Q. 34 - Calculus

Question

$$\begin{gathered} \mathrm{For} \mathrm{given} \mathrm{function} f \mathrm{and} \mathrm{interval} [\mathrm{a}, \mathrm{b}], \mathrm{it} \mathrm{is} \mathrm{possible} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{exact} \\ \mathrm{signed} \mathrm{area} \mathrm{between} \mathrm{the} \mathrm{graph} \mathrm{of} f \mathrm{and} \mathrm{the} \mathrm{x}-\mathrm{axis} \mathrm{on} [\mathrm{a}, \mathrm{b}] \mathrm{geometrically}. \\ \mathrm{Find} \mathrm{this} \mathrm{exact} \mathrm{area}, \mathrm{and} \mathrm{then} \mathrm{calculate} \mathrm{the} \mathrm{left}, \mathrm{right} \mathrm{and} \mathrm{trapezoid} \mathrm{sums} \\ \mathrm{with} n = 4. \\ f\left(x\right) = 5, \left[a, b\right] = \left[-2, 2\right] \end{gathered}$$

Step-by-Step Solution

Answer

left-sum = 20, right-sum= 20 , trapezoid sum = 20.

Step 1. Given Information

$\mathrm{The} \mathrm{function} \mathrm{is} f\left(x\right)= 5 \mathrm{and} \mathrm{the} \mathrm{interval} \mathrm{is} [a,b] =[-2,2].$

Step 2. Finding area.

The value of n = 4

Calculate the exact area using calculator 

Press Math and Select 9.

Step 3. Calculating Exact Sum

$$\begin{gathered} \mathrm{Enter} \mathrm{the} \mathrm{function} (\boxed{5}, \mathrm{X},-2,2) \\ \mathrm{Press} \mathrm{ENTER} \\ \mathrm{The} \mathrm{display} \mathrm{is} \mathrm{shown} \mathrm{below}, \end{gathered}$$

Therefore ,the exact sum is $\boxed{20}$  

Step 4. Calculating left-sum

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

Step 5. Calculating the right-sum

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

Step 6. calculating trapezoid sum

$$\begin{gathered} \mathrm{The} \mathrm{trapezoid} \mathrm{sum} \mathrm{is} \\ \frac{20 + 20}{2} \\ = 20 \\ \mathrm{Therefore} , \mathrm{the} \mathrm{trapezoid} \mathrm{sum} \mathrm{is} \boxed{20}. \end{gathered}$$