SPM High-Weightage Core Topic

Definite Integrals, Area Between Curves & Volume of Revolution

Integration serves as the reverse process of differentiation and the gateway to calculating continuous geometric areas and 3D volumes of revolution. Master finding constants of integration, computing areas bounded by curves and axes, subtracting upper and lower curves, and rotating cross-sections $360^\circ$ about the $x$- and $y$-axes.

1. Core Integration Formulas & Area/Volume Blueprints

Standard Indefinite Integrals

$\int ax^n\,dx = \frac{ax^{n+1}}{n+1} + c$
$\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$

Remember to divide by coefficient $a$ when integrating composite linear terms!

Area Under Curve

$\text{Area}_x = \int_a^b y\,dx$
$\text{Area}_y = \int_c^d x\,dy$

Between two curves: $\int_a^b (y_{\text{upper}} - y_{\text{lower}})\,dx$.

Volume of Revolution ($360^\circ$)

$V_x = \pi \int_a^b y^2\,dx$
$V_y = \pi \int_c^d x^2\,dy$

Always multiply by constant $\pi$ outside the integral bracket!

2. SPM Examiner Pitfalls in Integration

Trap 1: Squaring Before Subtracting in Volume of Revolution:

When calculating the volume of a shaded region between curve $y_1$ and line $y_2$ rotated about the $x$-axis, the formula is $\pi \int (y_1^2 - y_2^2)\,dx$. Writing $\pi \int (y_1 - y_2)^2\,dx$ is an algebraic tragedy that loses all method marks!

Trap 2: Ignoring Area Portions Below the $x$-axis:

Any area below the $x$-axis integrates to a negative value. If a shaded region spans above and below the $x$-axis (e.g. from $0$ to $4$ crossing $x=2$), integrating $\int_0^4 y\,dx$ directly cancels positive and negative portions! You must split the integral: $\int_0^2 y\,dx + |\int_2^4 y\,dx|$.

Trap 3: Rotating About the $y$-axis ($dy$):

When rotating about the $y$-axis, you MUST express $x^2$ in terms of $y$ and integrate with respect to $dy$ across $y$-limits ($c$ to $d$). Students repeatedly plug in $x$-limits by mistake.

3. Progressive Worked Examples with SPM Marking Rubrics

Example 1 • Gradient Function to Curve Equation [4 Marks]
The gradient function of a curve is given by $\frac{dy}{dx} = 3x^2 - 4x + k$, where $k$ is a constant. The tangent to the curve at point $(2, 5)$ is parallel to the $x$-axis.

(a) Find the value of $k$. [2 marks]

(b) Find the equation of the curve. [2 marks]

(a) Parallel to $x$-axis $\implies \frac{dy}{dx} = 0$ at $x = 2$. [1m: P1: Equate gradient to 0]
$3(2)^2 - 4(2) + k = 0 \implies 12 - 8 + k = 0 \implies 4 + k = 0 \implies k = -4$. [1m: N1: $k = -4$]
(b) $y = \int (3x^2 - 4x - 4)\,dx = x^3 - 2x^2 - 4x + c$. [1m: K1: Integrate gradient function]
Passes $(2, 5) \implies 5 = 2^3 - 2(2^2) - 4(2) + c \implies 5 = 8 - 8 - 8 + c \implies c = 13 \implies y = x^3 - 2x^2 - 4x + 13$. [1m: N1: Correct equation]
Example 2 • Paper 2 Section B (Area & Volume 10m Blueprint) [10 Marks]
The curve $y = x^2 - 4x + 3$ intersects the straight line $y = x - 1$ at points $A$ and $B$.

(a) Find the coordinates of points $A$ and $B$. [3 marks]

(b) Calculate the area of the shaded region enclosed between the curve and the line. [4 marks]

(c) Calculate the volume of revolution generated, in terms of $\pi$, when the region bounded by the curve, the line $y = 0$, and the lines $x = 0$ to $x = 1$ is rotated through $360^\circ$ about the $x$-axis. [3 marks]

(a) $x^2 - 4x + 3 = x - 1 \implies x^2 - 5x + 4 = 0 \implies (x - 1)(x - 4) = 0$. [1m: K1: Solve simultaneous equations]
At $x = 1$: $y = 1 - 1 = 0 \implies A(1, 0)$. [1m: N1: Point A]
At $x = 4$: $y = 4 - 1 = 3 \implies B(4, 3)$. [1m: N1: Point B]
(b) Between $x = 1$ and $x = 4$, line is above curve: $\text{Area} = \int_1^4 [(x - 1) - (x^2 - 4x + 3)]\,dx$. [1m: K1: Subtract curve from line]
$= \int_1^4 (-x^2 + 5x - 4)\,dx = \left[-\frac{1}{3}x^3 + \frac{5}{2}x^2 - 4x\right]_1^4$. [1m: K1: Integrate correctly]
At $x = 4$: $-\frac{64}{3} + 40 - 16 = \frac{8}{3}$. At $x = 1$: $-\frac{1}{3} + \frac{5}{2} - 4 = -\frac{11}{6}$. [1m: K1: Substitute limits]
$\text{Area} = \frac{8}{3} - \left(-\frac{11}{6}\right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = 4.5\text{ unit}^2$. [1m: N1: Correct area]
(c) $V = \pi \int_0^1 y^2\,dx = \pi \int_0^1 (x^2 - 4x + 3)^2\,dx = \pi \int_0^1 (x^4 - 8x^3 + 22x^2 - 24x + 9)\,dx$. [1m: K1: Expand and integrate $y^2$]
$= \pi \left[\frac{1}{5}x^5 - 2x^4 + \frac{22}{3}x^3 - 12x^2 + 9x\right]_0^1 = \pi \left(\frac{1}{5} - 2 + \frac{22}{3} - 12 + 9\right)$. [1m: K1: Substitute limit 1]
$= \pi \left(\frac{1}{5} + \frac{22}{3} - 5\right) = \pi \left(\frac{3 + 110 - 75}{15}\right) = \frac{38}{15}\pi\text{ unit}^3 \approx 2.533\pi\text{ unit}^3$. [1m: N1: Correct volume in terms of $\pi$]
Open Form 5 Chapter 3 Worksheet