval into subintervals of equal length,wheref(x) =
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The limitis the limit of a right Riemann sum for a certain definite integralusing a partition of the interval into subintervals of equal length,wheref(x) =
The limit
is the limit of a right Riemann sum for a certain definite integral
using a partition of the interval into subintervals of equal length,where
f(x) =
The limitis the limit of a right Riemann sum for a certain definite integralusing a partition of the interval into subintervals of equal length,wheref(x) =
[32*(4k/n)-20]/n
=[8*(4*k/n)-5]*(4/n)
所以可以把4/n看作h
即区间长度是4,分成n份
4*k/n的范围是0到4
所以积分区间可以看作(0,4)
x=4*k/n
积分就是积分 8x-5 dx
a=0
b=4
f(x)=8x-5
a和b是决定不了的, 只要b>a都可以, f(x)=4(32x-27a-5b)/(b-a)^2.
For any n and any k=1,2,...,n, we have
f(a+k*(b-a)/n)*(b-a)/n=(32*4k/n-20)/n, (1)
Let x=a+k(b-a)/n. Then k/n=(x-a)/(b-a).
So (1) become...
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a和b是决定不了的, 只要b>a都可以, f(x)=4(32x-27a-5b)/(b-a)^2.
For any n and any k=1,2,...,n, we have
f(a+k*(b-a)/n)*(b-a)/n=(32*4k/n-20)/n, (1)
Let x=a+k(b-a)/n. Then k/n=(x-a)/(b-a).
So (1) becomes
f(x)=(128(x-a)/(b-a)-20)/(b-a)=4(32x-27a-5b)/(b-a)^2.
根据右Riemann和的定义, 非要根据第一个和的形式来指定a和b是不合理的.
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