Compute the integral ${\displaystyle \underset{2}{\overset{5}{}}}|2x-6|dx{x}_i^{*}=a+ix$ where $ais$ the left endpoint $of$ the interval you are

integrating over$2,5$ becomes $2,b$ and $b,5$ and you compute the $limitof$ two sums.

$If$ not, enter $"no".If$ yes, enter which number you need $to$ break the interval $at.$

$b=$

$(b)(2$ points$)$ What $i\frac{s}{a}rex?$

$A.\frac{1\mathrm{.}5}{n},\frac{1\mathrm{.}5}{n}$

$B.\frac{1}{n},\frac{2}{n}$

$C.\frac{3}{n}$

$D.\frac{0\mathrm{.}5}{n},\frac{2\mathrm{.}5}{n}$

$E.\frac{5}{n}$

$(c)(4$ points$)$ After using summation properties and formulas $to$ simplify the sum$(s)$

before taking the $limit,$ you end $up$ with a formula that looks like ${\displaystyle \underset{2}{\overset{5}{}}}|2x-6|dx=$

$\underset{n}{lim}(c+\frac{d}{n}).$ What are $c$ and $d?$

$c=$

$d=$

$(d)(1$ point$)$ What $is$ the value $of$ the integral?

${\displaystyle \underset{2}{\overset{5}{}}}|2x-6|dx=$

$(e)(2$ points$)$ What other formula could you have used $to$ compute this integral?

$A.\frac{1}{2}(3)(4)$

$B.-\frac{1}{2}(1)(2)+\frac{1}{2}(2)(4)$

$C.\frac{1}{2}(3)(5)$

$D.-\frac{1}{2}(1)(3)+\frac{1}{2}(3)(5)$

$E.\frac{1}{2}(1)(2)+\frac{1}{2}(2)(4)$ The graph $of$ the function $f(x)is$ pictured below

$(a)(3$ points$)$ Where $is$ the function $F(x)={\displaystyle \underset{-4}{\overset{x}{}}}f(x)dx$ decreasing $on$ the interval $-4,2?$

$A.(-3,-1),(0,1)$

$B.(-3\mathrm{.}6,-2\mathrm{.}2),(-0\mathrm{.}6,0\mathrm{.}6),(1\mathrm{.}7,2)$

$C.(-2\mathrm{.}3,-0\mathrm{.}5),(0\mathrm{.}4,1\mathrm{.}2)$

$D.(-4,-3),(-1,0),(1,2)$

$(b)(2$ points$)$ Which value $do$ you expect ${\displaystyle \underset{-1}{\overset{1}{}}}f(x)dxtobe$ closest $to?(Estimate)$

$A.-1$

$B.0$

$C.0\mathrm{.}5$

$D.1$

$E.2$

$(c)(4$ points$)$ Given $F(-1)=-5$ and your answer $to$ part $(b),$ what $is$ the value $of{\displaystyle \underset{6}{\overset{11}{}}}f(7-x)dx(approximately)?$

${\displaystyle \underset{6}{\overset{11}{}}}f(7-x)dx=$ Mark $as$ true $or$ false. For parts $a-f,$ mark $as$ true $if$ you have use the given method $at$

any point $to$ evaluate the integral. Mark $as$ false $if$ there $is$ a way $to$ evaluate the integral

without ever using the given method.

$(a)(2$ points$)$ True

$(b)(2$ points$)$ True

$(c)(2$ points$)$ True

$(d)(2$ points$)$ True

$(e)(2$ points$)$ True

$(f)(2$ points$)$ True

$(g)(2$ points$)$ True

$(h)(2$ points$)$ True

$(i)(2$ points$)$ True

False

False

False

False

False

False

False

False

False

You must use substitution $to$ evaluate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}ln(\sqrt[3]{x})dx$

You must use trigonometric substitution $to$ evaluate

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{-1}(x)x\sqrt[\phantom{2}]{1-x^2}dx$

You must use integration $by$ parts $to$ evaluate

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{-1}(x)x\sqrt[\phantom{2}]{1-x^2}dx$

You must use trigonometric substitution $to$ evaluate

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^4+2x^2+x+1}{(x^2+1{)}^5}dx$

You must use integration $by$ parts $to$ evaluate

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^4+2x^2+x+1}{(x^2+1{)}^5}dx$

You must use some trigonometric identities $to$ eval$-$

wate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^4+2x^2+x+1}{(x^2+1{)}^5}dx$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^4+2x^2+x+1}{(x^2+1{)}^5}dx=-\frac{1}{8(x^2+1{)}^4}+{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^{4}xdx$

${\displaystyle \underset{0}{\overset{1}{}}}(si{n}^{-1}x{)}^{2}dx={\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}{}^{2}cosd$

${\displaystyle \underset{0}{\overset{1}{}}}x{e}^{1\mathrm{.}5x}dx={\displaystyle \underset{1}{\overset{e}{}}}u\sqrt[\phantom{2}]{u}lnudu$

$(25$ points$)$ Arrange the integrals $in$ increasing order

Note: $(a+b{)}^4=a^4+4a{b}^3+6a^2{b}^2+4a^{3}b+b^4$

$A={\displaystyle \underset{0}{\overset{1}{}}}xsi{n}^2(x)dx$

$B={\displaystyle \underset{0}{\overset{}{}}}$The following questions are about the function $g(t)={\displaystyle \underset{0}{\overset{1}{}}}\frac{t{x}^t}{1+x^3}dx$ where $g(t)is$ defined

for $t3.$

$(a)(1$ point$)$ What $isg(6)+2g(3)?$

$g(6)+2g(3)=$

$(b)(2$ points$)$ What method$(s)d$