Application of Calculus

In this section on applications of integration, we will explore how integration allows us to measure the space enclosed by curves and estimate the volume of solids formed by revolving those regions. The disc method provides a systematic way to compute volumes by slicing a solid into circular cross-sections and summing their contributions. This technique is valuable in fields such as engineering, architecture, and manufacturing, where precise volume calculations are essential. By mastering these ideas, you will gain the skills to apply definite integrals to real-world problems, deepen your understanding of three-dimensional geometry, and build a strong foundation for more advanced topics in calculus and applied mathematics.

KEY IDEAS

• Definite integrals – identifying these as areas to create an understanding of how integration measures the exact area between curves over a specified interval. • Setting up bounds – identifying intersection points of curves to establish correct limits of integration. • Solids of revolution – recognising that revolving a plane region around an axis generates a three-dimensional solid. • Disc method – slicing the solid perpendicular to the axis of revolution and expressing each cross-section’s area as to determine the volume of the solid.

In mathematics, area is not just about counting space; it tells a deeper story when functions rise above or fall below the x - axis. Whether a curve soars or dips, the regions it creates carry meaning: sometimes adding up, other times cancelling out. We will explore how these areas differ, and sort them into totals, nets, or signed values based on their position above or below the x-axis. In Years 1 and 2 we learnt how to represent graphs of linear and non-linear functions. We will now go through an activity to establish a few theories.

Activity 6.1 Exploring Area Above and Below the x - axis

In pairs, or in small groups, work through the following activity.

Step 1: On the same coordinate grid, carefully sketch the graphs of the following functions from $x = -4$ to $x = 4$ : a. $f(x) = 3$ b. $g(x) = -2$ c. $h(x) = x$ d. $k(x) = x^2 - 4$ Your sketch should look like this. [Figure] Figure 6.1: Graphical representations of functions in Step 1. Step 2: For each graph, shade the region between the line/curve and the $x -$ axis within the interval given. Use two different colours: a. One colour for parts of the graph above the $x -$ axis b. Another colour for parts below the $x -$ axis

[Figure] [Figure]

Figure 6.2: Graphical representation of Figure 6.3: Graphical representation of function $k(x) = x^2 - 4$ function $f(x) = 3$

[Figure] [Figure] Figure 6.4: Graphical representation of function g $(x) = -2$ Figure 6.5: Graphical representation of function h $(x) = x$

Step 3: For each function, write the following in your notebook: a. Which parts of the graph lie above the $x -$ axis? b. Which parts lie below the $x -$ axis?

Observations Function Part above x- axis (+ area) Part below x- axis (- area) $f(x) = 3$ $[-4, 4]$ None $g(x) = -2$ None $[-4, 4]$ $h(x) = x$ $[0, 4]$ $[-4, 0]$ $k(x) = x^2 - 4$ $[-4, -2]$ and $[2, 4]$ $[-2, 2]$

From Activity 6.1 , you have come across different kinds of functions. There are the constant functions, $(f(x) = 3, g(x) = -2)$ , a linear function $(h(x) = x)$ and a quadratic function $(k(x) = x^2 - 4)$ . To be able to determine the area covered by these functions and the x-axis, you need to find the definite integral of the functions based on the interval they cover. It is important to note that there are three different kinds of area to be determined, the total area, net area and signed area.

Total Area

In determining the total area, we are only interested in the physical area between the function and the $x -$ axis. We are not concerned if a portion is below or above the $x -$ axis. We treat all the areas as positive and add them up. Mathematically, we say a total area refers to the sum of the absolute values of all areas between the function and the $x -$ axis. It is always positive or zero and represents the actual physical area irrespective of the position relative to the $x -$ axis. It is denoted by $\int |f(x)|\,dx$ . Given an interval $[a, b]$ , total area $= \int_a^b |f(x)|\,dx$ or

1. Divide the interval into subintervals where $f(x)$ is entirely above or below the $x -$ axis.
2. Add the absolute values of all areas. Note If $f(x)$ is a trigonometric function, analyse how $f(x)$ behaves in specific quadrants and apply the signs needed to make them positive when calculating the total area.

Net Area

When it comes to the net area, we particularly observe whether a portion of the area is above or below the $x -$ axis. The portion above is treated as positive while the one below is treated as a negative area. We do not find the sum of the absolute values. Mathematically, net area is the algebraic sum of areas between the function and the $x -$ axis. It can be positive, negative or zero and represents the balance between positive and negative contributions.

Given an interval $[a, b]$ , net area $= \int_a^b f(x)\,dx$ . The shaded region below the $x -$ axis will have a negative area and will affect the final result.

Signed area

Signed area is equivalent to net area but emphasises that the result carries a sign indicating position relative to the $x -$ axis.

Example 6.1

Determine the total, net and signed area of the functions over the interval $x = -4$ to $x = 4$ . a. $f(x) = 3$ b. $g(x) = -2$ c. $h(x) = x$ d. $k(x) = x^2 - 4$

Solution

a. Total area $= \int_{-4}^{4} |3|\,dx$ $= \left|[3x]_{-4}^{4}\right|$ $= |3(4) - (3 \times (-4))|$ $= 24$ units squared Net Area $= \int_{-4}^{4} 3\,dx$ $= [3x]_{-4}^{4}$ $= 3(4) - (3 \times (-4))$ $= 24$ units squared Signed area $= \int_{-4}^{4} 3\,dx$ $= [3x]_{-4}^{4}$ $= 3(4) - (3 \times (-4))$ $= +24$ units squared b. Total area $= \int_{-4}^{4} |-2|\,dx$ $= \left|[-2x]_{-4}^{4}\right|$ $= |-2(4) - (-2 \times (-4))|$ $= 16$ units squared

Net Area $= \int_{-4}^{4} -2\,dx$ $= [-2x]_{-4}^{4}$ $= -2(4) - (-2 \times (-4))$ $= -16$ units squared Signed area $= \int_{-4}^{4} -2\,dx$ $= [-2x]_{-4}^{4}$ $= -2(4) - (-2 \times (-4))$ $= -16$ units squared c. Total area $= \int_{-4}^{0} |x|\,dx + \int_0^4 |x|\,dx$ $= \left|\left[\frac{x^2}{2}\right]_{-4}^{0}\right|$ $+ \left|\left[\frac{x^2}{2}\right]_0\right|$ $= \left|\left(\frac{0^2}{2} - \frac{(-4)^2}{2}\right)\right| + \left|\left(\frac{4^2}{2} - \frac{0^2}{2}\right)\right|$ $= |-8| + |8|$ $= 16$ Net Area $= \int_{-4}^{0} x\,dx + \int_0^4 x\,dx$ $= \left[\frac{x^2}{2}\right]_{-4}^{0} + \left[\frac{x^2}{2}\right]_0^4$ $= \left(\frac{0^2}{2} - \frac{(-4)^2}{2}\right) + \left(\frac{4^2}{2} - \frac{0^2}{2}\right)$ $= -8 + 8$ $= 0$ Signed area $= \int_{-4}^{0} x\,dx + \int_0^4 x\,dx$ $= \left[\frac{x^2}{2}\right]_{-4}^{0} + \left[\frac{x^2}{2}\right]_0^4$ $= \left(\frac{0^2}{2} - \frac{(-4)^2}{2}\right) + \left(\frac{4^2}{2} - \frac{0^2}{2}\right)$ $= -8 + 8$ $= 0$

d. Total Area $= \int_{-4}^{-2} |x^2 - 4|\,dx + \int_{-2}^{2} |x^2 - 4|\,dx + \int_2^4 |x^2 - 4|\,dx$ $= \left|\left[\frac{x^3}{3} - 4x\right]_{-4}^{-2}\right| + \left|\left[\frac{x^3}{3} - 4x\right]_{-2}^{2}\right| + \left|\left[\frac{x^3}{3} - 4x\right]_2^4\right| \\ = \left|\left[\left(\frac{(-2)^3}{3} - 4(-2)\right) - \left(\frac{(-4)^3}{3} - 4(-4)\right)\right]\right| + \left|\left[\left(\frac{2^3}{3} - 4(2)\right) - \left(\frac{(-2)^3}{3} - 4(-2)\right)\right]\right| + \left|\left[\left(\frac{4^3}{3} - 4(4)\right) - \left(\frac{2^3}{3} - 4(2)\right)\right]\right|$ $= \left|\frac{32}{3}\right| + \left|-\frac{32}{3}\right| + \left|\frac{32}{3}\right|$ $= 32$ units squared Net Area $= \int_{-4}^{-2} (x^2 - 4)\,dx + \int_{-2}^{2} (x^2 - 4)\,dx + \int_2^4 (x^2 - 4)\,dx$ $= \left[\frac{x^3}{3} - 4x\right]_{-4}^{-2} + \left[\frac{x^3}{3} - 4x\right]_{-2}^{2} + \left[\frac{x^3}{3} - 4x\right]_2^4$ $= \left[\left(\frac{(-2)^3}{3} - 4(-2)\right) - \left(\frac{(-4)^3}{3} - 4(-4)\right)\right] + \left[\left(\frac{2^3}{3} - 4(2)\right) - \left(\frac{(-2)^3}{3} - 4(-2)\right)\right] \\ + \left[\left(\frac{4^3}{3} - 4(4)\right) - \left(\frac{2^3}{3} - 4(2)\right)\right]$ $= \frac{32}{3} - \frac{32}{3} + \frac{32}{3}$ $= \frac{32}{3} \approx 10.\dot{6}$ units squared Signed Area $= \int_{-4}^{-2} (x^2 - 4)\,dx + \int_{-2}^{2} (x^2 - 4)\,dx + \int_2^4 (x^2 - 4)\,dx$ $= \left.\left(\frac{x^3}{3} - 4x\right)\right|_{-4}^{-2} + \left.\left(\frac{x^3}{3} - 4x\right)\right|_{-2}^{2} + \left.\left(\frac{x^3}{3} - 4x\right)\right|_2^4$ $= \left[\left(\frac{(-2)^3}{3} - 4(-2)\right) - \left(\frac{(-4)^3}{3} - 4(-4)\right)\right] + \left[\left(\frac{2^3}{3} - 4(2)\right) - \left(\frac{(-2)^3}{3} - 4(-2)\right)\right] \\ + \left[\left(\frac{4^3}{3} - 4(4)\right) - \left(\frac{2^3}{3} - 4(2)\right)\right]$ $= \frac{32}{3} - \frac{32}{3} + \frac{32}{3}$ $= +\frac{32}{3} \approx 10.\dot{6}$ units squared Ask yourself the following questions: Is it possible to have functions with the same net area but different total area? If such situations exist, what might be the possible explanations? To check your answers work through the following activity individually, or in pairs.

Activity 6.2 Functions with equal net areas but different total areas

Step 1 : On separate grids, plot the following functions over the given intervals: a. $f(x) = x$ for $-2 \leq x \leq 2$ b. $g(x) = 2x$ for $-2 \leq x \leq 2$ Step 2 : For each function shade the part of the graph a. above the $x -$ axis in green b. below the $x -$ axis in red. [Figure] Figure 6.6: Graphical representation of function $f(x) = x$ [Figure] Figure 6.7: Graphical representation of function $g(x) = 2x$

Step 3 : Calculate the area a. above the $x -$ axis. b. below the $x -$ axis (take the magnitude only). Step 4 : For each function calculate the total and net area Step 5 : Answer the following questions in your notebook: a. Are the net areas the same or different? Why? b. Are the total areas the same or different? Why? c. Which measure tells you “Overall effect” and which tells you “Total space covered”? Step 6 : Give a real-world situation where net area would be more useful. Step 7 : Give a real-world situation where total area would be more useful.

Area under constant function

From Activity 6.1, we observed $f(x) = 3$ and $g(x) = -2$ as constant functions because, they had same values as outputs regardless of the input in the domain. Such functions are usually represented by $f(x) = k$ where $k$ is a constant and $x$ is the independent variable. Note: If $f(x) = k$ where $k$ is a constant,

1. $\frac{d}{dx}(k) = 0$ .
2. $\int k\,dx = kx + C$ . To determine the area under a constant function $f(x) = k$ from $x = a$ to $x = b$ , the area will be; $\int_a^b f(x)\,dx = \int_a^b k\,dx$ $= k\int_a^b 1\,dx$ $= k(b - a)$ Note:
3. If $k > 0$ the area is above the $x\text{-axis}$
4. If $k < 0$ the area is below the $x\text{-axis}$ just as seen in Figure 6.3 and Figure 6.4 above.

Example 6.2

Determine the total and net areas between the following functions and the $x -$ axis. a. $f(x) = 4$ using interval $[-3, 1]$ b. $h(x) = 6$ using interval $[2, 5]$ c. $g(x) = -4$ using interval $[-3, 1]$ d. $r(x) = 5$ using interval $[0, 3]$ e. $k(x) = -7$ using interval $[-2, 3]$

Solution

a. Total area $= \int_{-3}^{1} |4|\,dx$ $= \left|4x|_{-3}^{1}\right|$ $= |4(1) - 4(-3)|$ $= 16$ units squared Net Area $= \int_{-3}^{1} 4\,dx$ $= 4x|_{-3}^{1}$ $= 4(1) - 4(-3)$ $= +16$ units squared b. Total area $= \int_2^5 |6|\,dx$ $= \left|6x|_2^5\right|$ $= |6(5) - 6(2)|$ $= 18$ units squared Net Area $= \int_2^5 6\,dx$ $= 6x|_2^5$ $= 6(5) - 6(2)$ $= 18$ units squared c. Total area $= \int_{-3}^{1} |-4|\,dx$ $= \left|-4x|_{-3}^{1}\right|$ $= |-4(1) - (-4(-3))|$ $= 16$ units squared Net Area $= \int_{-3}^{1} -4\,dx$ $= -4x|_{-3}^{1}$ $= -4(1) - \left(-4(-3)\right)$ $= -16$ units squared d. Total area $= \int_0^3 |5|\,dx$ $= \left|5x|_0^3\right|$ $= |5(3) - 5(0)|$ $= 15$ units squared Net Area $= \int_0^3 5\,dx$ $= 5x|_0^3$

$= 5(3) - 5(0)$ $= 15$ units squared e. Total area $= \int_{-2}^{3} |-7|\,dx$ $= \left|-7x|_{-2}^{3}\right|$ $= |-7(3) - (-7)(-2)|$ $= 35$ units squared Net Area $= \int_{-2}^{3} -7\,dx$ $= -7x|_{-2}^{3}$ $= -7(3) - (-7)(-2)$ $= -35$ units squared Note If you are required to find the area under a constant function and no clear indicator is given on the kind of area, calculate the total area .

Area under curves

If you need to find the area under a curve between two x -values a and b, the tool you use is the definite integral . You set up the integral of the function from a to b and evaluate it to get a number. Remember that the area can be Net area (regions below the x -axis count as negative) or Total area (every region is treated as positive, no matter where it sits).

Example 6.3

A drone’s vertical speed (in $m/s$ ) is given by $v(t) = 2t - t^2$ for $0 \leq t \leq 3.$ Determine the total area and net area under the curve created by the function of the vertical speed.

Solution

[Figure] Figure 6.8: Graphical representation of function $v(t) = 2t - t^2$ for $0 \leq t \leq 3$ .

Total Area $\int_0^3 |2t - t^2|\,dt = \int_0^2 |2t - t^2|\,dt + \int_2^3 |2t - t^2|\,dt$ $= \left|\left[t^2 - \frac{t^3}{3}\right]_0^2\right| + \left|\left[t^2 - \frac{t^3}{3}\right]_2^3\right|$ $= \left|\left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right)\right| + \left|\left(3^2 - \frac{3^3}{3}\right) - \left(2^2 - \frac{2^3}{3}\right)\right|$ $= \left|\frac{4}{3}\right| + \left|-\frac{4}{3}\right|$ $= \frac{8}{3}$ $= 2\frac{2}{3}$ units squared The total area under the curve created by the function of the vertical speed is $2\frac{2}{3}$ units squared . Net Area $\int_0^3 2t - t^2\,dt = \int_0^2 2t - t^2\,dt + \int_2^3 2t - t^2\,dt$ $= \left[t^2 - \frac{t^3}{3}\right]_0^2 + \left[t^2 - \frac{t^3}{3}\right]_2^3$ $= \left[\left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right)\right] + \left[\left(3^2 - \frac{3^3}{3}\right) - \left(2^2 - \frac{2^3}{3}\right)\right]$ $= \frac{4}{3} - \frac{4}{3}$ $= 0$ units squared The net area under the curve created by the function of the vertical speed is $0$ units squared.

Example 6.4

The Department of Marine and Fisheries Sciences in University of Ghana is studying wave-energy patterns during a powerful coastal storm. The instantaneous wave power (in kilowatts per metre of wave front) can be modelled over a 12-hour observation window by the function $P(t) = 20 + 12\sin\left(\frac{\pi t}{6}\right) + 6e^{-\frac{(t-12)^2}{40}}$ , where $t$ represents time in hours after midnight $(0 \leq t \leq 12).$ As a marine engineering student evaluating the safety of an offshore wind-farm platform, carry out the following analys es:

a. Estimate the total wave-energy impact (in kilowatt-hours per metre) during the 12- hour storm by evaluating; $\int_0^{12} P(t)\,dt.$ Use six equal sub-intervals and an appropriate numerical method to obtain your result. b. Determine the 2-hour period during which the wave power was, on average, the greatest. Justify your answer by sketching the graph of $P(t)$ . c. The platform is designed to withstand $280\ \mathrm{kWh\,m^{-1}}$ of wave energy during any 24- hour day. Based on your total-energy calculation, decide whether the recorded 12- hour storm would exceed the daily safety threshold if such conditions persisted for a full day. Identify all time intervals within $0 \leq t \leq 12$ when the wave power exceeded $30\mathrm{kW\ m^{-1}}$ , signalling a high-risk condition for equipment and personnel.

Solution

a. $\int_0^{12} |P(t)|\,dt = \int_0^{12} \left|20 + 12\sin\left(\frac{\pi t}{6}\right) + 6e^{-\frac{(t-12)^2}{40}}\right|dt$ $= \int_0^{12} |20|\,dt + \int_0^{12} \left|12\sin\left(\frac{\pi t}{6}\right)\right|dt + \int_0^{12} \left|6e^{-\frac{(t-12)^2}{40}}\right|dt$ $\int_0^{12} |20|\,dt = [20t]_0^{12} = 240$ Solving them separately: $\int_0^{12} \left|12\sin\left(\frac{\pi t}{6}\right)\right|dt = 12\int_0^{12} |\sin u|\,\frac{6}{\pi}\,du, \ u = \frac{\pi t}{6}$ . $\frac{du}{dt} = \frac{\pi}{6}, \ dt = \frac{6}{\pi}du$ . When $t = 0, u = 0$ and when $t = 12, u = 2\pi$ $\int_0^{12} \left|12\sin\left(\frac{\pi t}{6}\right)\right|dt = \frac{72}{\pi}\int_0^{2\pi} |\sin(u)|\,du$ $\frac{72}{\pi}\int_0^{2\pi} |\sin(u)|\,du = \frac{72}{\pi}[|-\cos(u)|]_0^{2\pi}$ $= \frac{72}{\pi}|-\cos(2\pi) + \cos(0)|$ $= \frac{72}{\pi}$ Using the trapezium rule and $h = \frac{12 - 0}{6} = 2$ ,

$\int_0^{12} \left|6e^{-\frac{(t-12)^2}{40}}\right|dt = \frac{2}{2}\left[\left(6e^{-\frac{(0-12)^2}{40}}\right) + \left(6e^{-\frac{(12-12)^2}{40}}\right) + 2\left(6e^{-\frac{(2-12)^2}{40}} + 6e^{-\frac{(4-12)^2}{40}} + 6e^{-\frac{(6-12)^2}{40}} + 6e^{-\frac{(8-12)^2}{40}} + 6e^{-\frac{(10-12)^2}{40}}\right)\right] \\ = 33.3524 \\ \int_0^{12} \left|20 + 12\sin\left(\frac{\pi t}{6}\right) + 6e^{-\frac{(t-12)^2}{40}}\right|dt = 240 + \frac{72}{\pi} + 33.3524 = 296.271\,\mathrm{KWh\,m^{-1}}$ a. The 2-hour period with highest intensity was $2 - 4$ hours after midnight. b. The platform tolerates $280$ kWh $m^{-1}$ in a 24-hour day. If the 12-hour storm conditions persist for 24 hours at the same average rate, total energy will be $2 \times 296.271 = 592.542$ kWh $m^{-1}$ which is well above 280. Hence the recorded storm, if sustained, would exceed the daily threshold so the current design is not adequate. c. From the graph, the values exceed $30$ kW $m^{-1}$ around $1.75 - 4.45$ hours where $1.75 \leq t \leq 4.45$ . [Figure] Figure 6.9: Graphical representation of function $P(t) = 20 + 12\sin\left(\frac{\pi t}{6}\right) + 6e^{-\frac{(t-12)^2}{40}}$

Example 6.5

Find the exact area of the region bounded by the curve $f(x) = -x^3 - 2x^2 + x + 1$ and the $x$ -axis between $x = -2$ and $x = 0.8$ .

Solution

To find the area $(A)$ , integrate the given function such that $A = \int_{-2}^{0.8} f|(x)|\,dx$ . From the graph in Figure 6.10, the area between $x = -2$ and $x = -0.55$ is below the $x$ -axis while the area between $x = -0.55$ and $x = 0.8$ is above the $x$ -axis. [Figure] Figure 6.10: Graphical representation of function $f(x) = -x^3 - 2x^2 + x + 1$ Hence, $\int_{-2}^{0.8} |f(x)|\,dx = \int_{-2}^{-0.55} |f(x)|\,dx + \int_{-0.55}^{0.8} |f(x)|\,dx$ $= \int_{-2}^{-0.55} |-x^3 - 2x^2 + x + 1|\,dx + \int_{-0.55}^{0.8} |-x^3 - 2x^2 + x + 1|\,dx$ $= \left[\left|-\frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + x\right|\right]_{-2}^{-0.55} + \left[\left|-\frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + x\right|\right]_{-0.55}^{0.8}$ $= \left|\left(-\frac{(-0.55)^4}{4} - \frac{2(-0.55)^3}{3} + \frac{(-0.55)^2}{2} + (-0.55)\right) - \left(-\frac{(-2)^4}{4} - \frac{2(-2)^3}{3} + \frac{(-2)^2}{2} + (-2)\right)\right| \\ + \left|\left(-\frac{(0.8)^4}{4} - \frac{2(0.8)^3}{3} + \frac{(0.8)^2}{2} + 0.8\right) - \left(-\frac{(-0.55)^4}{4} - \frac{2(-0.55)^3}{3} + \frac{(-0.55)^2}{2} + (-0.55)\right)\right|$ $= |-1.6440| + |0.9873|$ $= 2.6313$ units squared Therefore, the exact area bounded by the curve is $2.6313$ units squared.

Note When a question/activity requires you to determine the area under the curve without stating whether it is total area or net area, calculate the total area .

The area between two regions

So far we have explored areas between curves and the $x -$ axis. Now, we will take a look at determining the area between two curves. If $y = f(x)$ lies above $y = g(x)$ as illustrated in Figure 6.12 from $x = a$ to $x = b$ , the area of the region between $f(x)$ and $g(x)$ from $x = a$ to $x = b$ is; $A = \sum_{i=1}^{n}[f(c_i) - g(c_i)]h$ , where $h = b - a$ . $A = \int_a^b [f(x) - g(x)]\,dx.$

Activity 6.3: Calculating areas of specific regions using appropriate tech -

niques

Let us estimate the area of grass between a curved hedge and a cement walkway edge in your school using three simple steps: • Manual estimate • Simple calculation • Exact method. Materials Needed • Measuring tape/string/metre ruler • Chalk or sticks for marking • Graph paper and [Figure] Figure 6.11: Curved hedge and cement walkway

Note Red arrow – curved hedge edge Purple arrow- cement walkway edge Step 1:

1. Choose a straight base line about $6 - 8\,m$ along the hedge.

2. Mark every 1 m along this line (these are your x-values).

3. At each mark, measure the perpendicular distance to; a. the hedge edge (upper boundary) b. the walkway edge (lower boundary). c. Record these as two sets of $(x, y)$ pairs. Step 2: Choose one of these methods for manual area estimation

4. Counting Squares: a. Plot both sets of points on graph paper and shade the region. b. Count full squares inside and estimate halves on the edges. c. Multiply by the real area of one square to get a rough total.

5. Strip Averaging: a. For each $1m$ interval, find the average width (hedge distance – walkway distance) of the strip. b. Multiply by $1m$ to get each strip’s area. c. Add all strip areas. Step 3: Apply learnt techniques and compare results

6. Use the same measurements to calculate the area formally with the trapezium rule. I f the points look like a line or a U-shape, create an equation: a. Line: use 2 points to find $y = mx + c$ b. Quadratic: Use 3 points to solve for $y = ax^2 + bx + c$

7. Use $\int_{x_0}^{x_n} [y_{hedge} - y_{walk}]\,dx$ for an exact answer.

8. Check how close the manual , trapezium , and (if done) integral results are.

9. Think about why small differences appear and which method is easiest for your setting Observations Going through the activity, these are some realisations you might come to:

10. Comparison of Methods a. Estimating the area by counting squares or averaging strip widths gives a quick result but may differ slightly from the value obtained using the trapezium rule or definite integral. b. The trapezium rule usually provides a closer approximation because it uses the measured widths more systematically. c. When an equation of the curve is known, integration produces the most exact value.

11. Effect of Measurement a. Small measuring errors, such as uneven ground or a tape that is not straight, can affect all calculations. b. Using shorter intervals (more measuring points) improves the accuracy of both manual and trapezium estimates.

12. Nature of the Curves a. The hedge and walkway edges are not perfectly smooth. b. If the boundaries follow a regular pattern, it is easier to guess a simple linear or quadratic equation.

13. Practical Importance The same techniques can be applied to real tasks such as estimating the size of a lawn for cutting grass, planning a flower bed, or costing materials for a walkway.

14. Mathematical Insight The activity shows that integration and area approximation methods are not only classroom ideas but also tools for solving everyday problems in the school environment.

Example 6.6

Find the area of the region between $y = 2x^2 - 3x + 3$ and $y = -3x^2 + 4x$ from $x = 0$ to $x = 1$ .

Solution

$\int_0^1 [(2x^2 - 3x + 3) - (-3x^2 + 4x)]\,dx$ $= \left[\frac{2}{3}x^3 - \frac{3}{2}x^2 + 3x + \frac{3x^3}{3} - \frac{4x^2}{2}\right]_0^1$ $= \left[\frac{2}{3}x^3 - \frac{3}{2}x^2 + 3x + x^3 - 2x^2\right]_0^1$

$= \left[\frac{5}{3}x^3 - \frac{7}{2}x^2 + 3x\right]_0^1$ $= \frac{5}{3}(1)^3 - \frac{7}{2}(1)^2 + 3(1)$ $= \frac{7}{6}$ $= 1\frac{1}{6}$ units squared Note that the two functions can be subtracted first to obtain $\int_0^1 \left[\frac{5}{3}x^3 - \frac{7}{2}x^2 + 3x\right]dx$ . Then integrated to result in $1\frac{1}{6}$ units squared. [Figure] Figure 6.12: Graphical representation of area between $y = 2x^2 - 3x + 3$ and $y = -3x^2 + 4x$

Example 6.7

Determine the area of the region enclosed between $y = x^3 - 2x^2 + 4$ and $y = -2x^2 + 3x + 2$ from $x = -1$ to $x = 2$ .

Solution

[Figure] Figure 6.13: Graphical representation of area between $y = x^3 - 2x^2 + 4$ and $y = -2x^2 + 3x + 2$

From Figure 6.13 the curve of $y = x^3 - 2x^2 + 4$ lies on top for the given interval $[-1, 2]$ . Area integral: $A = \int_{-1}^{2} [(x^3 - 2x^2 + 4) - (-2x^2 + 3x + 2)]\,dx$ Simplify: $A = \int_{-1}^{2} (x^3 - 3x + 2)\,dx$ Integrate: $A = \left[\frac{1}{4}x^4 - \frac{3}{2}x^2 + 2x\right]_{-1}^{2}$ $= \left(\frac{1}{4}(2)^4 - \frac{3}{2}(2)^2 + 2(2)\right) - \left(\frac{1}{4}(-1)^4 - \frac{3}{2}(-1)^2 + 2(-1)\right)$ $= 2 - \left(-\frac{13}{4}\right)$ $= \frac{21}{4}$

Example 6.8

Determine the area of the region bounded by $y = 5 - x^2$ and $y = 2x + 1$ from $x = -2$ to $x = 1.$

Solution

Top curve on $[-2, 1]$ is $y = 5 - x^2.$ Area integral: $A = \int_{-2}^{1} [(5 - x^2) - (2x + 1)]\,dx$ Simplify: $A = \int_{-2}^{1} (-x^2 - 2x + 4)\,dx$ Integrate: $A = \left[-\frac{1}{3}x^3 - x^2 + 4x\right]_{-2}^{1}$ $= \left(-\frac{1}{3}(1)^3 - (1)^2 + 4(1)\right) - \left(-\frac{1}{3}(-2)^3 - (-2)^2 + 4(-2)\right)$ $= \frac{8}{3} - \left(-\frac{28}{3}\right)$ $= 12$

The area enclosed by two graphs

Sometimes, two curves may intersect at some point(s), creating a region enclosed (bounded) by such graphs. To determine the area enclosed by the curves let us go through this activity.

Activity 6.4: Exploring areas between enclosed curves

Step 1: Draw the graph of the first function $y = -x^2 + 4$ and shade the area between the function and the $x$ -axis.

[Figure]

Figure 6.14: Graphical representation of area between $y = -x^2 + 4$ and the $x$ -axis. Step 2 : On the same axes, draw the graph of the second function $y = x^2 + 2$ and shade the area between the function and the $x$ -axis with a different style. Step 3 : Find and label all points where the two parabolas intersect. Step 4 : Identify the different regions enclosed (bounded) between the curves. [Figure]

Figure 6.15: Graphical representation of area between $y = -x^2 + 4$ and $y = x^2 + 2$ Step 5 : For each region, note which curve is above and which one is below across that interval of $x$ . Step 6 : Noting the Area under each of the curves, explain how you think we can determine the area of the enclosed region.

Step 7 : Express the area under curve using definite integrals and use your explanation to express the area of the enclosed region as a definite integral (Show it to your teacher. You do not need to solve the integral at this point).

Generalisation

From Activity 6.4, it can be observed that the area enclosed by two curves can be expressed as; $A = \int_a^b |f(x) - g(x)|\,dx$ where $f(x)$ lies above $g(x)$ in the given interval, just as illustrated in Figures 6.15 and 6.16. [Figure] Figure 6.16: Graphical representation of area between $y = f(x)$ and $y = g(x)$ .

Example 6.9

Draw the graphs of the curves $y = \frac{x^2}{4} + 1$ and $y = \sqrt{x + 2}$ , where $-2 \leq x \leq 3$ on the same axes. From your graph: i. Determine their points of intersection. ii. Calculate the area enclosed by the two curves.

Solution

i. From the graphical illustrations, the curves intersect at $(-0.72, 1.13)$ and $(2, 2)$ within the interval $[-2, 3]$ .

ii. Enclosed Area $= \int_{-0.72}^{2} \left[\left(\sqrt{x + 2}\right) - \left(\frac{x^2}{4} + 1\right)\right]dx$ $= \left[\frac{2(x + 2)^{\frac{3}{2}}}{3} - \left(\frac{x^3}{12} + x\right)\right]_{-0.72}^{2}$ $= \left[\left(\frac{2(2 + 2)^{\frac{3}{2}}}{3}\right) - \left(\frac{2^3}{12} + 2\right)\right] - \left[\left(\frac{2(-0.72 + 2)^{\frac{3}{2}}}{3}\right) - \left(\frac{(-0.72)^3}{12} + (-0.72)\right)\right]$ $= 0.9501$ [Figure] Figure 6.17: Graphical illustration of area between $y = \frac{x^2}{4} + 1$ and $y = \sqrt{x + 2}$ .

We have familiarised ourselves with calculating areas using integration. Now we will explore its usefulness in real-life using these examples.

Example 6.10

A hot-air balloon is rising vertically, and its upward velocity is modelled by $v(t) = 20(1 - e^{-0.4t})\,m/s$ , where $t$ is the time in seconds after take-off. Find the total height of the balloon after the first $8$ seconds of ascent. [Clue: The total height is the area under the velocity-time graph].

Solution

$v(t) = 20(1 - e^{-0.4t})\,m/s$ , The height of the balloon can be determined by integrating the function $v(t)$ from 0 to 8 seconds. Hence height of balloon $= \int_0^8 v(t)\,dt$ $\int_0^8 v(t)\,dt = \int_0^8 20(1 - e^{-0.4t})\,dt$ $\int_0^8 v(t)\,dt = \int_0^8 20(1 - e^{-0.4t})\,dt$ $= 20\left(\int_0^8 1\,dt - \int_0^8 e^{-0.4t}\,dt\right)$ Solve $\int_0^8 e^{-0.4t}\,dt$ using $\int_a^b e^{kt}\,dt = \left[\frac{1}{k}(e^{kt})\right]_a^b$ where $k = -0.4, a = 0, b = 8.$ $= 20\left[t - \left(-\frac{1}{0.4}e^{-0.4t}\right)\right]_0^8$ $= 20\left[t + \frac{1}{0.4}e^{-0.4t}\right]_0^8$ $= 20\left[\left(8 + 2.5e^{-0.4(8)}\right) - \left(0 + 2.5e^{-0.4(0)}\right)\right]$ $= 20(5.5 + 2.5e^{-3.2})$ $= 110 + 50e^{-3.2}$ $= 112.04m$ Therefore, the balloon is about $112.04m$ high after $8$ seconds.

Example 6.11

Water is being pumped into and drained out of a swimming pool at different times of the day. The net rate of change of water in the pool is given by $r(t) = 150(t - 3)$ litres per minute, where $0 \leq t \leq 6$ .

1. Determine the time at which the water being pumped into the pool and the water being drained is equal.
2. If the pool initially contains $1\,000$ litres of water at $t = 0$ , find the total amount of water in the pool at $t = 6$ minutes.

Solution

1. Water level decreases if $r(t) < 0$ so,

$150(t - 3) < 0$ $150(t - 3) = 0$ $150t - 450 = 0$ $150t = 450$ $t = 3$ At $t = 3\ minutes, r(3) = 0$ , the water level is neither increasing nor decreasing. 2. Amount of water at time $t$ is $A(t) = A(0) + \int_0^t r(t)\,dt$ For $t = 6$ $A(6) = 1000 + \int_0^6 150(t - 3)\,dt$ $= 1000 + 150\left[\frac{t^2}{2} - 3t\right]_0^6$ $= 1000 + 150\left(\frac{6^2}{2} - 3(6)\right)$ $= 1000$ litres The pool still has $1000$ litres of water after $6$ seconds.

When two-dimensional figures are rotated about an axis. A solid is formed. For example, a rectangle on the $x$ -axis as shown in Figure 6.18 and Figure 6.19.

[Figure] [Figure] Figure 6.18: Representation of a rectangle Figure 6.19: solid of evolution of a rectangle

The width of the rectangle is given by w. Its height is represented by R. When the rectangle is rotated about the axis, it creates a cylinder whose volume, V can be found as the product of the area of the circle formed from the rotation $(\pi R^2)$ and the thickness of the cylinder (same as the width of the rectangle, $w$ ) i.e., $V = \pi R^2w$ .

Method of discs

A variety of interesting solids can be obtained from revolving the curve represented by $f(x)$ between a given interval of $x$ , say $[a, b]$ about the $x$ -axis or an interval of $y$ about the $y$ -axis. [Figure] [Figure] Figure 6.21: Resulting solid of evolution of Figure Figure 6.20: Graphical representation of $f(x)$ at 6.20 interval $[a, b]$ [Figure] Figure 6.22: 3-dimensional view of Figure 6.21 Since the curve may not be horizontal (for revolutions about the $x$ -axis) or vertical (for revolutions about the $y$ -axis), the heights of the curve at the extremities of the intervals i.e., $f(a)$ and $f(b)$ may be unequal i.e., $f(a) \neq f(b)$ and thus, $R$ cannot be used to represent the height. This means we cannot simply generalise that the volume of the solid obtained is $V = \pi R^2w$ , where, $w = b - a$ . A way to take care of this problem is to divide the interval, $[a, b]$ into $n$ smaller intervals of width $\Delta x$ or $\Delta y$ so small that the heights of the curve at the ends of those intervals ( $x_i$ , for $i = 0, 1, 2, \ldots, n$ or $y_i,$ for $i = 0, 1, 2, \ldots, n$ ) are the significantly the same. This would mean that many rectangles of height $f(x_i)$ , for $i = 0, 1, 2, \ldots, n$ and width, $\Delta x$ or $\Delta y$ are created. The revolutions of these rectangles would produce $n$ circles ( $c_i, \text{for } i = 0, 1, 2, \ldots, n$ ) of radii, $f(x_i)$ and thereby $n$ cylinders (discs) of thickness, $\Delta x$ or $\Delta y$ .

The volume of the resulting solid is thus the summation of all the volumes of the $n$ cylinders (discs) obtained i.e., $V = \sum_{i=1}^{n} [A(c_i)]\,\Delta x$ This process is referred to as the method of discs . Computing the definite integral over the interval $[a, b]$ removes the need for repeated calculations. In practical terms, the method of discs can be described as follows: Let $f(x) \geq 0$ be a continuous function in the interval $[a, b]$ , consider the region bounded by the curve $y = f(x)$ and the $x$ -axis, for $a \leq x \leq b$ . When this region is revolved about the $x$ - axis, it generates a solid (Figure 6.23 and Figure 6.24). The volume of this solid can be obtained by slicing it perpendicular to the $x\text{-axis}$ noting that each cross section is a circular disk of radius $r = f(x)$ . The volume of the solid is therefore given by: $V = \sum_{i=1}^{n} [A(c_i)]\,\Delta x = \int_a^b \pi[f(x)]^2\,dx$ [Figure] [Figure] Figure 6.24: Graph of the resulting solid of revolution Figure 6.23: Graphical representation of $y = f(x)$ rotated about the $x$ -axis of $y = f(x)$ [Figure] Figure 6.25: 3-dimensional view of Figure 6.24

In a 3-dimensional plane, this is what the rotated curve looks like and the purple circle within the green solid shade shows the cross section is a circular disk

Example 6.12

Revolve the region under the curve $y = \sqrt{x}$ on the interval $[1, 5]$ about the $x\text{-axis}$ and find the volume of the resulting solid of revolution.

Solution

[Figure] [Figure] Figure 6.26: Graph of $y = \sqrt{x}$ when rotated about the Figure 6.27: The resulting solid of revolution of $x$ -axis $y = \sqrt{x}$ [Figure] Figure 6.28: 3-dimensional view of Figure 6.27 $\text{The circular area of a disc} = \pi y^2$ $= \pi\left(\sqrt{x}\right)^2 \text{ squared units}$

$\text{Volume of solid} = \int_1^5 \pi\left(\sqrt{x}\right)^2\,dx$ $= \pi\int_1^5 \left(\sqrt{x}\right)^2\,dx$ $= \pi\int_1^5 x\,dx$ $= \pi\left[\frac{x^2}{2}\right]_1^5$ $= \pi\left[\frac{5^2}{2} - \frac{1^2}{2}\right]$ $= 12\pi$ cubic units

Revolving about the y-axis

It is important to note that when the shape revolves around the y -axis, then the integration is done with respect to y( i.e., dy) and not with respect to x.

Example 6.13

Determine the volume of the solid resulting from revolving the region bounded by the curve $y = 2 - x^2$ from $y = 1$ to $y = 3$ .

Solution

$y = 2 - x^2$ , $y = 1$ , $y = 3$ By solving for $x,$ we have; $x = \sqrt{2 - y}$ Volume $= \int_1^3 \pi\left(\sqrt{2 - y}\right)^2\,dy$ $= \int_1^3 \pi(2 - y)\,dy$ $= \pi\left[2y - \frac{y^2}{2}\right]_1^3$ $= \pi\left[(8 - 4.5) - \left(2 - \frac{1}{2}\right)\right]$ $= 2\pi$ cubic units

Example 6. 14

Determine the volume of the solid resulting from revolving the portion of the curve $y = 1 - \frac{x^2}{3}$ from $x = 0$ to $x = 2$ about the $y\text{-axis}$ .

Solution

$y = 1 - \frac{x^2}{3}$ $\Rightarrow x = \sqrt{3 - 3y}$ When $x = 0$ , $y = 1$ When $x = 2$ , $y = -\frac{1}{3}$ Note that area and volume is a scalar quantity and hence it would not matter which limit is used as the lower or upper limit. $\int_1^{-\frac{1}{3}} \pi\left(\sqrt{3 - 3y}\right)^2\,dy = \int_{-\frac{1}{3}}^{1} \pi\left(\sqrt{3 - 3y}\right)^2\,dy$ $\int_{-\frac{1}{3}}^{1} \pi\left(\sqrt{3 - 3y}\right)^2\,dy = \pi\int_{-\frac{1}{3}}^{1} (3 - 3y)\,dy$ $= \pi\left[3y - \frac{3y^2}{2}\right]_{-\frac{1}{3}}^{1}$ $= \pi\left[\left(3(1) - \frac{3(1)^2}{2}\right) - \left(3(0) - \frac{3(0)^2}{2}\right)\right]$ $= 1.5\pi$ cubic units