# Applications to kinematics, areas and volumes

> Edexcel International GCSE Further Pure Mathematics · 4PM1 (2016 spec)
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-applications-to-kinematics-areas-and/

This guide teaches you to apply Edexcel IGCSE Further Pure Maths calculus to linear kinematics problems, area under curve calculations, and volumes of revolution about the x and y coordinate axes, with exam-aligned examples.

**Prerequisites:** Differentiation of polynomial, sin(kx) and cos(kx) functions; Indefinite and definite integration of $x^n$ (n≠-1), sin(kx) and cos(kx) functions; Basic kinematics definitions (displacement, velocity, acceleration)

## Learning objectives

- Use differentiation and integration to solve linear kinematics problems relating displacement, velocity and acceleration
- Calculate the area under a curve using definite integration
- Compute volumes of revolution about the x and y coordinate axes using integration

## Calculus Applications to Linear Kinematics

**Kinematics Calculus Relationships** — For motion in a straight line: $v = \frac{ds}{dt}$, $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$, $s = \int v \, dt$, $v = \int a \, dt$. Constants of integration represent initial displacement or initial velocity.

When solving kinematics problems, pay attention to whether you are given displacement, velocity or acceleration, and use the appropriate derivative or integral to find the unknown quantity. Remember to substitute initial conditions to find the value of any constant of integration.

**Worked example:** A particle moves in a straight line with velocity $v = 6t - 2t^2$ m/s, where t is time in seconds. At t=0, displacement s = 0. Find the displacement of the particle after 3 seconds, and the acceleration at t=2 seconds.

1. Step 1: Calculate displacement by integrating velocity with respect to time:

   $$s = \int v \, dt = \int (6t - 2t^2) dt = 3t^2 - \frac{2}{3}t^3 + c$$
2. Step 2: Use initial condition s=0 at t=0 to find c: 0 = 0 - 0 + c → c=0, so $s = 3t^2 - \frac{2}{3}t^3$
3. Step 3: Substitute t=3 to find displacement at 3s:

   $$s = 3(3)^2 - \frac{2}{3}(3)^3 = 3(9) - \frac{2}{3}(27) = 27 - 18 = 9 \, m$$
4. Step 4: Calculate acceleration by differentiating velocity with respect to time:

   $$a = \frac{dv}{dt} = 6 - 4t$$
5. Step 5: Substitute t=2 to find acceleration at 2s:

   $$a = 6 - 4(2) = 6 - 8 = -2 \, m/s^2$$

> **tip**
>
> Negative acceleration means the particle is decelerating, or accelerating in the opposite direction to the positive displacement axis. Always include units in your final answer.

*Calculator:* allowed

## Calculating Area Under a Curve Using Definite Integration

**Area Under a Curve** — The area bounded by a curve $y = f(x)$, the x-axis, and the lines $x=a$ and $x=b$ is given by the definite integral $\int_a^b f(x) dx$, provided the curve lies above the x-axis over the interval $[a,b]$. If the curve is below the x-axis, take the absolute value of the integral.

When calculating total area across an interval where the curve crosses the x-axis, split the integral into parts at each crossing point, calculate the absolute value of each integral, and add them together to get the total area.

**Worked example:** Find the area bounded by the curve $y = 4x - x^2$, the x-axis, $x=1$ and $x=3$.

1. Step 1: First confirm the curve is above the x-axis between x=1 and x=3: roots at x=0 and x=4, so all values are positive in this interval.
2. Step 2: Set up the definite integral for the area:

   $$Area = \int_1^3 (4x - x^2) dx$$
3. Step 3: Integrate the function term by term:

   $$\int (4x - x^2) dx = 2x^2 - \frac{1}{3}x^3$$
4. Step 4: Evaluate at the upper and lower limits and subtract:

   $$\left[2x^2 - \frac{1}{3}x^3\right]_1^3 = \left(2(9) - \frac{1}{3}(27)\right) - \left(2(1) - \frac{1}{3}(1)\right) = 9 - \frac{5}{3} = \frac{22}{3} \, square units$$

> **warning**
>
> Never leave an area as a negative value. If your definite integral gives a negative result, the curve is below the x-axis over that interval, so take the absolute value to get the correct area.

*Calculator:* allowed

## Volumes of Revolution About Coordinate Axes

**Volume of Revolution Formulae** — For a curve $y = f(x)$ rotated 360° about the x-axis between $x=a$ and $x=b$, volume is $V = \pi\int_a^b y^2 dx$. For a curve $x = g(y)$ rotated 360° about the y-axis between $y=c$ and $y=d$, volume is $V = \pi\int_c^d x^2 dy$.

Always expand $y^2$ or $x^2$ correctly before integrating, and remember to include the $\pi$ term in your final answer. You must state the axis of revolution and limits in your working to get full marks.

**Worked example:** Find the volume of the solid formed when the curve $y = x^2 + 1$ is rotated 360° about the x-axis between $x=0$ and $x=2$.

1. Step 1: Write down the volume formula for rotation about the x-axis:

   $$V = \pi\int_0^2 y^2 dx$$
2. Step 2: Substitute $y = x^2 + 1$ and expand $y^2$:

   $$y^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1$$
3. Step 3: Set up the integral with expanded terms:

   $$V = \pi\int_0^2 (x^4 + 2x^2 + 1) dx$$
4. Step 4: Integrate term by term:

   $$\int (x^4 + 2x^2 + 1) dx = \frac{1}{5}x^5 + \frac{2}{3}x^3 + x$$
5. Step 5: Evaluate at upper and lower limits:

   $$\left[\frac{1}{5}x^5 + \frac{2}{3}x^3 + x\right]_0^2 = \frac{32}{5} + \frac{16}{3} + 2 = \frac{206}{15}$$
6. Step 6: Multiply by $\pi$ to get the final volume:

   $$V = \frac{206}{15}\pi \, cubic units$$

> **tip**
>
> If you are asked to give a numerical value, use the $\pi$ function on your calculator and round to 3 significant figures unless told otherwise.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to add the constant of integration when solving indefinite integrals for kinematics.
  - Why it fails: The constant represents initial displacement or velocity, so omitting it gives incorrect values for s or v at non-zero time.
  - Correct: Always add +c after integrating, then substitute given initial conditions to find the value of c.
- **Wrong:** Leaving an area calculation as a negative value.
  - Why it fails: Area is a scalar quantity, so negative values indicate the curve is below the x-axis, not a negative area.
  - Correct: Take the absolute value of any definite integral that returns a negative value when calculating area.
- **Wrong:** Forgetting to square y or x when calculating volumes of revolution.
  - Why it fails: The volume formula explicitly requires the square of the function value to account for the circular cross-section of the solid of revolution.
  - Correct: Always write the formula first, substitute the function, expand the squared term, then integrate.
- **Wrong:** Using the wrong limits when rotating about the y-axis.
  - Why it fails: When rotating about the y-axis, limits must be y-values, not x-values, as you are integrating with respect to y.
  - Correct: Rearrange the function to make x the subject, use the corresponding y-values as limits, and integrate $x^2 dy$.
- **Wrong:** Omitting the $\pi$ term in the final volume of revolution answer.
  - Why it fails: The volume formula includes $\pi$ from the area of a circular cross-section ($A = \pi r^2$).
  - Correct: Always include $\pi$ in your answer, unless explicitly asked to give a numerical value by substituting $\pi \approx 3.14$ or using a calculator.

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Kinematics: Velocity from Displacement | $v = \frac{ds}{dt}$ | Differentiate displacement with respect to time |
| Kinematics: Acceleration from Velocity | $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ | Differentiate velocity once or displacement twice |
| Kinematics: Displacement from Velocity | $s = \int v \, dt + c$ | c = initial displacement at t=0 |
| Kinematics: Velocity from Acceleration | $v = \int a \, dt + c$ | c = initial velocity at t=0 |
| Area under curve (x-axis between a,b) | $\int_a^b y \, dx$ | Take absolute value if curve is below x-axis |
| Volume of revolution (x-axis between a,b) | $V = \pi\int_a^b y^2 dx$ | Rotate 360° about x-axis; limits are x-values |
| Volume of revolution (y-axis between c,d) | $V = \pi\int_c^d x^2 dy$ | Rotate 360° about y-axis; limits are y-values |

## What's next

Now that you have mastered core calculus applications for Edexcel IGCSE Further Pure Maths, you can move on to more advanced calculus topics and exam practice. The next subtopic covers rates of change using calculus, which builds on the differentiation and integration skills you have used here. You should also practice past paper questions on this topic to familiarize yourself with exam phrasing and common question structures, as this topic appears frequently in both Paper 1 and Paper 2 of the 4PM1 exam. Make sure you memorize the volume of revolution formulae, as these are not provided on the formula sheet, and practice applying them to a range of curve types including polynomials and trigonometric functions.

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