Approximating the area beneath a curve

So how do we find the area of the region from $a$ to $b$ beneath the graph of a function $f$ ? Let's denote this area by

A_f

There is actually a generic method to find $A_f$ . The idea is to first find an approximation of the area, then we form better and better approximations until we see see toward which values these approximations converge. And this value has to be the area we are looking for. The analogues step in differential calculus was to find better and better approximations of the slope of the tangent using the secants. The slope of the tangent was then the number towards which the slope of the secants converged.

To keep things concrete, let's return to our example where we want to find the area from $a=0.1$ to $b=1.5$ beneath the graph of the function

f(x)=\frac{1}{4} x^2+\frac{1}{2}

Thus, $A_f$ is the area of the shaded region in the figure below. The figure also shows how we can find an approximation of this area. We simply cover the area by many vertical bars of equal width and sum the areas of these bars. If we take $n$ bars, and the bar areas are denoted by $A_1, ..., A_n$ , we thus have

\boxed{A_f \approx \sum_{k_1}^n A_k = A_1+A_2+...+A_n}

We call the sum on the right the sum of bar areas. Let's now discuss how to find the bar areas $A_1, ..., A_n$ .

We first need to discuss how we actually constructed the bars, so that we can find their width and heights. Only then can we proceed to calculate the areas of these bars.

We want each bar to have equal width, which we denote by $\boxed{\Delta x}$
The width of the bars is determined by the number of bars $n$ we choose to use. This is so because the bars have to start at $a$ and have to end at $b$ , which is a length of $b-a$ . Thus, we have to divide this length in $n$ equal parts, each of which forms the base of a bar. Thus, the width of each bar is given by the formula $\boxed{\Delta x =\frac{b-a}{n}}$ In the example,the $4$ bars have to fit into the interval of length $3-0.1$ , and for the width $\Delta x$ of each bar it has to be true that $4\cdot \Delta x=3-0.1$ and thus we have $\Delta x =\frac{3-0.1}{4}=0.725$
The height of a bar is determine by its right border. This border has to end on the graph of $f$ . Note that we know where on the $x$ -axis each bar has its right border: $\boxed{\begin{array}{lll} \text{bar } 1 : & x_1 = a+1\cdot \Delta x\\ \text{bar } 2 : & x_2 = a+2\cdot \Delta x\\ ... & \\ \text{bar } k : & x_k = a+k\cdot \Delta x\\ ... & \\ \text{bar } n : & x_n = a+n\cdot \Delta x = b\\ \end{array}}$ For the example, we have $\begin{array}{l} \text{bar } 1 : & x_1 = 0.1+1\cdot 0.725 =0.825\\ \text{bar } 2 : & x_2 = 0.1+2\cdot 0.725 = 1.55\\ \text{bar } 3 : & x_3 = 0.1+3\cdot 0.725 = 2.275\\ \text{bar } 4 : & x_4 = 0.1+4\cdot 0.725 = 3\\ \end{array}$ We therefore can also calculate the height of each bar: $\boxed{\begin{array}{lll} \text{bar } 1 : & h_1 = f(x_1)\\ \text{bar } 2 : & h_2 = f(x_2)\\ ... & \\ \text{bar } k : & h_k = f(x_k)\\ ... & & \\ \text{bar } n : & h_n = f(x_n)=f(b)\\ \end{array}}$ For the example, we have $\begin{array}{l} \text{bar } 1: & h_1 = f(0.825)=\frac{1}{4}\cdot0.825^2+\frac{1}{2}=0.670\\ \text{bar } 2 : & h_2 = f(1.55)=\frac{1}{4}\cdot1.55^2+\frac{1}{2}=1.100\\ \text{bar } 3 : & h_3 = f(2.275)=\frac{1}{4}\cdot2.275^2+\frac{1}{2}=1.794\\ \text{bar } 4 : & h_4 = f(3)=\frac{1}{4}\cdot 3^2+\frac{1}{2}=2.25\\ \end{array}$
Thus we can determine the area of each bar: $\boxed{\begin{array}{lll} \text{bar } 1 : & A_1 = f(x_1)\cdot \Delta x\\ \text{bar } 2 : & A_2 = f(x_2)\cdot \Delta x\\ ... & \\ \text{bar } k : & A_k = f(x_k)\cdot \Delta x\\ ... & & \\ \text{bar } n : & A_n = f(x_n)\cdot \Delta x=f(b)\cdot \Delta x\\ \end{array}}$ For the example we have $\begin{array}{lll} \text{bar } 1: & A_1 = f(0.825)\cdot \Delta x = 0.67 \cdot 0.725=0.485\\ \text{bar } 2: & A_2 = f(1.55)\cdot \Delta x = 1.10 \cdot 0.725=0.798\\ \text{bar } 3: & A_3 = f(2.275)\cdot \Delta x = 1.79 \cdot 0.725=1.300\\ \text{bar } 4: & A_4 = f(3)\cdot \Delta x = 2.25 \cdot 0.725=1.994\\ \end{array}$
Finally, we can form the sum of the bar areas, which is an approximation of $A_f$ . $\boxed{\begin{array}{lll} A_f & \approx \sum_{k=1}^n \underbrace{f(x_k)\Delta x}_{A_k}\\ A_f & \approx \underbrace{f(x_1)\Delta x}_{A_1}+ ... + \underbrace{f(x_n)\Delta x}_{A_n}\\ \end{array}}$ For the example, we have $\begin{array}{lll} A_f & \approx f(0.825)\cdot \Delta x + f(1.55)\cdot \Delta x + f(2.275)\cdot \Delta x + f(3)\cdot \Delta x\\ & = 0.485+0.798+1.300+1.994\\ & = 4.577 \end{array}$

Exercise 1

Use $n$ bars to approximate the area of the region from $a$ to $b$ under the graph of the function $f$ . Furthermore, where possible, calculate the exact area and compare with the approximation (see also section 27). Always create a figure and indicate the bars.
1. $f(x)=3$ for all $x$ (constant function), $a=0.5, b=4, n=7$ .
2. $f(x)=0.5x+1, a=-1, b=3, n=4$ .
3. $f(x)=\sqrt{1-x^2}, a=-1, b=1, n=5$ .
4. $f(x)=x^2$ , $a=0$ , $b=2$ , $n=3$
5. $f(x)=\frac{2}{x}$ , $a=1$ , $b=2$ , $n=2$
Use $100$ bars to approximate the area from $a=0$ to $b=1$ beneath the graph of the function $f(x)=x^2$ . Hint: You need a sum formula we discussed in section 2 of integral calculus.
A car moves along a straight line. The instantaneous speed at time $x$ can be described by some function $f(x)$ . We approximate the area from time $a$ to time $b$ beneath the graph of $f$ using $n$ bars. In this context, what is the meaning of the sum of bar areas?
$\sum_{k=1}^n f(x_k)\Delta x$

Solution

The graphs are shown below.
1. $\Delta x =\frac{4-0.5}{7}=0.5$ $x_1=1, x_2=1.5, x_3=2, x_4=2.5, x_5=3, x_6=3.5, x_7=4$ $\begin{array}{lll} \sum_{k=1}^7 f(x_k)\Delta x & = f(1)\cdot 0.5+ f(1.5)\cdot 0.5+...+f(4)\cdot 0.5\\ & = 1.5+1.5+1.5+1.5+1.5+1.5+1.5\\ & = \underline{10.5}\\ \end{array}$ $A_f=3.5\cdot 3=\underline{10.5}$ (see exercise in previous section)
2. $\Delta x =\frac{3-(-1)}{4}=1$ $x_1=0, x_2=1, x_3=2, x_4=3$ $\begin{array}{lll} \sum_{k=1}^4 f(x_k)\Delta x & = f(0)\cdot 1+ f(1)\cdot 1+f(2)\cdot 1+f(3)\cdot 1\\ & =1+1.5+2+2.5\\ & =\underline{7}\\ \end{array}$ $A_f=\underline{6}$ (see exercise in previous section).
3. $\Delta x =\frac{1-(-1)}{5}=0.4$ $x_1=-0.6, x_2=-0.2, x_3=0.2, x_4=0.6, x_5=1$ $\begin{array}{lll} \sum_{k=1}^5 f(x_k)\Delta x & = f(-0.6)\cdot 0.4+ f(-0.2)\cdot 0.4+f(0.2)\cdot 0.4+f(0.6)\cdot 0.4+f(1)\cdot 0.4\\ & = 0.32+0.39+0.39+0.32+0\\ & =\underline{1.42}\\ \end{array}$ $A_f=\frac{\pi}{2}=\underline{1.57...}$ (see exercise in previous section).
4. $\Delta x =\frac{2-0}{3}=\frac{2}{3}$ $x_1=\frac{2}{3}, x_2=\frac{4}{3}, x_3=2$ $\begin{array}{lll} \sum_{k=1}^3 f(x_k)\Delta x & = f(\frac{2}{3})\cdot \frac{2}{3}+ f(\frac{4}{3})\cdot \frac{2}{3}+f(2)\cdot \frac{2}{3}\\ & =\frac{8}{27}+\frac{32}{27}+\frac{72}{27}\\ & =\frac{112}{27}=\underline{4.15}\\ \end{array}$ ( $A_f=\underline{2.\overline{6}}$ , but you do not know how to calculate this yet).
5. $\Delta x =\frac{2-1}{2}=0.5$ $x_1=1.5, x_2=2$ $\begin{array}{lll} \sum_{k=1}^2 f(x_k)\Delta x & = f(1.5)\cdot 0.5+ f(2)\cdot 0.5\\ & =\frac{2}{3}+\frac{1}{2}=\underline{1.1\overline{6}}\\ \end{array}$ ( $A_f=\underline{1.386...}$ , but you do not know how to calculate this yet)
We have $\Delta x = \frac{1-0}{100}=\frac{1}{100}$ Thus, $\begin{array}{lll} x_1 =& \frac{1}{100} \\ x_2 =& \frac{2}{100} \\ x_3 =& \frac{3}{100} \\ ...& \\ x_{100} =& \frac{100}{100}=1 \\ \end{array}$ and therefore $\begin{array}{lll} \sum_{k=1}^n f(x_k)\Delta x & =& f(x_1)\Delta x+f(x_2)\Delta x +f(x_3)\Delta x + ...+ f(x_n)\Delta x \\ & = & \left(\frac{1}{100}\right)^2 \frac{1}{100}+\left(\frac{2}{100}\right)^2 \frac{1}{100}+\left(\frac{3}{100}\right)^2 \frac{1}{100}+...+\left(\frac{100}{100}\right)^2 \frac{1}{100}\\ & = & \frac{1^2}{100^3} + \frac{2^2}{100^3}+\frac{3^2}{100^3}+...+\frac{100^2}{100^3}\\ & = & \frac{1}{100^3} (1^2 +2^2+3^2+...+100^2)\\ \end{array}$ Note that we have a formula for finding the first $n$ square numbers (see section 25): $1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$ For $n=100$ , we get $1^2+2^2+3^2+...+100^2=\frac{100\cdot 101 \cdot 201}{6}=338350$ Thus, the sum of bars is $\sum_{k=1}^n f(x_k)\Delta x = \frac{338350}{100^3}=\underline{0.33835}$ The exact area, by the way, is $\frac{1}{3}$ , but you do not know how to calculate this presently.
Each bar area is the product of a speed $f(x)$ and a time $\Delta x$ , thus each bar area is the distance travelled by the car during the time $\Delta x$ (at least an approximation of it). The sum of bars between the times $a$ and $b$ is therefore the approximate distance travelled in the time interval from $a$ to $b$ .