problems

pset #2

$f(x)=(2-x)(x+1)$
from $x=-2$ to $x=2$

$A\approx L_n=f(x_0)\Delta x+f(x_1)\Delta x+\cdots +f(x_n-1)\Delta x={\sum }_{i=1}^{n}f(x_{i-1})\Delta x$

$L=f(-2)0.5+f(-1.5)0.5+f(-1)0.5+f(-0.5)0.5+f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5$
$R=f(-1.5)0.5+f(-1)0.5+f(-0.5)0.5+f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5$

$L=1.5,R=3.5$

${\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(a+\frac{(i-1)(b-a)}{n})\cdot \frac{(b-a)}{n}$

$L={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(-2+\frac{4(i-1)}{n})\cdot \frac{4}{n}$
$R={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(-2+\frac{4i}{n})\cdot \frac{4}{n}$

$L=R$
${\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(-2+\frac{4(i-1)}{n})\cdot \frac{4}{n}={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(-2+\frac{4i}{n})\cdot \frac{4}{n}$

$L-R={\lim }_{n\to \infty }[\frac{4}{n}\cdot {\sum }_{i=1}^n[f(-2+\frac{4(i-1)}{n})-f(-2+\frac{4i}{n})]]$
cancels. as ${\sum }_{i=1}^{n}f(x_{i-1})-f(x_i)=f(0)-f(n)$,
so ${\sum }_{i=1}^n[f(-2+\frac{4(i-1)}{n})-f(-2+\frac{4i}{n})]=f(-2)-f(-2+4)$
and $L-R={\lim }_{n\to \infty }\frac{4}{n}\cdot (f(-2)-f(-2+4))$

when taking the limit, $\frac{4}{n}\cdot (f(-2)-f(-2+4))=0$
so $L=R$, since $L-R=0$

range is from $-2$ to $2$, meaning range is $4$
using $\Delta x>4$ means ‘width’ will exceed range
so $0<\Delta x\le 4$ for having true area bounded between $L$ and $R$

for each, $|L-R|<0.5\cdot 10^{-3}$

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