Q. 35 - Calculus

Question

$$\begin{gathered} \mathrm{For} \mathrm{each} \mathrm{function} \mathrm{f} \mathrm{and} \mathrm{interval} [\mathrm{a}, \mathrm{b}] \mathrm{in} \mathrm{Exercises} 34–38, \\ \mathrm{it} \mathrm{is} \mathrm{possible} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{exact} \mathrm{signed} \mathrm{area} \mathrm{between} \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{f} \mathrm{and} \mathrm{the} \mathrm{x}-\mathrm{axis} \mathrm{on} [\mathrm{a}, \mathrm{b}] \\ \mathrm{geometrically} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{circles}, \mathrm{triangles}, \mathrm{and} \mathrm{rectangles}. \mathrm{Find} \mathrm{this} \mathrm{exact} \mathrm{area}, \\ \mathrm{and} \mathrm{then} \mathrm{calculate} \mathrm{the} \mathrm{left}, \mathrm{right}, \mathrm{midpoint}, \mathrm{upper}, \mathrm{lower}, \mathrm{and} \mathrm{trapezoid} \mathrm{sums} \mathrm{with} \mathrm{n} = 4. \\ \mathrm{Which} \mathrm{approximation} \mathrm{rule} \mathrm{is} \mathrm{most} \mathrm{accurate}? \\ \mathrm{f}\left(\mathrm{x}\right)= 3\mathrm{x}+1 , [\mathrm{a},\mathrm{b}] =[3,5] \end{gathered}$$.

Step-by-Step Solution

Answer

left-sum = 24.5, right-sum= 27.5 , trapezoid sum = 26.

Step 1. Given information is .

$\mathrm{The} \mathrm{function} \mathrm{is} \mathrm{f}\left(\mathrm{x}\right)= 3\mathrm{x}+1 \mathrm{and} \mathrm{the} \mathrm{interval} \mathrm{is} [\mathrm{a},\mathrm{b}] =[3,5].$

Step 2 . Finding area .

Step 3. Calculation the exact sum .

$$\begin{gathered} \mathrm{Enter} \mathrm{the} \mathrm{function} (\boxed{3\mathrm{x}+5}, \mathrm{X},3,5) \\ \mathrm{Press} \mathrm{ENTER} \\ \mathrm{The} \mathrm{display} \mathrm{is} \mathrm{shown} \mathrm{below}, \end{gathered}$$

Therefore ,the exact sum is $\boxed{26}$.

Step 4. Calculating left- sum .

$$\begin{gathered} \mathrm{The} \mathrm{left}-\mathrm{sum} \mathrm{defined} \mathrm{for} \mathrm{n} \mathrm{rectangles} \mathrm{on} [\mathrm{a},\mathrm{b}] \mathrm{is} \sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left({\mathrm{x}}_{\mathrm{k}-1}\right)∆\mathrm{x}. \\ \mathrm{Where} , ∆\mathrm{x}= \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},{\mathrm{x}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}. \\ \mathrm{Now}, ∆\mathrm{x}= \frac{5-3}{4} =\frac{1}{2}. \\ \mathrm{So}, {\mathrm{x}}_{\mathrm{k}} =3 +\mathrm{k}\left(\frac{1}{2}\right) =\frac{6+\mathrm{k}}{2}. \\ \mathrm{In} \mathrm{the} \mathrm{left} \mathrm{sum}, {\mathrm{x}}_{\mathrm{k}-1}, \mathrm{is} \mathrm{the} \mathrm{leftmost} \mathrm{point} \mathrm{in} \mathrm{the} \mathrm{interval} [{\mathrm{x}}_{\mathrm{k}-1, } {\mathrm{x}}_{\mathrm{k}}] . \\ \mathrm{So}, {\mathrm{x}}_{\mathrm{k}}= \frac{5+\mathrm{k}}{2}. \\ \mathrm{The} \mathrm{left} \mathrm{sum} \mathrm{is}, \\ \sum _{\mathrm{k}=1 }^4\left(3\left(\frac{5+\mathrm{k}}{2}\right)+1\right)\frac{1}{2} =\frac{1}{2}\sum _{\mathrm{k}=1}^4\left(\frac{17+3\mathrm{k}}{2}\right) \\ = \frac{1}{2}\left(\frac{17+3}{2}+\frac{17+6}{2}+\frac{17+9}{2}+\frac{17+12}{2}\right)= 24.5 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{left} \mathrm{sum} \mathrm{is} \boxed{24.5} . \end{gathered}$$

Step 5. Calculating the right-sum .

$$\begin{gathered} \mathrm{The} \mathrm{Right}-\mathrm{sum} \mathrm{defined} \mathrm{for} \mathrm{n} \mathrm{rectangles} \mathrm{on} [\mathrm{a},\mathrm{b}] \mathrm{is} \sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left({\mathrm{x}}_{\mathrm{k}}\right)∆\mathrm{x}. \\ \mathrm{Where} , ∆\mathrm{x}= \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},{\mathrm{x}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}. \\ \mathrm{Now}, ∆\mathrm{x}= \frac{5-3}{4} =\frac{1}{2}. \\ \mathrm{So}, {\mathrm{x}}_{\mathrm{k}} =3 +\mathrm{k}\left(\frac{1}{2}\right) =\frac{6+\mathrm{k}}{2}. \\ \mathrm{The} \mathrm{right} \mathrm{sum} \mathrm{is}, \\ \sum _{\mathrm{k}=1 }^4\left(3\left(\frac{6+\mathrm{k}}{2}\right)+1\right)\frac{1}{2} =\frac{1}{2}\sum _{\mathrm{k}=1}^4\left(\frac{20+3\mathrm{k}}{2}\right) \\ = \frac{1}{2}\left(\frac{20+3}{2}+\frac{20+6}{2}+\frac{20+9}{2}+\frac{20+12}{2}\right)= 27.5 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{right} \mathrm{sum} \mathrm{is} \boxed{27.5} . \end{gathered}$$

Step 6. Calculating trapezoid sum .

$$\begin{gathered} \mathrm{The} \mathrm{trapezoid} \mathrm{sum} \mathrm{is} , \\ \frac{24.5+27.5}{2}= 26. \\ \mathrm{Therefore} , \mathrm{the} \mathrm{trapezoid} \mathrm{sum} \mathrm{is} \boxed{26}. \end{gathered}$$

Step 7. Calculating midpoint sum .

$$\begin{gathered} \mathrm{The} \mathrm{midpoint} \mathrm{sum} \mathrm{defined} \mathrm{for} \mathrm{n} \mathrm{rectangles} \mathrm{on} [\mathrm{a},\mathrm{b}] \mathrm{is} \sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left({\mathrm{x}}_{\mathrm{k}}^{*}\right)∆\mathrm{x}. \\ \mathrm{Where} , ∆\mathrm{x}= \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},{\mathrm{x}}_{\mathrm{k}}^{*}=\frac{{\mathrm{x}}_{\mathrm{k}-1 }+{\mathrm{x}}_{\mathrm{k}}}{2}. \\ \mathrm{Now}, \mathrm{the} \mathrm{midpoint} \mathrm{term} \mathrm{is} {\mathrm{x}}_{\mathrm{k}}^{*}= \frac{{\displaystyle \frac{\mathrm{k}+6}{2}}+{\displaystyle \frac{\mathrm{k}+5}{2}}}{2} =\frac{2\mathrm{k}+11}{4}. \\ \mathrm{The} \mathrm{midpoint} \mathrm{sum} \mathrm{is}, \\ \sum _{\mathrm{k}=1 }^4\left(3\left(\frac{11+2\mathrm{k}}{2}\right)+1\right)\frac{1}{2} =\frac{1}{2}\sum _{\mathrm{k}=1}^4\left(\frac{35+6\mathrm{k}}{4}\right) \\ = \frac{1}{2}\left(\frac{35+6}{2}+\frac{35+12}{2}+\frac{35+18}{2}+\frac{35+24}{2}\right)= 26 \\ \mathrm{Therefore}, \mathrm{the} midpoint \mathrm{sum} \mathrm{is} \boxed{26} . \end{gathered}$$