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In the following exercises, express the limits as integrals.
$\underset{n\to \infty}{\text{lim}}{\displaystyle \sum _{i=1}^{n}\left({x}_{i}^{*}\right)\text{\Delta}x}$ over $\left[1,3\right]$
$\underset{n\to \infty}{\text{lim}}{\displaystyle \sum _{i=1}^{n}\left(5{\left({x}_{i}^{*}\right)}^{2}-3{\left({x}_{i}^{*}\right)}^{3}\right)\text{\Delta}x}$ over $\left[0,2\right]$
${\int}_{0}^{2}\left(5{x}^{2}-3{x}^{3}\right)dx$
$\underset{n\to \infty}{\text{lim}}{\displaystyle \sum _{i=1}^{n}{\text{sin}}^{2}\left(2\pi {x}_{i}^{*}\right)\text{\Delta}x}$ over $\left[0,1\right]$
$\underset{n\to \infty}{\text{lim}}{\displaystyle \sum _{i=1}^{n}{\text{cos}}^{2}\left(2\pi {x}_{i}^{*}\right)\text{\Delta}x}$ over $\left[0,1\right]$
${\int}_{0}^{1}{\text{cos}}^{2}\left(2\pi x\right)dx$
In the following exercises, given L _{n} or R _{n} as indicated, express their limits as $n\to \infty $ as definite integrals, identifying the correct intervals.
${L}_{n}=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}\frac{i-1}{n}}$
${R}_{n}=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}\frac{i}{n}}$
${\int}_{0}^{1}xdx$
${L}_{n}=\frac{2}{n}{\displaystyle \sum _{i=1}^{n}\left(1+2\frac{i-1}{n}\right)}$
${R}_{n}=\frac{3}{n}{\displaystyle \sum _{i=1}^{n}\left(3+3\frac{i}{n}\right)}$
${\int}_{3}^{6}xdx$
${L}_{n}=\frac{2\pi}{n}{\displaystyle \sum _{i=1}^{n}2\pi \frac{i-1}{n}\text{cos}\left(2\pi \frac{i-1}{n}\right)}$
${R}_{n}=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\text{log}\left({\left(1+\frac{i}{n}\right)}^{2}\right)}$
${\int}_{1}^{2}x\phantom{\rule{0.1em}{0ex}}\text{log}\left({x}^{2}\right)dx$
In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x -axis.
In the following exercises, evaluate the integral using area formulas.
${\int}_{0}^{3}\left(3-x\right)dx$
${\int}_{2}^{3}\left(3-x\right)dx$
The integral is the area of the triangle, $\frac{1}{2}$
${\int}_{\mathrm{-3}}^{3}\left(3-\left|x\right|\right)dx$
${\int}_{0}^{6}\left(3-\left|x-3\right|\right)dx$
The integral is the area of the triangle, 9.
${\int}_{\mathrm{-2}}^{2}\sqrt{4-{x}^{2}}dx$
${\int}_{1}^{5}\sqrt{4-{\left(x-3\right)}^{2}}dx$
The integral is the area $\frac{1}{2}\pi {r}^{2}=2\pi .$
${\int}_{0}^{12}\sqrt{36-{\left(x-6\right)}^{2}}dx$
${\int}_{\mathrm{-2}}^{3}\left(3-\left|x\right|\right)dx$
The integral is the area of the “big” triangle less the “missing” triangle, $9-\frac{1}{2}.$
In the following exercises, use averages of values at the left ( L ) and right ( R ) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.
$\left\{\left(0,0\right),\left(2,1\right),\left(4,3\right),\left(5,0\right),\left(6,0\right),\left(8,3\right)\right\}$ over $\left[0,8\right]$
$\left\{\left(0,2\right),\left(1,0\right),\left(3,5\right),\left(5,5\right),\left(6,2\right),\left(8,0\right)\right\}$ over $\left[0,8\right]$
$L=2+0+10+5+4=21,R=0+10+10+2+0=22,\frac{L+R}{2}=21.5$
$\left\{\left(\mathrm{-4},\mathrm{-4}\right),\left(\mathrm{-2},0\right),\left(0,\mathrm{-2}\right),\left(3,3\right),\left(4,3\right)\right\}$ over $\left[\mathrm{-4},4\right]$
$\left\{\left(\mathrm{-4},0\right),\left(\mathrm{-2},2\right),\left(0,0\right),\left(1,2\right),\left(3,2\right),\left(4,0\right)\right\}$ over $\left[\mathrm{-4},4\right]$
$L=0+4+0+4+2=10,R=4+0+2+4+0=10,\frac{L+R}{2}=10$
Suppose that ${\int}_{0}^{4}f\left(x\right)dx}=5$ and ${\int}_{0}^{2}f\left(x\right)dx}=\mathrm{-3},$ and ${\int}_{0}^{4}g\left(x\right)dx}=\mathrm{-1$ and ${\int}_{0}^{2}g\left(x\right)dx}=2.$ In the following exercises, compute the integrals.
${\int}_{0}^{4}\left(f\left(x\right)+g\left(x\right)\right)dx$
${\int}_{2}^{4}\left(f\left(x\right)+g\left(x\right)\right)dx$
${\int}_{2}^{4}f\left(x\right)dx}+{\displaystyle {\int}_{2}^{4}g\left(x\right)dx}=8-3=5$
${\int}_{0}^{2}\left(f\left(x\right)-g\left(x\right)\right)dx$
${\int}_{2}^{4}\left(f\left(x\right)-g\left(x\right)\right)dx$
${\int}_{2}^{4}f\left(x\right)dx}-{\displaystyle {\int}_{2}^{4}g\left(x\right)dx}=8+3=11$
${\int}_{0}^{2}\left(3f\left(x\right)-4g\left(x\right)\right)dx$
${\int}_{2}^{4}\left(4f\left(x\right)-3g\left(x\right)\right)dx$
$4{\displaystyle {\int}_{2}^{4}f\left(x\right)dx}-3{\displaystyle {\int}_{2}^{4}g\left(x\right)dx}=32+9=41$
In the following exercises, use the identity ${\int}_{\text{\u2212}A}^{A}f\left(x\right)dx}={\displaystyle {\int}_{\text{\u2212}A}^{0}f\left(x\right)dx}+{\displaystyle {\int}_{0}^{A}f\left(x\right)dx$ to compute the integrals.
${\int}_{\text{\u2212}\pi}^{\pi}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{1+{t}^{2}}dt$ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\text{\u2212}t\right)=\text{\u2212}\text{sin}\left(t\right)\text{)}$
${\int}_{\text{\u2212}\sqrt{\pi}}^{\sqrt{\pi}}\frac{t}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}t}dt$
The integrand is odd; the integral is zero.
${\int}_{1}^{3}\left(2-x\right)dx$ ( Hint: Look at the graph of f .)
${\int}_{2}^{4}{\left(x-3\right)}^{3}dx$ ( Hint: Look at the graph of f .)
The integrand is antisymmetric with respect to $x=3.$ The integral is zero.
In the following exercises, given that ${\int}_{0}^{1}xdx}=\frac{1}{2},{\displaystyle {\int}_{0}^{1}{x}^{2}dx}=\frac{1}{3},$ and ${\int}_{0}^{1}{x}^{3}dx}=\frac{1}{4},$ compute the integrals.
${\int}_{0}^{1}\left(1+x+{x}^{2}+{x}^{3}\right)dx$
${\int}_{0}^{1}\left(1-x+{x}^{2}-{x}^{3}\right)dx$
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{7}{12}$
${\int}_{0}^{1}{\left(1-x\right)}^{2}dx$
${\int}_{0}^{1}{\left(1-2x\right)}^{3}dx$
${\int}_{0}^{1}\left(1-2x+4{x}^{2}-8{x}^{3}\right)dx}=1-1+\frac{4}{3}-2=-\frac{2}{3$
${\int}_{0}^{1}\left(6x-\frac{4}{3}{x}^{2}\right)dx$
${\int}_{0}^{1}\left(7-5{x}^{3}\right)dx$
$7-\frac{5}{4}=\frac{23}{4}$
In the following exercises, use the comparison theorem .
Show that ${\int}_{0}^{3}\left({x}^{2}-6x+9\right)dx}\ge 0.$
Show that ${\int}_{\mathrm{-2}}^{3}\left(x-3\right)\left(x+2\right)dx}\le 0.$
The integrand is negative over $\left[\mathrm{-2},3\right].$
Show that ${\int}_{0}^{1}\sqrt{1+{x}^{3}}dx}\le {\displaystyle {\int}_{0}^{1}\sqrt{1+{x}^{2}}dx}.$
Show that ${\int}_{1}^{2}\sqrt{1+x}dx}\le {\displaystyle {\int}_{1}^{2}\sqrt{1+{x}^{2}}dx}.$
$x\le {x}^{2}$ over $\left[1,2\right],$ so $\sqrt{1+x}\le \sqrt{1+{x}^{2}}$ over $\left[1,2\right].$
Show that ${\int}_{0}^{\pi \text{/}2}\text{sin}\phantom{\rule{0.1em}{0ex}}tdt}\ge \frac{\pi}{4}.$ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\ge \frac{2t}{\pi}$ over $\left[0,\frac{\pi}{2}\right]\text{)}$
Show that ${\int}_{\text{\u2212}\pi \text{/}4}^{\pi \text{/}4}\text{cos}\phantom{\rule{0.1em}{0ex}}tdt}\ge \pi \sqrt{2}\text{/}4.$
$\text{cos}\left(t\right)\ge \frac{\sqrt{2}}{2}.$ Multiply by the length of the interval to get the inequality.
In the following exercises, find the average value f _{ave} of f between a and b , and find a point c , where $f\left(c\right)={f}_{\text{ave}}.$
$f\left(x\right)={x}^{2},a=\mathrm{-1},b=1$
$f\left(x\right)={x}^{5},a=\mathrm{-1},b=1$
${f}_{\text{ave}}=0;c=0$
$f\left(x\right)=\sqrt{4-{x}^{2}},a=0,b=2$
$f\left(x\right)=\left(3-\left|x\right|\right),a=\mathrm{-3},b=3$
$\frac{3}{2}$ when $c=\pm \frac{3}{2}$
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