Approximating area with rectangles

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We introduce the basic idea of using rectangles to approximate the area under a curve.

Rectangles and areas

We can calculate areas of many different shapes: rectangles, triangles, trapezoids, circles, ... etc. What about the area of something like the region under the graph of a function?

We want to compute the area between the curve $y=f(x)$ and the horizontal axis on the interval $[a,b]$ :

One way to do this would be to approximate the area with rectangles. With one rectangle we get a rough approximation:

Two rectangles might make a better approximation:

With even more, we get a closer, and closer, approximation:

If we are approximating the area between a curve and the $x$ -axis on $[a,b]$ with $n$ rectangles of width $\Delta x$ , then $\Delta x = \frac {b-a}{n}.$

Suppose we wanted to approximate area between the curve $y=x^2+1$ and the $x$ -axis on the interval $[-1,1]$ , with $8$ rectangles. What is $\Delta x$ ? $\Delta x = \answer {1/4}$

As we add rectangles, we are more closely approximating the area we are interested in:

Let’s setup some notation to help with these calculuations:

When approximating an area with $n$ rectangles, the grid points $x_0,x_1,x_2,\dots ,x_n$ are the $x$ -coordinates that determine the edges of the rectangles. In the graph below, we’ve numbered the rectangles to help you see the relation between the indices of the grid points and the $k$ th rectangle.

Note, if we are approximating the area between a curve and the horizontal axis on $[a,b]$ with $n$ rectangles, then it is always the case that $x_0=a\qquad \text {and}\qquad x_n = b.$

If we are approximating the area between a curve and the horizontal axis with $11$ rectangles, how many grid points will we have?

You can draw it!

We’ll have $\answer {12}$ grid points.

But which set of rectangles?

When we use $n$ rectangles to compute the area under a curve, the width of each rectangle is given by $\Delta x=\frac {b-a}{n}$ . It is clear that $\Delta x=x_k-x_{k-1}$ , for $k=1,\dots , n$ .

But how do we determine the height of the rectangle?

We choose a sample point $x_k^*$ and evaluate the function at that point. The value $f(x_k^*)$ determines the height of a rectangle.

When approximating an area with rectangles, a sample point is the $x$ -coordinate that determines the height of $k^{th}$ rectangle. For $k=1,\dots , n$ , we denote a sample point as: $x_k^*$ and the value $f( x_k^*)$ is the height of the $k^{th}$ rectangle.

What is the area of the $k^{th}$ rectangle shown in the figure above?

$A=\Delta x$ $A=f(x)\Delta x$ $A= f( x_k^*)x_k^*$ $A= f( x_k^*)\Delta x$ $A=f(x)x$ $A=k\Delta k$ $A= f( x_k^*)(x_k-x_{k-1})$

Here are three options for sample points that we consider:

Rectangles defined by left-endpoints

We can set the rectangles up so that the left-endpoint touches the curve.

In the graph above, the $k^{th}$ rectangle’s left-endpoint is touching the curve.

Rectangles defined by right-endpoints

We can set the rectangles up so that the right-endpoint touches the curve.

In the graph above, the $k^{th}$ rectangle’s right-endpoint is touching the curve.

Rectangles defined by midpoints

We can set the rectangles up so that the midpoint of one of the horizontal sides touches the curve.

In the graph above, the midpoint of the horizontal side of the $k^{th}$ rectangle is touching the curve.

Riemann sums and approximating area

Once we know how to identify our rectangles, we can compute some approximate areas. If we are approximating area with $n$ rectangles, then the area of the $k$ th rectangle (for $k$ between $1$ and $n$ ) is given by: $(\text {height of $k$th rectangle}) \times (\text {width of $k$th rectangle}) = f(x_k^*) \Delta x$ The area of the region is approximately: $\text {Area} \approx f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x + \dots + f(x_n^*)\Delta x.$

A sum of the form: $f(x_1^*)\Delta x + f(x_2^*)\Delta x + \dots + f(x_n^*)\Delta x$ is called a Riemann sum, pronounced “ree-mahn” sum.

A Riemann sum computes an approximation of the area between a curve and the $x$ -axis on the interval $[a,b]$ . It can be defined several different ways. In our class, it will be defined via left-endpoints, right-endpoints, or midpoints. Here we see the explicit connection between a Riemann sum defined by left-endpoints and the area between a curve and the $x$ -axis on the interval $[a,b]$ :

and here is the associated Riemann sum $f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x + f(x_4^*)\Delta x + f(x_5^*)\Delta x.$

Left Riemann sums

Consider $f(x) = x^3/8-x+2$ . Approximate the area between $f$ and the $x$ -axis on the interval $[-1,3]$ using a left-endpoint Riemann sum with $n=5$ rectangles.

First note that the width of each rectangle is $\Delta x = \frac {3-(-1)}{5} = \answer [given]{4/5}.$ The grid points define the edges of the rectangle and are seen below:

On the other hand, the sample points identify which endpoints we use:

It is helpful to collect all of this data into a table: $\begin {array}{c|c|c|c} k & x_k & x^*_k & f(x^*_k) \\ \hline 0 & \answer [given]{-1} & \text {NA} & \text {NA} \\ 1 & \answer [given]{-0.2} & \answer [given]{-1} & \answer [given]{2.875} \\ 2 & \answer [given]{0.6} & \answer [given]{-0.2} & \answer [given]{2.199} \\ 3 & \answer [given]{1.4} & \answer [given]{0.6} & \answer [given]{1.427} \\ 4 & \answer [given]{2.2} & \answer [given]{1.4} & \answer [given]{0.943} \\ 5 & \answer [given]{3} & \answer [given]{2.2} & \answer [given]{1.131} \\ \end {array}$ Now we may write a left Riemann sum and approximate the area $f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x +f(x_4^*)\Delta x+ f(x_5^*)\Delta x,$ which evaluates to $\begin{align*} = 2.875 \cdot (4/5) &+ 2.199 \cdot(4/5) + 1.427 \cdot(4/5)\\ &+ 0.943 \cdot(4/5) + 1.131 \cdot(4/5) \end{align*}$

and we find $= \answer [given]{6.86}.$

Right Riemann sums

Consider $f(x) = x^3/8-x+2$ . Approximate the area between $f$ and the $x$ -axis on the interval $[-1,3]$ using a right-endpoint Riemann sum with $n=5$ rectangles.

First note that the width of each rectangle is $\Delta x = \frac {3-(-1)}{5} = \answer [given]{4/5}.$ The grid points define the edges of the rectangle and are seen below:

On the other hand, the sample points identify which endpoints we use:

It is helpful to collect all of this data into a table: $\begin {array}{c|c|c|c} k & x_k & x^*_k & f(x^*_k) \\ \hline 0 & \answer [given]{-1} & \text {NA} & \text {NA} \\ 1 & \answer [given]{-0.2} & \answer [given]{-0.2} & \answer [given]{2.199} \\ 2 & \answer [given]{0.6} & \answer [given]{0.6} & \answer [given]{1.427} \\ 3 & \answer [given]{1.4} & \answer [given]{1.4} & \answer [given]{0.943} \\ 4 & \answer [given]{2.2} & \answer [given]{2.2} & \answer [given]{1.131} \\ 5 & \answer [given]{3} & \answer [given]{3} & \answer [given]{2.375} \\ \end {array}$ Now we may write a right Riemann sum and approximate the area $f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x +f(x_4^*)\Delta x+ f(x_5^*)\Delta x,$ which evaluates to $\begin{align*} = 2.199 \cdot (4/5) &+ 1.427 \cdot(4/5) + 0.943 \cdot(4/5)\\ &+ 1.131 \cdot(4/5) + 2.375 \cdot(4/5) \end{align*}$

and we find $= \answer [given]{6.46}.$

Midpoint Riemann sums

Consider $f(x) = x^3/8-x+2$ . Approximate the area between $f$ and the $x$ -axis on the interval $[-1,3]$ using a midpoint Riemann sum with $n=5$ rectangles.

First note that the width of each rectangle is $\Delta x = \frac {3-(-1)}{5} = \answer [given]{4/5}.$ The grid points define the edges of the rectangle and are seen below:

On the other hand, the sample points identify which endpoints we use:

It is helpful to collect all of this data into a table: $\begin {array}{c|c|c|c} k & x_k & x^*_k & f(x^*_k) \\ \hline 0 & \answer [given]{-1} & \text {NA} & \text {NA} \\ 1 & \answer [given]{-0.2} & \answer [given]{-0.6} & \answer [given]{2.573} \\ 2 & \answer [given]{0.6} & \answer [given]{0.2} & \answer [given]{1.801} \\ 3 & \answer [given]{1.4} & \answer [given]{1} & \answer [given]{1.125} \\ 4 & \answer [given]{2.2} & \answer [given]{1.8} & \answer [given]{0.929} \\ 5 & \answer [given]{3} & \answer [given]{2.6} & \answer [given]{1.597} \\ \end {array}$ Now we may write a midpoint Riemann sum and approximate the area $f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x+ f(x_4^*)\Delta x + f(x_5^*)\Delta x,$ which evaluates to $\begin{align*} = 2.573 \cdot (4/5) &+ 1.801\cdot(4/5) + 1.125 \cdot(4/5) \\ &+ 0.929 \cdot(4/5) + 1.597\cdot(4/5) \end{align*}$

and we find $= \answer [given]{6.42}.$

Summary

Riemann sums approximate the area between curves and the $x$ -axis via rectangles. When computing this area via rectangles, there are several things to know:

What interval are we on? In our discussion above we call this $[a,b]$ .
How many rectangles will be used? In our discussion above we called this $n$ .
What is the width of each individual rectangle? In our discussion above we called this $\Delta x$ .
What points will determine the height of the rectangle? In our discussion above we called these sample points, $x_k^*$ , and they can be left-endpoints, right-endpoints, or midpoints.
What is the actual height of the rectangle? This will always be $f(x_k^*)$ .