Question

How do you use the midpoint rule to estimate the area?

Answer 1

Hint: This rule refers to breaking off a graph into a finite number of smaller rectangles and using each of their midpoints to get the area of that rectangle. The greater their number, the more accurate the estimation, and the most accurate of them all is the process of integration. The rule though focuses on a much lesser number of rectangles, preferably 4 to 8.

Formulas used:
Width formula: $\Delta x = \dfrac{{b - a}}{n}$
Area additive formula: ${A_m} = \Delta x \cdot \sum\limits_{i = 1}^n {f\left( {{x_i}} \right)} $

Complete step by step solution:
In this estimation method, we use rectangles to denote a graph, and the midpoint of each rectangle is used as a reference for height whereas the width is given by the division of total domain by a number of rectangles in consideration. So, for moving forward, let us consider a function $f\left( x \right) = {x^3}$ for the domain $x \in \left[ {2,4} \right]$.
Let’s first get the area via integration for keeping a reference of the same.
\[\
  A = \int\limits_2^4 {f\left( x \right)dx} \\
   = \int\limits_2^4 {{x^3}dx} \\
   = \left( {\dfrac{{{x^4}}}{4}} \right)_2^4 \\
   = \dfrac{{{4^4}}}{4} - \dfrac{{{2^4}}}{4} \\
   = 64 - 4 \\
   = 60 \\
\ \]
Now by midpoint estimation rule, let us begin solving by dividing the domain into 4 parts and representing them on a number line. So, the width of the parts can be given by the width formula which will be
$\
  \Delta x = \dfrac{{b - a}}{n} \\
   = \dfrac{{4 - 2}}{4} \\
   = 0.5 \\
\ $
As such our points on the number line will be 2, 2.5, 3.0, 3.5, and 4 which we get after adding width to the domain. They are marked below the number line and our midpoints or midpoints of each domain are marked above the number line which we will use in the formula for area.

So, the area can be given by,
$\
  {A_m} = \Delta x \cdot \sum\limits_{i = 1}^n {f\left( {{x_i}} \right)} \\
   = \Delta x \cdot \left[ {f\left( {{x_1}} \right) + f\left( {{x_2}} \right) + f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right] \\
   = 0.5\left[ {{{\left( {2.25} \right)}^3} + {{\left( {2.75} \right)}^3} + {{\left( {3.25} \right)}^3} + {{\left( {3.75} \right)}^3}} \right] \\
   = 0.5\left[ {11.3906 + 20.7969 + 34.3281 + 52.7344} \right] \\
   = 0.5 \times 119.25 \\
   = 59.625 \\
\ $
Here we can see that the area is approximately equal to the area found by integration but not exactly equal and that is because we approximated the area into boxes and the bigger the boxes, the more the area of the curve gets omitted. Increase the number of boxes by making them smaller and less area will be omitted in the end.

Note:
There are left-hand and right-hand rules of the same method we used where the right hand gives a bigger approximation and the left hand gives a smaller approximation. In this method though, using midpoint gives the most accurate results. Also, more number of divisions in the method gives more accurate results which culminate in integration where we imagine the piece divided to be minuscule.