3.3 Approximating Area with \(n\) Rectangular Regions

3.3.1 General Set up

 

Goal. Given a function \(f\) such that \(f(x)\geq 0\) for all \(x\in [a,b]\), our aim here is to develop a method to compute \(R(f,n)\) for an arbitrary number} of rectangular regions \(n\).

 

The method will require that we deal with three variables:

\(x\): an arbitrary number in the interval \([a,b]\).

\(n\): number of rectangular regions used to approximate the area.

\(i\): an index, \(i=0,1,2,3,\dots, n\).

3.3.2 General Method to Compute the Exact Area

 

Step 1. Partition of the interval into \(n\) subintervals of equal length, \(\Delta x\).

 

Step 2. Determination of the height, \(h_i=f(x_i)\), and area, \(A_i = f(x_i)\Delta x\), of each rectangular region.

 

Step 3. Add and simplify the sum of the areas of the \(n\) rectangles, \(\displaystyle \sum_{i=1}^{n}f(x_i)\Delta x\).

 

Step 4. Compute the limit when \(n\to \infty\): \(\displaystyle \lim_{n\to \infty} \sum_{i=1}^{n}f(x_i)\Delta x\).

3.3.3 Area under the Graph of a Linear Function: a Trapezoid

 

The following example outlines the general procedure to compute the area under the graph of a linear function. In this case, it is possible to compute such an area using elementary geometry. In addition, the computations are relatively simple, thus making it easier to understand the general principles involved in the approximation method.

The method involves four steps, and you are strongly advised to learn well the four steps and how they are executed.

Example and General Procedure. Given the function \(f(x)=3x+4, x \in [-1,3]\), let \(\mathcal{R}\) be the region enclosed by its graph and the \(x\)-axis.

(a) Sketch \(\mathcal{R}\).

(b) Use elementary geometry to find the area of \(\mathcal{R}\).

(c) Compute \(R(f,n)\), for \(n=4\), and draw the rectangular approximation.

(d) Find a formula for \(R(f,n)\).

(e) Rewrite the formula for \(R(f,n)\) in (d) to compute \(\displaystyle \lim_{n\to \infty}\sum _{i=1}^n f\left(x_i\right)\Delta x\). Compute the limit.

Confirm that your answers in (b) and (e) are compatible.

Solution. (a) The graph of \(f\) is the line segment connecting the points \((-1,f(-1))=(-1,1)\) and
\((3,f(3))= (3, 13)\). Thus, \(\mathcal{R}\) is a trapezoid (see Figure 3.7(a)).

(b) Recall that the area of a trapezoid can be computed via the formula \(\displaystyle A =h\frac{b_1+b_2}{2}\), where \(h\) is the height, and \(b_1\) and \(b_2\) are the bases of the trapezoid. Thus, in this case, we have,
\[A = 4\cdot \frac{1+13}{2}=28.\](c) Computation of \(R(f,4)\).

(1) Partition the interval into four subintervals of equal length, \(\displaystyle \Delta x=\frac{3-(-1)}{4}=1\).

 

Points of the partition: \[x_0=-1,\;\; x_1=-1+1=0,\;\; x_2=-1+2(1)=1,\;\; x_3=-1+3(1)=2,\;\; x_4=-1+4(1)=3.\]

 

(2) Height and area of each rectangular region.
\[
\renewcommand{\arraystretch}{1.75}
\begin{array}{c|l|l}
\text{Rectangle} & \hspace{.6in}\text{Height} & \hspace{.5in}\text{Area}\\ \hline
R_1 &
h_1=f\left(x_1\right)=f\left(0\right)=4&
A_1=f\left(x_1\right)\Delta x=4\cdot 1\\
R_2 &
h_2=f\left(x_2\right)=f(1)=7&
A_2=f\left(x_2\right)\Delta x=7\cdot 1\\
R_3 &
h_3=f\left(x_3\right)=f\left(2\right)=10&
A_3=f\left(x_3\right)\Delta x=10\cdot 1\\
R_4 &
h_4=f\left(x_4\right)=f\left(3\right)=4&
A_4=f\left(x_4\right)\Delta x=13\cdot 1
\end{array}
\]

 

(3) Sum of the areas of the rectangular regions: \(R(f,4)=4\cdot 1 +7\cdot 1+10\cdot 1+13\cdot 1 = 34\).

 

(d) and (e) General Method: Four Steps

 

Here are the four steps to compute \({\displaystyle \lim_{n\to \infty}}R(f,n)\). Make sure you learn them well, and are able to carry them out.

 

Step 1. Partition of the interval into subintervals of equal length.

 

Length of each subinterval \(\displaystyle = \Delta x=\frac{3-(-1)}{n}=\frac{4}{n}\)

\[ x_0=-1, x_1=-1+\frac{4}{n}, x_2=-1+2\cdot\frac{4}{n}, x_3=-1+3\cdot\frac{4}{n}, \dots ,x_i=-1+i\cdot\frac{4}{n}, \dots , x_n=-1+n \frac{4}{n}=3.\]

The generic \(i\)-th point is thus given by \(\displaystyle x_i=-1+\frac{4i}{n}\), \(i=0, 1, 2,\dots, n\).

 

The generic \(i\)-th interval is give by \(\displaystyle I_i=\left[x_{i-1}, x_i\right]=\left[-1+\frac{4(i-1)}{n},-1+\frac{4i}{n}\right]\), \(i=1,2,3,\dots,
n\).

 

Step 2. Determination of the height and area of each rectangle.

 

The height and area of the \(i\)-th rectangle are given by \( h_i = f\left(x_i\right)\), and \(A_i=\text{height}\times \text{base}=f\left(x_i\right)\Delta x\). And since \(\displaystyle x_i=-1+\frac{4i}{n}\) then,
\[ h_i = f\left(x_i\right)=f\left(-1+\frac{4i}{n}\right), \;\; \; \text{and}\;\; A_i=f\left(x_i\right)\Delta x=f\left(-1+\frac{4i}{n}\right)\frac{4}{n}.\]
Computation of the height:
\[h_i=f\left(x_i\right)=f\left(-1+\frac{4i}{n}\right)=3\left(-1+\frac{4i}{n}\right)+4 = 1+\frac{12i}{n}.\]
Conclusion: the height and the area of the \(i\)-th rectangular region are given by
\[h_i=f\left(x_i\right)=1+\frac{12i}{n}, \;\; \text{and}\;\;
A_i= h_i b_i = f\left(x_i\right)\Delta x =\left(1+\frac{12i}{n}\right)\frac{4}{n}.\]

 

Step 3. Adding and simplifying the sum of the areas of the rectangles.

 

\begin{align*}
\sum _{i=1}^n f\left(x_i\right)\Delta x
&= \sum _{i=1}^n \left(1+\frac{12i}{n}\right)\frac{4}{n}
=\frac{4}{n}\sum_{i=1}^n\left(1+\frac{12i}{n}\right)
=\frac{4,}{n}\left(\sum _{i=1}^n1+\sum _{i=1}^n \frac{12i}{n}\right)=\\
&=\frac{4}{n}\left(\sum _{i=1}^n1+ \frac{12}{n}\sum _{i=1}^ni\right)
=\frac{4}{n}\left(n + \frac{12}{n}\frac{n(n+1)}{2}\right)
=4\left(1+6\frac{n+1}{n} \right).
\end{align*}

Remarks.

(1) Note that the common factor \(\Delta x=\frac{4}{n}\) can be taken out of the summation sign.

(2) Note that the final answer to each sum is left in factored form. This is convenient for two reasons, simplification and ability to take the limit as \(n\to \infty\).

 

Step 4. Computing the limit when the number of rectangles grows to infinity?

 

The last expression can be rewritten as follows (make sure to justify each step):
\[\sum _{i=1}^n f\left(x_i\right)\Delta x
= 4\left(1+6\frac{n+1}{n} \right) = 4\left(1+6\left(1+\frac{1}{n}\right) \right)
=4\left(7+6\frac{1}{n} \right).\]

The reason to take this extra step to rewrite the expression for \(\displaystyle \sum _{i=1}^n f\left(x_i\right)\Delta x\) is because the approximations to the exact area of the region improve as the number of rectangular regions increases. As \(n\) increases, the terms \(\displaystyle \frac{1}{n}\) decrease and become negligible, that is \(\displaystyle \frac{1}{n}\to 0\) as \(n \to \infty\), thus, the sum gets closer and closer to the actual area as follows:
\[\sum _{i=1}^n f\left(x_i\right)\Delta x = 4\left(7+6\frac{1}{n} \right)\overset{n\to \infty }{\to }4\left(7+6(0) \right) = 28.\]
Hence,
\[\text{Area of the region under the graph of } f =\lim_{n\to \infty}\sum _{i=1}^n f\left(x_i\right)\Delta x =28.\]

3.3.4 Area under the Graph of a Polynomial Function of Degree 3

Example and General Procedure. Find a formula to compute
\[R(f,n), \;\; \text{ for } \;\; f(x)=x^3-x^2+2, \;\; x \in [-1,2],\]
and compute the limit of \(R(f,n)\) as \(n \to \infty\).

 

Step 1. Partition of the interval into subintervals of equal length.

 

Length of each subinterval \(\displaystyle = \Delta x=\frac{2-(-1)}{n}=\frac{3}{n}\)

\[ x_0=-1, x_1=-1+\frac{3}{n}, x_2=-1+2\cdot\frac{3}{n}, x_3=-1+3\cdot\frac{3}{n}, \dots ,x_i=-1+i\cdot\frac{3}{n}, \dots , x_n=-1+n \frac{3}{n}=2.\]
The generic \(i\)-th point is thus given by \(\displaystyle x_i=-1+\frac{3i}{n}\), \(i=0, 1, 2,\dots, n\).

 

The generic \(i\)-th interval is give by \(\displaystyle I_i=\left[x_{i-1}, x_i\right]=\left[-1+\frac{3(i-1)}{n},-1+\frac{3i}{n}\right]\), \(i=1,2,3,\dots,
n\).

 

Step 2. Determination of the height and area of each rectangle.

The height and area of the \(i\)-th rectangle are given by \( h_i = f\left(x_i\right)\), and \(A_i=\text{height}\times \text{base}=f\left(x_i\right)\Delta x\). And since \(\displaystyle x_i=-1+\frac{3i}{n}\) then,
\[ h_i = f\left(x_i\right)=f\left(-1+\frac{3i}{n}\right), \;\; \; \text{and}\;\; A_i=f\left(x_i\right)\Delta x=f\left(-1+\frac{3i}{n}\right)\frac{3}{n}.\]
Computation of the height:
\begin{align*}
h_i=f\left(x_i\right)&=f\left(-1+\frac{3i}{n}\right)=\left(-1+\frac{3i}{n}\right)^3-\left(-1+\frac{3i}{n}\right)^2+2
=\\
&=-1+\frac{9i}{n}-\frac{27i^2}{n^2}+\frac{27i^3}{n^3}-\left(1-\frac{6i}{n}+\frac{9i^2}{n^2}\right)+2
=\frac{27i^3}{n^3}-\frac{36i^2}{n^2}+\frac{15i}{n}.
\end{align*}
Conclusion:
\[\displaystyle h_i=f\left(x_i\right)=\frac{27i^3}{n^3}-\frac{36i^2}{n^2}+\frac{15i}{n}, \;\; \text{and}\;\;
A_i= h_i b_i = f\left(x_i\right)\Delta x =\left(\frac{27i^3}{n^3}-\frac{36i^2}{n^2}+\frac{15i}{n}\right)\frac{3}{n}.\]

 

Step 3. Adding and simplifying the sum of the areas of the rectangles

 

Justify the following steps (see 3.2.2).
\begin{align*}
\sum _{i=1}^n f\left(x_i\right)\Delta x &
= \sum _{i=1}^n \left(\frac{27i^3}{n^3}-\frac{36i^2}{n^2}+\frac{15i}{n}\right)\frac{3}{n}=\frac{3}{n}\sum_{i=1}^n \left(\frac{27i^3}{n^3}-\frac{36i^2}{n^2}+\frac{15i}{n}\right)=\\
&=\frac{3}{n}\left(\sum _{i=1}^n \frac{27i^3}{n^3}-\sum _{i=1}^n \frac{36i^2}{n^2}+\sum
_{i=1}^n \frac{15i}{n}\right)
=\frac{3}{n}\left(\frac{27}{n^3}\sum _{i=1}^n i^3-\frac{36}{n^2}\sum _{i=1}^n i^2+\frac{15}{n}\sum _{i=1}^n i\right) =\\
&=\frac{3}{n}\left(\frac{27}{n^3}\left(\frac{n(n+1)}{2}\right)^2-\frac{36}{n^2}\frac{n(n+1)(2n+1)}{6}+\frac{15}{n}\frac{n(n+1)}{2}\right).
\end{align*}

 

Step 4. Computing the limit when the number of rectangles tends to infinity?

 

The last expression can be rewritten as follows (make sure to justify each step):
\begin{align*}
\sum _{i=1}^n f\left(x_i\right)\Delta x
&\overset{?}{=} 3\left(\frac{27}{4}\frac{n^2(n+1)^2}{n^4}-6\frac{n(n+1)(2n+1)}{n^3}+\frac{15}{2}\frac{n(n+1)}{n^2}\right)=\\
&\overset{?}{=} 3\left(\frac{27}{4}\frac{(n+1)^2}{n^2}-6\frac{(n+1)(2n+1)}{n^2}+\frac{15}{2}\frac{n+1}{n}\right)=\\
&\overset{(*)}{=} 3\left(\frac{27}{4}\left(1+\frac{1}{n}\right)^2-6\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)+\frac{15}{2}\left(1+\frac{1}{n}\right)\right)
\end{align*}

\((*)\) (1) Note that none of the products was expanded. (2) The reason to take this extra step to rewrite the expression for \(\displaystyle \sum _{i=1}^n f\left(x_i\right)\Delta x\) is because the approximations to the exact area of the region improve as the number of rectangular regions increases. As \(n\) increases, the terms \(\displaystyle \frac{1}{n}\) decrease and become negligible, that is \(\displaystyle \frac{1}{n}\to 0\) as \(n \to \infty\), thus, the sum gets closer and closer to the actual area as follows:
\begin{align*}
\sum _{i=1}^n f\left(x_i\right)\Delta x
&= 3\left(\frac{27}{4}\left(1+\frac{1}{n}\right)^2-6\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)+\frac{15}{2}\left(1+\frac{1}{n}\right)\right)\overset{n\to \infty }{\to }\\
&\overset{n\to \infty }{\to } 3\left(\frac{27}{4}(1+0)^2-6(1+0)(2+0)+\frac{15}{2}(1+0)\right) =\\
&=3\left(\frac{27}{4}-12+\frac{15}{2}\right)=3\left(\frac{27-48+30}{4}\right)=3\left(\frac{9}{4}\right)=\frac{27}{4}
\end{align*}
Hence,

Area of the region under the graph of \(\displaystyle f =\lim_{n\to \infty}\sum _{i=1}^n f\left(x_i\right)\Delta x =\frac{27}{4}\).

 

On Your Own 1. Find a formula to compute \(R(f,n)\) where \(f(x)=x^2+1\), \(x \in [-1,2]\), and compute \(\displaystyle \lim_{n\to \infty} R(f,n)= \lim_{n\to \infty} \sum _{i=1}^n f\left(x_i\right)\Delta x \).