Chapter 5.9 Approximating Area in the Plane

In this section, we explore ways to approximate and calculate the area of regions that are represented by the graphs of certain functions. We learn how to calculate the exact area of specific regions that are either squares, rectangles, triangles, or circles. Then, we discuss an approximation method to determine the area of more complicated regions. Finally, we discuss certain properties of the approximation methods that can aid in calculating and estimating the areas.

Area of Regions Bounded by Simple Regions

Back in the early parts of Chapter 5, we discussed how to calculate the area of a polygon \(P\). There were two ways we did it:

Construct a triangulation for \(P\) and summing up the area of triangle in the triangulation.
Use the shoelace formula on \(P.\)

The limitation to this method is that it only works for polygons. However, in Chapter 5.4, we discussed an approximation scheme that can be used to approximate the area of area of a circle, which is to construct polygons that “cover” more of the circle, but the gabs of the circle get smaller and smaller.

This idea can be used to calculate more complicated areas. For example, the region given below:

This region is not a triangle because the “hypothenus” is not a straight line. We will come back to this shape in the next subsection.

For now we focus on regions \(R\) whose boundaries \(\partial R\) can be described by a piecewise function of the form

\[c(t)=\begin{cases}(t,f(t))&\text{if }a\leq t\leq b\\ (b,f(b)-(t-b)f(b))&\text{if }b\leq t\leq b+1\\ ((t-b-1)(a-b)+b,0)&\text{if }b+1\leq t\leq b+2\\ (a,(t-b-2)f(a))&\text{if }b+2\leq t\leq b+3\end{cases}\]

where \(f\) is a continuous, non-negative function on \([a,b].\)

Although the formula looks complicated, all it is that the boundary is describing the region \(R\) that is form by bounding

the \(x\)-axis,
the vertical line \(x=a\)
the vertical line \(x=b\)
the function \(f\).

In the first example, we study a region made up of just vertical and horizontal lines. For this kind of region, we can calculate its area.

Example 1

Take \(R\) to be the region formed by bounding \(f(x)=4\), \(x=1\), and \(x=5\) and the \(x\)-axis. Determine if the area of \(R\) can be calculated exactly.

The region \(R\) looks like this:

This region is a rectangle with width equal to 4 and length equal to 4. Its area is thus \(\mathcal{A}(R)=4\cdot 4=16.\)

What happens if the region \(R\) is constructed with a function that is negative or below the \(x\)-axis? If that case, when the parameterization of the boundary is constructed, the orientation will be negative.

In that case, we will refer to it as signed area. In the next example, calculate the area of the following region.

Example 2

Take \(R\) to be the region formed by bounding \(f(x)=-6\), \(x=2\), and \(x=4\) and the \(x\)-axis. Determine if the area of \(R\) can be calculated exactly.

The region \(R\) looks like this:

This region is a rectangle below the \(x\)-axis with width equal to 2 and length equal to 6. Its signed area is thus \(\mathcal{A}(R)=-2\cdot 6=-12.\)

In the next example, pay attention to how we decompose the region \(R\) into two sub-regions that we can calculate the area of.

Example 3

Take \(R\) to be the region formed by bounding \(f(x)=x-4\), \(x=1\), and \(x=5\) and \(y=0\). Determine if the signed area of \(R\) can be calculated exactly.

The region \(R\) looks like this:

Decompose \(R\) into two regions, \(R_1\) and \(R_2\) like this:

Both regions are triangles, so \[\mathcal{A}(R_1)=-\frac{1}{2}\cdot 3\cdot 3=-\frac{9}{2}\quad\text{and}\quad\mathcal{A}(R_2)=\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}.\] Therefore, the area of \(R\) is \[\mathcal{A}(R)=\mathcal{A}(R_1)+\mathcal{A}(R_2)=-\frac{9}{2}+\frac{1}{2}=-4.\]

In the next example, we determine the area by graphing the region and identifying the geometric shape.

Example 4

Take \(R\) to be the region formed by bounding \(f(x)=\sqrt{4-x^2}\), \(x=-2\), and \(x=2\) and \(y=0\). Determine if the area of \(R\) can be calculated exactly.

The region \(R\) looks like this:

The region \(R\) looks like half a circle. To confirm, rewrite the formula like this: \[\begin{align*} f(x)&=\sqrt{4-x^2}\\ y&=\sqrt{4-x^2}\\ y^2&=4-x^2\\ x^2+y^2&=4. \end{align*}\] Indeed, the region is the upper half of the circle with center \((0,0)\) and radius \(r=2\).

Therefore, the area of \(R\) is \[\mathcal{A}(R)=\frac{1}{2}\pi\cdot r^2=\frac{1}{2}\pi\cdot 2^2=2\pi.\]

In this next example, we also determine the area by identifying the geometric shape.

Example 5

Take \(R\) to be the region formed by bounding \(f(x)=3\sqrt{1-x^2}\), \(x=-1\), and \(x=1\) and \(y=0\). Determine if the area of \(R\) can be calculated exactly.

The region \(R\) looks like this:

The region \(R\) looks like half an ellipse. To confirm, rewrite the formula like this: \[\begin{align*} f(x)&=3\sqrt{1-x^2}\\ y&=3\sqrt{1-x^2}\\ \frac{y}{3}&=\sqrt{1-x^2}\\ \frac{y^2}{9}&=1-x^2\\ x^2+\frac{y^2}{9}&=1\\ \frac{x^2}{1}+\frac{y^2}{9}&=1.\\ \end{align*}\] Indeed, the region is the upper half of the ellipse that is the circle with center \((0,0)\) and radius \(1\) asymmetrically scaled in \(y\) by \(3\) (so \(a=1\) and \(b=3\)).

Therefore, the area of \(R\) is \[\mathcal{A}(R)=\frac{1}{2}\pi\cdot a\cdot b=\frac{1}{2}\pi\cdot 1\cdot 3=\frac{3}{2}\pi.\]

In the above example, we use the fact that the ellipse whose equation is given by \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] has area equal to \(\pi \cdot ab.\)

Approximating Area

Now we return to the region from the start of the chapter.

This is actually the graph of \(f(x)=x^2\) on \([0,1].\) Because this is not a geometric shape we are familar with, we cannot evaluate the area exactly.

However, we can approximate it. There are many ways to approximate it. For example, we can select points \((x,f(x))\) to construct a triangulation of the figure. However, to simplify the process, rather than triangulate, we create rectangles with the same width, but varying heights determined by the function \(f\).

In this next example, we look at six approximations.

Example 6

Take \(R\) to be the region formed by bounding \(f(x)=x^2\) and the \(x\)-axis on \([0,1]\). Approximate \(R\) using six different approximations.

First, here are two approximations of the area using one rectangle. The first approximation uses \(f(0)=0\) as the height of the rectangle with a width of \(1\) while the second approximation uses \(f(1)=1\) as the height of the rectangle with a width of \(1\).

The area of the first approximation is \(0\) while the area of the second approximation is \(1\).

The area \(\mathcal{A}(R)\) is therefore between \(0\) and \(1\).

For the next approximation, construct two rectangles of an equal width of \(\frac{1}{2}\). The first approximation uses \(f(0)=0\) and \(f(\frac{1}{2})=\frac{1}{4}\) as the heights of the two rectangles each with a width \(\frac{1}{2}\) while the second approximation uses \(f(1)=1\) and \(f(\frac{1}{2})=\frac{1}{4}\) as the heights of the two rectangles each with a width \(\frac{1}{2}.\)

The area of the first approximation is \(0+\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}\) while the area of the second approximation is \(\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{2}\cdot 1=\frac{5}{8}\).

The area \(\mathcal{A}(R)\) is therefore between \(\frac{1}{8}\) and \(\frac{5}{8}\).

For the next approximation, construct four rectangles of an equal width of \(\frac{1}{4}\). The first approximation uses \(f(0)=0\), \(f(\frac{1}{4})=\frac{1}{16}\), \(f(\frac{1}{2})=\frac{1}{4}\) and \(f(\frac{3}{4})=\frac{9}{16}\) as the heights of the four rectangles each with a width \(\frac{1}{4}\) while the second approximation uses \(f(1)=1\), \(f(\frac{1}{4})=\frac{1}{16}\), \(f(\frac{1}{2})=\frac{1}{4}\) and \(f(\frac{3}{4})=\frac{9}{16}\) as the heights of the four rectangles each with a width \(\frac{1}{4}.\)

The area of the first approximation is \(0\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{16}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{9}{16}=\frac{7}{32}\) while the area of the second approximation is \(\frac{1}{4}\cdot\frac{1}{16}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot\frac{9}{16}+\frac{1}{4}\cdot 1=\frac{15}{32}\).

The area \(\mathcal{A}(R)\) is therefore between \(\frac{7}{32}\) and \(\frac{15}{32}\).

Notice that the more rectangles we used, the more the rectangles “closely covered” the region \(R\). Is it possible that creating more rectangles and summing up the area will produce a better approximation for the area of the region \(R\)? In certain situations, the answer is yes.

However, before we go over under what conditions the approximation will get better, we first need to generalize the process.

Left Hand and Right Hand Riemann Sums

To approximate the area of the region bounded by \(y=f(x)\), \(x\)-axis, \(x=a\) and \(x=b\) with \(n\) rectangles with equal width:

determine the width: \(\Delta x=\frac{b-a}{n}\)
determine the heights of the rectangle by decomposing the interval \([a,b]\): \([x_0,x_1],[x_1,x_2],[x_2,x_3],\dots,[x_{n-1},x_n]\) where \(x_i=a+i\cdot\Delta x\) where \(i\in\{0,1,2,3,\dots,n-1,n\}.\)

The left-hand Riemann sum is the sum of these \(n\) rectangles: \[\sum_{i=0}^{n-1}f(x_i)\Delta x=\underbrace{f(x_0)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\underbrace{f(x_1)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\dots+\underbrace{f(x_{n-1})}_{\text{height}}\underbrace{\Delta x}_{\text{width}}.\] This is read as “the sum from i equals 0 to n minus 1 of f of x sub i delta x.”

The right-hand Riemann sum is the sum of these \(n\) rectangles: \[\sum_{i=1}^{n}f(x_i)\Delta x=\underbrace{f(x_1)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\underbrace{f(x_2)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\dots+\underbrace{f(x_{n})}_{\text{height}}\underbrace{\Delta x}_{\text{width}}.\] This is read as “the sum from i equals 1 to n of f of x sub i delta x.”

In this next example, calculate the Riemann sums with six rectangles.

Example 7

Take \(R\) to be the region formed by bounding \(f(x)=x^2\) and the \(x\)-axis on \([0,1]\). Approximate \(R\) using six rectangles (Riemann sums) to approximate the area.

For this problem the left-end point is \(a=0\), the right-end point is \(b=1\) and the number of rectangles is \(n=6\). The width of each rectangle is \[\Delta x=\frac{b-a}{n}=\frac{1}{6}.\] The \(x\)-values that determine the heights of each rectangle are given by the formula \[x_i=a+i\Delta x=0+i\cdot\frac{1}{6}\] for \(i\in\{0,1,2,3,4,5,6\}.\) The \(x\)-values are

\(x_0=0\)
\(x_1=0+1\cdot\frac{1}{6}=\frac{1}{6}\)
\(x_2=0+2\cdot\frac{1}{6}=\frac{2}{6}\)
\(x_3=0+3\cdot\frac{1}{6}=\frac{3}{6}\)
\(x_4=0+4\cdot\frac{1}{6}=\frac{4}{6}\)
\(x_5=0+5\cdot\frac{1}{6}=\frac{5}{6}\)
\(x_6=0+6\cdot\frac{1}{6}=1.\)

The heights are

\(f(x_0)=f(0)=0\)
\(f(x_1)=f(\frac{1}{6})=\frac{1}{36}\)
\(f(x_2)=f(\frac{2}{6})=\frac{1}{9}\)
\(f(x_3)=f(\frac{3}{6})=\frac{1}{4}\)
\(f(x_4)=f(\frac{4}{6})=\frac{4}{9}\)
\(f(x_5)=f(\frac{5}{6})=\frac{25}{36}\)
\(f(x_6)=f(1)=1.\)

The left-hand approximation is

\[\begin{align*} \sum_{i=0}^{5}f(x_i)\Delta x&=f(x_0)\Delta x+f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x+f(x_5)\Delta x\\ &=0\cdot \frac{1}{6}+\frac{1}{36}\cdot\frac{1}{6}+\frac{1}{9}\cdot\frac{1}{6}+\frac{1}{4}\cdot\frac{1}{6}+\frac{4}{9}\cdot\frac{1}{6}+\frac{25}{36}\cdot\frac{1}{6}\\ &=\frac{55}{216}. \end{align*}\]

The right-hand approximation is

\[\begin{align*} \sum_{i=1}^{6}f(x_i)\Delta x&=f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x+f(x_5)\Delta x+f(x_6)\Delta x\\ &=\frac{1}{36}\cdot\frac{1}{6}+\frac{1}{9}\cdot\frac{1}{6}+\frac{1}{4}\cdot\frac{1}{6}+\frac{4}{9}\cdot\frac{1}{6}+\frac{25}{36}\cdot\frac{1}{6}+1\cdot \frac{1}{6}\\ &=\frac{91}{216}. \end{align*}\]

Here is a graph of the left-hand Riemann sum approximation in Example 7:

Here is a graph of the right-hand Riemann sum approximation in Example 7:

Remember that we decompose the interval \([a,b]\) into \(n\) sub-intervals: \([x_0,x_1],[x_1,x_2],\dots,[x_{n-1},x_n].\)

The right-hand approximation samples the right endpoints, so we use \(x_1,x_2,\dots,x_n\). The left-hand approximation samples the left endpoints, so we use \(x_0,x_1,\dots,x_{n-1}.\)

However, those are not the only kind of sampling we can do. Instead of picking the endpoints, we can pick any real number in each of the sub-intervals. For example, the midpoints.

Midpoint Riemann Sum

To approximate the area of the region bounded by \(y=f(x)\), \(x\)-axis, \(x=a\) and \(x=b\) with \(n\) rectangles with equal width using midpoints:

determine the width: \(\Delta x=\frac{b-a}{n}\)
decomposing the interval \([a,b]\): \([x_0,x_1],[x_1,x_2],[x_2,x_3],\dots,[x_{n-1},x_n]\) where \(x_i=a+i\cdot\Delta x\) where \(i\in\{0,1,2,3,\dots,n-1,n\}.\)
calculate the midpoint of each interval: given the interval \([x_i,x_{i+1}]\) the midpoint is \(\frac{x_i+x_{i+1}}{2}.\)

The midpoint Riemann sum is the sum of these \(n\) rectangles: \[\sum_{i=0}^{n-1}f\left(\frac{x_i+x_{i+1}}{2}\right)\Delta x=\underbrace{f\left(\frac{x_0+x_1}{2}\right)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\underbrace{f\left(\frac{x_1+x_2}{2}\right)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}+\dots+\underbrace{f\left(\frac{x_{n-1}+x_{n}}{2}\right)}_{\text{height}}\underbrace{\Delta x}_{\text{width}}.\]

In this next example, calculate the midpoint Riemann sums with six rectangles.

Example 8

Take \(R\) to be the region formed by bounding \(f(x)=x^2\) and the \(x\)-axis on \([0,1]\). Approximate \(R\) using six midpoint Riemann sums to approximate the area.

For this problem the left-end point is \(a=0\), the right-end point is \(b=1\) and the number of rectangles is \(n=6\). The width of each rectangle is \[\Delta x=\frac{b-a}{n}=\frac{1}{6}.\] The \(x_i\)-values are given by the formula \[x_i=a+i\Delta x=0+i\cdot\frac{1}{6}\] for \(i\in\{0,1,2,3,4,5,6\}.\) The \(x_i\)-values are

\(x_0=0\)
\(x_1=0+1\cdot\frac{1}{6}=\frac{1}{6}\)
\(x_2=0+2\cdot\frac{1}{6}=\frac{2}{6}\)
\(x_3=0+3\cdot\frac{1}{6}=\frac{3}{6}\)
\(x_4=0+4\cdot\frac{1}{6}=\frac{4}{6}\)
\(x_5=0+5\cdot\frac{1}{6}=\frac{5}{6}\)
\(x_6=0+6\cdot\frac{1}{6}=1.\)

The midpoints are

\(\frac{x_0+x_1}{2}=\frac{0+\frac{1}{6}}{2}=\frac{1}{12}\)
\(\frac{x_1+x_2}{2}=\frac{\frac{1}{6}+\frac{1}{3}}{2}=\frac{3}{12}\)
\(\frac{x_2+x_3}{2}=\frac{\frac{1}{3}+\frac{1}{2}}{2}=\frac{5}{12}\)
\(\frac{x_3+x_4}{2}=\frac{\frac{1}{2}+\frac{2}{3}}{2}=\frac{7}{12}\)
\(\frac{x_4+x_5}{2}=\frac{\frac{2}{3}+\frac{5}{6}}{2}=\frac{9}{12}\)
\(\frac{x_5+x_6}{2}=\frac{\frac{5}{6}+1}{2}=\frac{11}{12}\)

The heights are

\(f\left(\frac{x_0+x_1}{2}\right)=f\left(\frac{1}{12}\right)=\frac{1}{144}\)
\(f\left(\frac{x_1+x_2}{2}\right)=f(\frac{3}{12})=\frac{1}{16}\)
\(f\left(\frac{x_2+x_3}{2}\right)=f(\frac{5}{12})=\frac{25}{144}\)
\(f\left(\frac{x_3+x_4}{2}\right)=f(\frac{7}{12})=\frac{49}{144}\)
\(f\left(\frac{x_4+x_5}{2}\right)=f(\frac{9}{12})=\frac{9}{16}\)
\(f\left(\frac{x_5+x_6}{2}\right)=f(\frac{11}{12})=\frac{121}{144}\)

The midpoint approximation is

\[\begin{align*} \sum_{i=0}^{n-1}f\left(\frac{x_i+x_{i+1}}{2}\right)\Delta x&=f\left(\frac{x_0+x_1}{2}\right)\Delta x+f\left(\frac{x_1+x_2}{2}\right)\Delta x+f\left(\frac{x_2+x_3}{2}\right)\Delta x+f\left(\frac{x_3+x_4}{2}\right)\Delta x+f\left(\frac{x_4+x_5}{2}\right)\Delta x+f\left(\frac{x_5+x_6}{2}\right)\Delta x\\ &=\frac{1}{144}\cdot \frac{1}{6}+\frac{1}{16}\cdot\frac{1}{6}+\frac{25}{144}\cdot\frac{1}{6}+\frac{49}{144}\cdot\frac{1}{6}+\frac{9}{16}\cdot\frac{1}{6}+\frac{121}{144}\cdot\frac{1}{6}\\ &=\frac{143}{432}. \end{align*}\]

Here is a graph of the midpoint Riemann sum approximation in Example 8:

Notice that compared to the left-hand and right-hand approximation, the midpoint approximation appears to have less “spacing.” In a later section, we will discuss the error of all these different approximation schemes.

Determing Area Exactly

The more rectangles we used, the closer the rectangles matched the shape of the region described by \(f(x)=x^2\) on \([0,1].\) It turns out that taking the limit will give us a finite number and that is how we define the area.

Area of Region R as a Limit of Riemann Sums

Take \(f\) to be a continuous function on \([a,b].\) Take \(R\) to be the region bounded by \(f\) and the \(x\)-axis on \([a,b].\) Then the area of \(R\) is defined and \[\mathcal{A}(R)=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x\] where \(\Delta x=\frac{b-a}{n}\) and \(x_i=a+i\Delta x\).

To compute the limit, we first need to determine \(\sum_{i=1}^nf(x_i)\Delta x\). This involves finding a formula that is no longer a sum.

The following formulas will be useful.

Summation Formulas

Take \(n\) to be a natural number, \(c\) be a real number, and \(a_i\) and \(b_i\) to be sequences of real numbers. Then the following formulas.

\(\displaystyle\sum_{i=1}^ni=1+2+3+\dots+n=\frac{n(n+1)}{2}\)
\(\displaystyle\sum_{i=1}^ni^2=1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}\)
\(\displaystyle\sum_{i=1}^ni^3=1^3+2^3+3^3+\dots+n^3=\left(\frac{n(n+1)}{2}\right)^2\)
\(\displaystyle\sum_{i=1}^n c=\underbrace{c+c+c+\dots+c}_{n\text{ copies}}=n\cdot c\)
\(\displaystyle\sum_{i=1}^n ca_i=c\sum_{i=1}^n a_i\)
\(\displaystyle\sum_{i=1}^n (a_i\pm b_i)=\sum_{i=1}^n a_i\pm\sum_{i=1}^n b_i\)

We will use these formulas to compute the area exactly of this region:

Example 9

Take \(R\) to be the region formed by bounding \(f(x)=x^2\) and the \(x\)-axis on \([0,1]\). Calculate the area of \(R\).

For this problem \(a=0\) and \(b=1\) and \(f(x)=x^2\). For each natural number \(n\), the widths of the rectangle are \[\Delta x=\frac{1-0}{n}=\frac{1}{n}.\] The \(x_i\) are \[x_i=a+i\Delta x=0+i\frac{1}{n}=\frac{i}{n}.\] The heights are \[f(x_i)=\left(\frac{i}{n}\right)^2=\frac{i^2}{n^2}.\] The Riemann sums look like this: \[\begin{align*} \sum_{i=1}^n f(x_i)\Delta x&=\sum_{i=1}^n\frac{i^2}{n^2}\frac{1}{n}\\ &=\sum_{i=1}^n\frac{i^2}{n^3}\\ &=\sum_{i=1}^n\frac{1}{n^3}i^2.\\ \end{align*}\] The only thing changing in the sum is anything with an \(i\). The \(n\) is a constant, so the sum can be rewritten like this: \[\begin{align*} \sum_{i=1}^n f(x_i)\Delta &=\sum_{i=1}^n\frac{1}{n^3}i^2.\\ &=\frac{1}{n^3}\left[\sum_{i=1}^ni^2\right]\\ &=\frac{1}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]&\text{use the formula for sum}\\ &=\frac{(n+1)(2n+1)}{6n^2}. \end{align*}\] The Riemann sum formula simplifies down to \[\sum_{i=1}^n f(x_i)\Delta x=\frac{(n+1)(2n+1)}{6n^2}.\]

Take the limit as \(n\) goes to infinity to get

\[\begin{align*} \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x=&\lim_{n\to\infty}\frac{(n+1)(2n+1)}{6n^2}\\ =&\lim_{n\to\infty}\frac{2n^2+3n+1}{6n^2}\\ =&\frac{1}{3}. \end{align*}\]

The area of \(S\) is thus \[\mathcal{A}(S)=\frac{1}{3}.\]

Here is another example of computing the sum exactly.

Example 10

Take \(R\) to be the region formed by bounding \(f(x)=x^2+1\) and the \(x\)-axis on \([4,5]\). Calculate the area of \(R\).

For this problem \(a=4\) and \(b=5\) and \(f(x)=x^2+1\). For each natural number \(n\), the widths of the rectangle are \[\Delta x=\frac{5-4}{n}=\frac{1}{n}.\] The \(x_i\) are \[x_i=a+i\Delta x=4+i\frac{1}{n}=4+\frac{i}{n}.\] The heights are \[\begin{align*}f(x_i)&=\left(4+\frac{i}{n}\right)^2+1\\ &=\frac{i^2}{n^2}+8\frac{i}{n}+16+1\\ &=\frac{i^2}{n^2}+8\frac{i}{n}+17. \end{align*}\] The Riemann sums look like this: \[\begin{align*} \sum_{i=1}^n f(x_i)\Delta x&=\sum_{i=1}^n\left(\frac{i^2}{n^2}+8\frac{i}{n}+17\right)\frac{1}{n}\\ &=\sum_{i=1}^n\left(\frac{i^2}{n^3}+8\frac{i}{n^2}+\frac{17}{n}\right)\\ &=\sum_{i=1}^n\frac{1}{n^3}i^2+\sum_{i=1}^n8\frac{i}{n^2}+\sum_{i=1}^n\frac{17}{n}.\\ \end{align*}\] The only thing changing in the sum is anything with an \(i\). The \(n\) is a constant, so the sum can be rewritten like this: \[\begin{align*} \sum_{i=1}^n f(x_i)\Delta &=\sum_{i=1}^n\frac{1}{n^3}i^2+\sum_{i=1}^n8\frac{i}{n^2}+\sum_{i=1}^n\frac{17}{n}.\\ &=\frac{1}{n^3}\sum_{i=1}^ni^2+\frac{8}{n^2}\sum_{i=1}^ni+\frac{17}{n}\sum_{i=1}^n1. \end{align*}\] Use the Summation Formulas to get \[\begin{align*} \sum_{i=1}^n f(x_i)\Delta &=\frac{1}{n^3}\sum_{i=1}^ni^2+\frac{8}{n^2}\sum_{i=1}^ni+\frac{17}{n}\sum_{i=1}^n1.\\ &=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{8}{n^2}\cdot\frac{n(n+1)}{2}+\frac{17}{n}n\\ &=\frac{(n+1)(2n+1)}{6n^2}+\frac{8(n+1)}{2n}+17\\ \end{align*}\] The Riemann sum formula simplifies down to \[\sum_{i=1}^n f(x_i)\Delta x=\frac{(n+1)(2n+1)}{6n^2}+\frac{8n(n+1)}{2n^2}+17.\]

Take the limit as \(n\) goes to infinity to get

\[\begin{align*} \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x&=\lim_{n\to\infty}\left(\frac{(n+1)(2n+1)}{6n^2}+\frac{8(n+1)}{2n}+17\right)\\ &=\lim_{n\to\infty}\frac{2n^2+3n+1}{6n^2}+\lim_{n\to\infty}\frac{8n+8}{2n}+\lim_{n\to\infty}17\\ &=\frac{1}{3}+4+17\\ &=\frac{64}{3} \end{align*}\]

The area of \(S\) is thus \[\mathcal{A}(S)=\frac{64}{3}.\]

This is the region we calculated the area of in Example 10.

Definite Integrals

For future sections, we will use the following notation called the definite integral.

Take \(f\) to be a bounded function. We say it is Riemann Integrable on \([a,b]\) if there exists a real number \(L\) so that the Riemann sum converges to \(L\). In such case, we define the definite integral to be that number \(L\). The notation is \[\int_a^bf(x)\,\mathrm{d}x=L,\] or \[\int_a^bf(x)\,\mathrm{d}x=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x.\] The notation is read as “the integral from a to b of f of x dx”. Sometimes, the function \(f\) is referred to as the integrand.

One interpretation of the definite integral is the (signed) area of \(f\) on the interval \([a,b].\)

Use this interpretation to complete the following example.

Example 11

Compute the following by interpreting the definite integral as signed area.

\(\displaystyle\int_0^1 x^2\,\mathrm{d}x\)
\(\displaystyle\int_0^4 (x-3)\,\mathrm{d}x\)
\(\displaystyle\int_{-3}^3 \sqrt{9-x^2}\,\mathrm{d}x\)

This is computing the signed area of \(f(x)=x^2\) on \([0,1]\). In Example 9, we calculated it and got \(\frac{1}{3}\). So \[\int_{0}^1x\,\mathrm{d}x=\frac{1}{3}.\]
This is computing the signed area of \(f(x)=x-3\) on \([0,4]\). This is similar to Example 3. Graph the function \(f\) to determine the region, decompose it into two parts, and sum up the area to get \(-4\). So \[\int_{0}^4(x-3)\,\mathrm{d}x=-4.\]
This is computing the signed area of \(f(x)=\sqrt{9-x^2}\) on \([-3,3]\). This is similar to Example 4. Graph the function \(f\) to determine the region. It is half a circle with radius \(r=3\). So \[\int_{-3}^3\sqrt{9-x^2}\,\mathrm{d}x=\frac{9}{2}\pi.\]

As with limits, there are ways to break up complicated integrals into simpler ones. Here is a list of properties:

Definite Integral Properties

Take \(a\),\(b\) and \(c\) to be real numbers with \(a<b\). Take \(f\) and \(g\) to be two integrable functions with \[\int_a^bf(x)\,\mathrm{d}x=L\quad\text{and}\quad\int_a^bg(x)\,\mathrm{d}x=M.\] Then the following properties apply

\(\displaystyle\int_a^b(f(x)\pm g(x))\,\mathrm{d}x=\int_a^bf(x)\,\mathrm{d}x+\int_a^bg(x)\,\mathrm{d}x\)
\(\displaystyle\int_a^bcf(x)\,\mathrm{d}x=c\int_a^bf(x)\,\mathrm{d}x\)
\(\displaystyle\int_a^bc\,\mathrm{d}x=c\cdot (b-a).\)

The first property states that if we have a definite integral of a sum or difference of functions, then that definite integral is equal to the sum or difference of the definite integral of the individual function.

The second property states that if we have a definite integral of a function that is scaled by a real number \(c\), then that definite integral is equal to the definite integral of the function scaled by \(c\).

The third property states that if we have a definite integral of a constant function \(c\) on \([a,b]\), then the definite integral is \(c\dot (b-a)\). This is because the region described is a rectangle with width \(b-a\) and (signed) height \(c.\)

Here is an example in which we use the properties.

Example 12

Suppose that \(f\) and \(g\) are integrable with \[\int_2^7f(x)\,\mathrm{d}x=3\quad\text{and}\quad\int_2^7g(x)\,\mathrm{d}x=-2.\]

Calculate \[\int_2^7\left(f(x)+2g(x)+3\right)\,\mathrm{d}x.\]

The integrand \(f(x)+2g(x)+3\) is the sum of three functions, so use the Definite Integral Properties to get that \[\int_2^7\left(f(x)+2g(x)+3\right)\,\mathrm{d}x=\int_2^7f(x)\,\mathrm{d}x+\int_2^72g(x)\,\mathrm{d}x+\int_2^73\,\mathrm{d}x.\]

The integrand of the definite integral \(\int_2^72g(x)\,\mathrm{d}x\) is the function \(g\) scaled by \(2\), so use the Definite Integral Properties to get that \[\int_2^72g(x)\,\mathrm{d}x=2\int_2^7g(x)\,\mathrm{d}x.\]

The integrand of the definite integral, \(\int_2^73\,\mathrm{d}x\) is the constant function \(h(x)=3\), so the integral is equal to the area of the rectangle with height \(3\) and width \(7-2=5\): \[\int_2^723\,\mathrm{d}x=3\cdot 5=15.\]

The definite integral is thus \[\begin{align*} \int_2^7\left(f(x)+2g(x)+3\right)\,\mathrm{d}x&=\int_2^7f(x)\,\mathrm{d}x+\int_2^72g(x)\,\mathrm{d}x+\int_2^73\,\mathrm{d}x\\ &=\int_2^7f(x)\,\mathrm{d}x+2\int_2^7g(x)\,\mathrm{d}x+\int_2^73\,\mathrm{d}x\\ &=3+2(-2)+3\cdot 5\\ &=3-4+15\\ &=14. \end{align*}\]

The final property we have is a way to bound the definite integral of a function assuming we have an upper and lower bound for the function.

Upper and Lower Bound for Definite Integral

Take \(f\) to be an integrable function on \([a,b]\) with \[m\leq f(x)\leq M.\] Then we have the following bound on the definite integral of \(f\) on \([a,b]\): \[m(b-a)\leq \int_a^bf(x)\,\mathrm{d}x\leq M(b-a).\]

The above theorem tell us that if an integrable function \(f\) is bounded below by \(m\) and bounded above by \(M\) on \([a,b]\) then its definite integral is greater than or equal to the area of a rectangle with height \(m\) and width \(b-a\) but less than or equal to the area of a rectangle with height \(M\) and width \(b-a\).

Use this property to estimate the definite integral in the next example.

Example 13

Suppose that \(f\) is an integrable function with \[-1\leq f(x)\leq 2\quad\text{on }[-1,4].\]

Determine an upper and lower bound for the definite integral \[\int_{-1}^4f(x)\,\mathrm{d}x.\]

By Upper and Lower Bound for Definite Integral we have that \[\begin{align*} -1\cdot(4-(-1))&\leq \int_{-1}^4f(x)\,\mathrm{d}x\leq 2\cdot(4-(-1))\\ -1\cdot(5)&\leq \int_{-1}^4f(x)\,\mathrm{d}x\leq 2\cdot(5)\\ -5&\leq \int_{-1}^4f(x)\,\mathrm{d}x\leq 10. \end{align*}\]

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