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Questions:

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my solutions:

4) $\sum_{i=1}^{n} f(x_i)\Delta x - \sum_{i=1}^{n} f(x_{i-1})\Delta x$ (By definition)

=$\sum_{i=1}^{n} (f(x_i)\Delta x - f(x_{i-1})\Delta x)$ (Sigma notation properties)

= $\sum_{i=1}^{n} (f(x_i) - f(x_{i-1}))\Delta x$ (Factored $\Delta x$)

= $\Delta x (f(b) - f(a))$ (Yeah, I don't know what to say about this.)

= $\frac{b-a}{n} (f(b)-f(a))$ ($\Delta x$ definition) \

5)

(a) If $f(x)$ is positive and increasing, and $f(x) \geq 0, \forall x \in [a,b]$, then

$L_n \leq A \leq R_n$. Let n be any positive integer and let $\Delta x = \frac{b-a}{n}$, such that $x_i = a + i\Delta x$ for i = $0,1,2,\cdots, n$. Since the function is increasing we know that $f(x_{i-1}) \leq f(x_i)$, and also: $$\sum_{i=1}^{n} f(x_{i-1})\Delta x \leq \sum_{i=1}^{n} f(x_{i})$$, $\forall i$ in n rectangles \

(b) If $f(x)$ is positive and decreasing, and $f(x) \geq 0, \forall x \in [a,b]$, then

$R_n \leq A \leq L_n$. Let n be any positive integer and let $\Delta x = \frac{b-a}{n}$, such that $x_i = a + i\Delta x$ for i = $0,1,2,\cdots, n$. Since the function is decreasing we know that $f(x_{i}) \leq f(x_{i-1})$, and also: $$\sum_{i=1}^{n} f(x_{i})\Delta x \leq \sum_{i=1}^{n} f(x_{i-1})$$, $\forall i$ in n rectangles

Are they correct? I also have a question for question 6, does it really matter if it is positive or negative? If the function was negative, then the observation would still be the same, correct?

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closed as unclear what you're asking by Vladhagen, Daniel W. Farlow, JonMark Perry, user91500, user223391 Jan 19 '17 at 22:51

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  • $\begingroup$ Yeah, I don't know = Telescoping sum $\endgroup$ – user305860 Jan 17 '17 at 21:14
  • $\begingroup$ Tinler, I know where these questions are from, I go to same university as you LOL! #The6ix $\endgroup$ – K Split X Jan 17 '17 at 22:49

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