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Calculate $L(f,P)$ and $U(f,P)$ for the following: $$f(x) = 2x, x \in [0,1];~ P = \{0,\frac{1}{4},\frac{1}{2},1\}.$$

By definition I know that

$$L(f,P) = \sum_{i=1}^{n} m_i(x_{i}-x_{i-1}) \text{ and } U(f,P) = \sum_{i=1}^{n} M_i(x_{i} - x_{i-1})$$ where $m_i = inf\{f(x)|x \in[x_{i-1},x_i]\}$ and $M_i = sup\{f(x)|x\in[x_{i-1},x_{i}]\}$.

I think since the function is densed so we can say that $M_i = 2(1)$ and $m_i = 0$ ?

Then $$L(f,P) = \sum_{i=1}^{n} 0(x_{i}-x_{i-1}) \text{ and } U(f,P) = \sum_{i=1}^{n} 2(x_{i} - x_{i-1})$$

Therefore $$L(f,P) = 0(1-0)=0 \text{ and } U(f,P) = 2(1-0)=2.$$

Am I right?

user349557
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1 Answers1

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where $m_i = inf\{f(x)|x \in[x_{i-1},x_i]\}$ and $M_i = sup\{f(x)|x\in[x_{i-1},x_{i}]\}$.

I think since the function is densed so we can say that $M_i = 2(1)$ and $m_i = 0$ ?

This would be correct if you take the entire interval $[0,1]$, which would be the case if $P$ was given as $\{0,1\}$. You are however given a set $P$ with four elements:

$P = \{0,\frac{1}{4},\frac{1}{2},1\}.$

which defines a partition of $[0,1]$ into three sub-intervals. So in your case $n=3$ and you need to find $m_i$ and $M_i$ for $n=1,2,3$, corresponding to the three intervals: $$[0,\tfrac{1}{4}] \,\cup\, [\tfrac{1}{4},\tfrac{1}{2}] \,\cup\,[\tfrac{1}{2},1] \; = \; [0,1]$$

For example, for $n=2$, you need to find the minimum and maximum of $f(x)=2x$ on the interval $[\tfrac{1}{4},\tfrac{1}{2}]$, so you have $m_2 = 2\cdot \tfrac{1}{4} = \tfrac{1}{2}$ and $M_2 = 2\cdot \tfrac{1}{2} = 1$. Can you take it from here?

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StackTD
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  • Thank you, I get it now. – user349557 May 15 '17 at 18:32
  • Should $L(f,P)$ and $U(f,P)$ be the same since we know that $f(x)= 2x$ is integrable? I'm asking since I found that $L(f,P)= 5/8$ and $U(f,P)= 11/8$. Am I solving wrong? – Nour May 19 '17 at 08:06
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    @Nour The upper and lower integrals (i.e. the limit of the respective sums) will be equal for an integrable function. The (finite) sums won't be (unless $f$ is constant!) and if $I$ is the integral, then for any $n$ you have $L_n(f,P) \le I \le U_n(f,P)$. If $f$ is integrable, you have for $n \to \infty$ that $L(f,P)=U(f,P)=I$. – StackTD May 19 '17 at 08:09