To calculate the area for the upper (left) graph, we can use x = 1, 2, 3 and 4 to find the upper limit of each rectangle:
![\begin{gathered} f(1)=(3)/(1)+3=6\\ \\ f(2)=(3)/(2)+3=4.5\\ \\ f(3)=(3)/(3)+3=4\\ \\ f(4)=(3)/(4)+3=3.75 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s7yohuuiesgma0wkpib60mem8ohllnqad3.png)
Since the x-interval of each rectangle is 1 unit, the area of each rectangle is given by its y-value, so we have:
![\begin{gathered} A=f(1)+f(2)+f(3)+f(4)\\ \\ A=6+4.5+4+3.75=18.25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/aud1ksowsxmzub83rw2fk2sx354y14uvww.png)
Now, for the bottom (right) graph, the limits of the rectangles are x = 2, 3, 4 and 5.
So, let's find the value of f(5):
![f(5)=(3)/(5)+3=3.6](https://img.qammunity.org/2023/formulas/mathematics/college/wmq6iy84hmo7m34mlqjf42coaiuzvmxpxg.png)
So the area is given by:
![\begin{gathered} A=f(2)+f(3)+f(4)+f(5)\\ \\ A=4.5+4+3.75+3.6=15.85 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gal7qgmuyej75djcz2emreiqxih7e1bygg.png)