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Find a lower sum for $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval $\left[a,b\right]=\left[0,\frac{\pi}{2}\right];$ let $n=6.$
Let’s first look at the graph in [link] to get a better idea of the area of interest.
The intervals are $\left[0,\frac{\pi}{12}\right],\left[\frac{\pi}{12},\frac{\pi}{6}\right],\left[\frac{\pi}{6},\frac{\pi}{4}\right],\left[\frac{\pi}{4},\frac{\pi}{3}\right],\left[\frac{\pi}{3},\frac{5\pi}{12}\right],$ and $\left[\frac{5\pi}{12},\frac{\pi}{2}\right].$ Note that $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ is increasing on the interval $\left[0,\frac{\pi}{2}\right],$ so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum $\sum _{i=0}^{5}\text{sin}\phantom{\rule{0.1em}{0ex}}{x}_{i}\left(\frac{\pi}{12}\right)}.$ We have
Using the function $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval $\left[0,\frac{\pi}{2}\right],$ find an upper sum; let $n=6.$
$A\approx 1.125$
State whether the given sums are equal or unequal.
a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting $j=i-1.$ d. They are equal; the first sum factors the terms of the second.
In the following exercises, use the rules for sums of powers of integers to compute the sums.
$\sum _{i=5}^{10}i$
Suppose that $\sum _{i=1}^{100}{a}_{i}=15$ and $\sum _{i=1}^{100}{b}_{i}=\mathrm{-12}}.$ In the following exercises, compute the sums.
$\sum _{i=1}^{100}\left({a}_{i}+{b}_{i}\right)$
$\sum _{i=1}^{100}\left({a}_{i}-{b}_{i}\right)$
$15-\left(\mathrm{-12}\right)=27$
$\sum _{i=1}^{100}\left(3{a}_{i}-4{b}_{i}\right)$
$\sum _{i=1}^{100}\left(5{a}_{i}+4{b}_{i}\right)$
$5\left(15\right)+4\left(\mathrm{-12}\right)=27$
In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.
$\sum _{k=1}^{20}100\left({k}^{2}-5k+1\right)$
$\sum _{j=1}^{50}\left({j}^{2}-2j\right)$
$\sum _{j=1}^{50}{j}^{2}}-2{\displaystyle \sum _{j=1}^{50}j}=\frac{\left(50\right)\left(51\right)\left(101\right)}{6}-\frac{2\left(50\right)\left(51\right)}{2}=40,\text{}375$
$\sum _{j=11}^{20}\left({j}^{2}-10j\right)$
$\sum _{k=1}^{25}\left[{\left(2k\right)}^{2}-100k\right]$
$4{\displaystyle \sum _{k=1}^{25}{k}^{2}}-100{\displaystyle \sum _{k=1}^{25}k}=\frac{4\left(25\right)\left(26\right)\left(51\right)}{9}-50\left(25\right)\left(26\right)=\mathrm{-10},\text{}400$
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