Applications of Definite Integrals – Area, Volume, Work, and Probability

1. Core Definitions

Fundamental Theorem of Calculus (FTC)
If $F$ is an antiderivative of $f$ on $[a,b]$, then
$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

---

2. Area Between a Curve and the Axis

The signed area is obtained directly from the integral. For regions that cross the x‑axis, split the interval at the zeros.

Worked Example 1 – Area of a Park

A rectangular park has a curved northern boundary described by $y=4-\frac{x^2}{4}$ for $-4\le x\le 4$. The southern boundary is the x‑axis. Find the total area.

$$A={\int }_{-4}^4\left(4-\frac{x^2}{4}\right)dx$$

$$\begin{array}{rl}A& =[4x-\frac{x^3}{12}{]}_{-4}^4\\ & =\left(4\cdot 4-\frac{4^3}{12}\right)-\left(4(-4)-\frac{(-4{)}^3}{12}\right)\\ & =\left(16-\frac{64}{12}\right)-\left(-16+\frac{64}{12}\right)\\ & =32-\frac{128}{12}=32-\frac{32}{3}=\frac{64}{3}{\text{m}}^2.\end{array}$$

Interpretation: The park covers $21.33{\text{ m}}^2$.

---

3. Volumes of Solids of Revolution

3.1 Disk/Washer Method

When rotating about the x‑axis, the radius is the function value $y=f(x)$. If there is a hole (inner radius $g(x)$), use washers.

$$V=\pi {\int }_a^b[R(x{)}^2-r(x{)}^2]dx.$$

3.2 Shell Method

Rotating about the y‑axis (or any vertical line) often leads to shells:

$$V=2\pi {\int }_a^{b}r(x)h(x)dx,$$

where $r(x)$ is the distance from the axis and $h(x)$ the height of the shell.

Worked Example 2 – Daraz Delivery Drone Battery Housing

A drone’s battery compartment is formed by rotating the region bounded by $y=\sqrt{x}$, the x‑axis, and $x=4$ about the x‑axis. Compute the housing volume.

Disk method:
Radius $R(x)=\sqrt{x}$.

$$V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx=\pi [\frac{x^2}{2}{]}_0^4=\pi \frac{16}{2}=8\pi {\text{cm}}^3.$$

Interpretation: The housing holds about $25.13{\text{ cm}}^3$ of volume, enough for a standard Li‑Po cell.

---

4. Work Done by a Variable Force

Work is the integral of force over distance. If the force varies with position, $W={\int }_a^{b}F(x)dx$.

Worked Example 3 – Lifting a Water Tank (NTC)

A water tank of mass $m=500$ kg is lifted vertically from the ground to a height of 12 m. The lifting rope exerts a force equal to the weight plus a frictional component $F_f(x)=5x$ (N) where $x$ is the height in meters. Compute the total work.

Force at height $x$: $F(x)=mg+5x=500\cdot 9.8+5x=4900+5x$ N.

$$W={\int }_0^{12}(4900+5x)dx=[4900x+\frac{5x^2}{2}{]}_0^{12}=4900(12)+\frac{5(144)}{2}=58800+360=59160\text{J}.$$

Interpretation: NTC’s maintenance crew must supply about $59$ kJ of energy.

---

5. Average Value of a Function

The average value of $f$ on $[a,b]$ is

$$\overline{f}=\frac{1}{b-a}{\int }_a^{b}f(x)dx.$$

Worked Example 4 – Average Traffic Speed (Kathmandu)

Traffic speed on a 5 km stretch varies as $v(t)=60-10\sin \left(\frac{\pi t}{2}\right)$ km/h, where $t$ is time in hours from 0 to 2. Find the average speed.

$$\overline{v}=\frac{1}{2-0}{\int }_0^2(60-10\sin (\frac{\pi t}{2}))dt=\frac{1}{2}[60t+\frac{20}{\pi }\cos (\frac{\pi t}{2}){]}_0^2$$

$$=\frac{1}{2}[120+\frac{20}{\pi }(\cos \pi -\cos 0)]=\frac{1}{2}[120+\frac{20}{\pi }(-1-1)]=60-\frac{20}{\pi }\approx 53.64\text{km/h}.$$

---

6. Probability with Continuous Distributions

If a density function $f(x)\ge 0$ satisfies ${\int }_{-\infty }^{\infty }f(x)dx=1$, then for any interval $[a,b]$,

$$P(a\le X\le b)={\int }_a^{b}f(x)dx.$$

Worked Example 5 – eSewa Transaction Amount

Assume transaction amounts (in thousands of NPR) follow the density $f(x)=\frac{2}{9}x$ for $0\le x\le 3$ (and 0 elsewhere). Find the probability that a random transaction is between 1 k and 2 k NPR.

$$P(1\le X\le 2)={\int }_1^2\frac{2}{9}xdx=\frac{2}{9}[\frac{x^2}{2}{]}_1^2=\frac{2}{9}(\frac{4}{2}-\frac{1}{2})=\frac{2}{9}\cdot \frac{3}{2}=\frac{1}{3}.$$

Interpretation: One‑third of eSewa’s daily transactions lie in the 1–2 k NPR range.

---

7. Comparison of Methods for Volume

| Method | Best when… | Typical axis | Integral form |
|--------|------------|--------------|---------------|
| Disk/Washer | Region described as \(y = f(x)\) and rotated about a horizontal axis | Horizontal (x‑axis) or vertical (y‑axis) | \(\pi\int (R^{2}-r^{2})\,dx\) |
| Shell | Region easier to describe in terms of the distance from a vertical axis | Vertical (y‑axis) or any line parallel to it | \(2\pi\int r\,h\,dx\) |
| Cross‑section (Cavalieri) | Complex shapes where a simple cross‑section area is known | Any | \(\int A_{\text{cross}}(x)\,dx\) |

Advantages / Disadvantages

---

8. Real‑World Applications

In the real world

• Daraz order queue: The total number of orders processed in a day can be modeled by $N={\int }_0^{24}\lambda (t)dt$ where $\lambda (t)$ (orders per hour) follows a sinusoidal pattern peaking at 6 pm. The integral gives the daily workload for logistics planning.
• NTC tower construction: The volume of concrete needed for a cylindrical tower of varying radius $r(z)=0.5+0.02z$ (metres) over height $0\le z\le 50$ m is $V=\pi {\int }_0^{50}r(z{)}^{2}dz$. This yields the exact material estimate used in project budgeting.
• Google YouTube buffering: The amount of data buffered over time $t$ seconds when the download rate varies as $R(t)=5+2\sin (\frac{\pi t}{30})$ Mbps is $D={\int }_0^{T}R(t)dt$. The integral predicts when playback will stall, informing adaptive bitrate algorithms.

Worked real situation – Bank loan interest (Khalti) A personal loan of NPR 500,000 is repaid over 5 years with a continuously decreasing interest rate $r(t)=0.12e^{-0.05t}$ per year. The total interest paid is

$$I={\int }_0^{5}500,000r(t)dt=500,000{\int }_0^{5}0.12e^{-0.05t}dt=500,000[-\frac{0.12}{0.05}{e}^{-0.05t}{]}_0^5=500,000(\frac{0.12}{0.05}(1-e^{-0.25}))\approx 500,000\cdot 2.4\cdot 0.221=265,200\text{NPR}.$$

Thus Khalti’s loan calculator must integrate the rate to display the exact interest amount.

---

9. Common Exam Question Types

---

10. Exam tip

• Read the limits first. The limits of integration are dictated by the physical boundaries (height of a tank, time interval, etc.). Write them clearly before manipulating the integrand.
• Choose the simplest method. For rotations, decide between disk/washer and shell by checking which gives a single‑variable expression without solving for inverse functions.
• Check units. Convert all quantities to consistent units (meters, seconds, newtons) before integrating; the final answer’s unit often reveals a set‑up error.
• Integral test shortcut: For $\sum _{n=1}^{\infty }\frac{1}{n^2+1}$, compare with ${\int }_1^{\infty }\frac{dx}{x^2+1}=[\arctan x{]}_1^{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$. Since the integral converges, the series converges. Remember to state the test conditions (positive, decreasing, continuous).

By mastering the set‑up of each application and practicing the visualisation of the region before writing the integral, you will secure full marks in Unit 6.