Given:
![f(x)=(5)/(x),x_0=1,x_n=9](https://img.qammunity.org/2023/formulas/mathematics/high-school/an86u2rezyidkc0gb5ycpu7gx09zazhzqm.png)
Step-by-step explanation:
To find: The area when n = 2 (since, 2 rectangles)
Finding the given change in x,
![\begin{gathered} \Delta x=(b-a)/(n) \\ =(9-1)/(2) \\ =4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/cpj1w98t9goxqga3b4f0tkn2btpng8on24.png)
Defining the partition intervals,
![x:\lbrace(1,5),(5,9)\rbrace](https://img.qammunity.org/2023/formulas/mathematics/high-school/bdlx4n4yvevyvzzfflv1yzvepe8micrspo.png)
Choose the midpoint in each interval,
![x_m=\lbrace3,7\rbrace](https://img.qammunity.org/2023/formulas/mathematics/high-school/1ll2kz71byhu8j9693otsf3ys78l3aog07.png)
Finding the value of the function at the point,
![\begin{gathered} f(3)=(5)/(3) \\ f(3)\approx1.6667 \\ f(7)=(5)/(7) \\ f(7)\approx0.7143 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/rcm9jgws8u0qq7jtozl6hhg4edi8i72hvk.png)
Using the midpoint formula,
![\begin{gathered} A=\Delta x(f(x_(m_1))+f(x_(m_2))) \\ =4(1.6667+0.7143) \\ A\approx9.5238\text{ square units} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/p1vwt4pdhe6v881maegmrzdxm6ybx3agv9.png)
Final answer: The area under 2 rectangles is 9.5238 square units.
To find: The area when n = 4 (4 rectangles)
Finding the given change in x,
![\begin{gathered} \Delta x=(b-a)/(n) \\ =(9-1)/(4) \\ =2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vxhxd17z6ffzwickghhq1q1x489diema9m.png)
Defining the partition intervals,
![x:\lbrace(1,3),(3,5),(5,7),(5,9)\rbrace](https://img.qammunity.org/2023/formulas/mathematics/high-school/kk7kn2tx5rn8sptkqlwf8be92n7jy0gi4o.png)
Choose the midpoint in each interval,
![x_m=\lbrace2,4,6,8\rbrace](https://img.qammunity.org/2023/formulas/mathematics/high-school/pza7jg1yqx7bqwu48tv9sn7z4qgqxj89ai.png)
Finding the value of the function at the point,
![\begin{gathered} f(2)=(5)/(2)\Rightarrow f\mleft(2\mright)=2.5 \\ f(4)=(5)/(4)\Rightarrow f\mleft(4\mright)=1.25 \\ f(6)=(5)/(6)\operatorname{\Rightarrow}f(6)=0.8333 \\ f(8=(5)/(8)\operatorname{\Rightarrow}f(8)=0.625 \end{gathered}]()
Using the midpoint formula,
![\begin{gathered} A=\Delta x(f(x_(m_1))+f(x_(m_2))+f(x_(m_3))+f(x_(m_4))) \\ =2(2.5+1.25+0.833+0.625) \\ =2(5.2083) \\ A\approx10.4167\text{ square units} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xk2fhxi9pg01463oapmak4miu7i8pnr0g4.png)
Final answer: The area under 4 rectangles is 10.4167 square units.