The diagram below represents the area approximated using rectangular elements,
The interval width when an interval [a,b] is divided into 'n' rectangles is given by,
[tex]\Delta x=\frac{b-a}{n}[/tex]
According to the given problem,
[tex]\begin{gathered} f(x)=x^2+3x \\ a=1 \\ b=5 \\ n=4 \end{gathered}[/tex]
Then the width of each interval will be,
[tex]\begin{gathered} \Delta x=\frac{5-1}{4} \\ \Delta x=\frac{4}{4} \\ \Delta x=1 \end{gathered}[/tex]
The left end-points of the approximating rectangles are,
[tex]1,2,3,4[/tex]
The value of the function at that left end-point gives the height of that rectangle. So the height of each rectangle will be,
[tex]\begin{gathered} f(1)=(1)^2+3(1)=1+3=4 \\ f(2)=(2)^2+3(2)=4+6=10 \\ f(3)=(3)^2+3(3)=9+9=18 \\ f(4)=(4)^2+3(4)=16+12=28 \end{gathered}[/tex]
Then the approximate area (A) under the curve between the limits is given by the sum of the areas of all these 4 approximating rectangles,
[tex]\begin{gathered} A=\mleft\lbrace\Delta x\cdot f(1)\}+\mright?\lbrace\Delta x\cdot f(2)\}+\lbrace\Delta x\cdot f(3)\}+\lbrace\Delta x\cdot f(4)\} \\ A=\Delta x\cdot\mleft\lbrace f(1)+f(2)+f(3)+f(4)\}\mright? \end{gathered}[/tex]
Substitute the values and simplify,
[tex]\begin{gathered} A=1\cdot(4+10+18+28) \\ A=60 \end{gathered}[/tex]
Thus, the approximate area under the curve between the given limits is 60 square units.
