What does integration and its applications cover in O-Level A-Maths 4049?

It covers four outcomes in the Calculus strand of syllabus 4049: integration as the reverse of differentiation, including the constant of integration and integrating functions of a linear expression; integrating the exponential, reciprocal and trigonometric functions and their linear composites; definite integrals and the area under a curve; and the area enclosed between two curves or a curve and a line.

Why does indefinite integration need a constant of integration?

Differentiation removes a constant term, since the derivative of any constant is zero. So when you reverse the process there is a whole family of functions differing by a constant that all have the same derivative. The constant of integration $+ C$ records that ambiguity. A definite integral, evaluated between limits, cancels the constant, so it gives a single number.

How do you integrate a function of a linear expression like $(2x + 1)^5$?

Raise the power and divide by the new power, then divide again by the coefficient of $x$ inside the bracket: $\displaystyle\int (ax + b)^n\,dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + C$ for $n \neq -1$. So $\displaystyle\int (2x + 1)^5\,dx = \dfrac{(2x + 1)^6}{12} + C$. The same dividing-by-$a$ idea applies to the linear composites of $e^{ax+b}$, $\sin(ax+b)$ and similar.

How does a definite integral give an area?

The definite integral $\displaystyle\int_a^b f(x)\,dx$ equals the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. Above the axis the area counts as positive and below it as negative, so where a region dips below the axis you must split the integral at the $x$-axis crossings and add the absolute values to get the true geometric area.

How is the Calculus strand assessed in syllabus 4049?

Both Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, weighted 50 percent) draw on the whole syllabus, and candidates answer all questions. Integration questions range from short 'integrate this' tasks to definite integrals and area problems that combine finding intersection points with evaluating an integral, so the algebra and the calculus must both be secure.

SingaporeAdditional Mathematics

Integration and its applications in O-Level Additional Mathematics (4049): integration as the reverse of differentiation, integrating standard functions, definite integrals and area under a curve, and the area between curves

An O-Level Additional Mathematics (4049) overview of integration and its applications in the Calculus strand. How integration reverses differentiation, the integrals of powers and standard functions including linear composites, definite integrals and the area under a curve, and the area between two curves, with links to every dot point.

Generated by Claude Opus 4.87 min readSEAB-4049 CalculusUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

Differentiation run backwards
Integration as the reverse of differentiation
Integrating the special functions
Definite integrals and area under a curve
Area between two curves
How this module is examined
Check your knowledge

Differentiation run backwards

Integration is the second half of the Calculus strand of Additional Mathematics (SEAB 4049), and the cleanest way to understand it is as the reverse of differentiation. Where differentiation finds a gradient, integration recovers the original function from its gradient, and its great application is finding areas. The student who pairs each integral with the derivative it undoes learns the standard integrals quickly.

This module covers four outcomes that build in sequence: indefinite integration as the reverse of differentiation, integrating the standard special functions, definite integrals and area under a curve, and the area between curves. This overview ties the four dot points together; each has its own worked answers and practice.

Integration as the reverse of differentiation

The integration as the reverse of differentiation outcome establishes the basic rule. To reverse the power rule you add one to the index and divide by the new index: $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$ , where $+C$ is the constant of integration. For a function of a linear expression you also divide by the coefficient of $x$ inside the bracket.

Integrating the special functions

The integration of special functions outcome gives the standard integrals that mirror the standard derivatives: $\displaystyle\int e^x\,dx = e^x + C$ , $\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + C$ , $\displaystyle\int \cos x\,dx = \sin x + C$ and $\displaystyle\int \sin x\,dx = -\cos x + C$ . Each has a linear-composite version, where you divide by the coefficient of $x$ , for example $\displaystyle\int e^{ax}\,dx = \dfrac{1}{a}e^{ax} + C$ .

Definite integrals and area under a curve

The definite integrals and area under a curve outcome evaluates an integral between limits. You find an antiderivative $F(x)$ and compute $F(b) - F(a)$ ; the constant of integration cancels. This signed value is the area between the curve and the $x$ -axis, with regions below the axis counting as negative, so for true geometric area you split at the crossings.

Area between two curves

The area between curves outcome finds the region enclosed between two curves, or a curve and a line. You first find the intersection points, which become the limits, then integrate the upper function minus the lower function between them.

Find the area enclosed between

y = x + 2

and

y = x^2

step Find the intersection points

Set the curves equal: $x^2 = x + 2$ , so $x^2 - x - 2 = 0$ , giving $(x - 2)(x + 1) = 0$ and $x = -1$ or $x = 2$ . These are the limits.

step Integrate upper minus lower

Between the limits the line $y = x + 2$ is above the parabola $y = x^2$ , so

\text{Area} = \int_{-1}^{2} \left[(x + 2) - x^2\right]\,dx = \int_{-1}^{2} \left(x + 2 - x^2\right)\,dx.

step Evaluate the definite integral

The antiderivative is $\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}$ . At $x = 2$ it is $2 + 4 - \dfrac{8}{3} = \dfrac{10}{3}$ ; at $x = -1$ it is $\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}$ . The area is

\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.

How this module is examined

Both papers, all questions. Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, 50 percent) cover the full syllabus, and you answer every question.
Mind the constant and the limits. Include $+C$ for indefinite integrals; for definite integrals show the substitution $F(b) - F(a)$ clearly.
Decide upper minus lower from a sketch. For area between curves, a quick sketch tells you which function is on top, so the integrand is positive over the region.

Integration and its applications is the second half of the Calculus strand in O-Level A-Maths (SEAB 4049), best understood as differentiation reversed. The power rule reverses to $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ ( $n \neq -1$ ), with a $+C$ for every indefinite integral and an extra division by $a$ for a linear composite $(ax + b)$ . The standard integrals of $e^x$ , $\dfrac{1}{x}$ , $\cos x$ and $\sin x$ mirror the standard derivatives. A definite integral $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ gives the signed area under a curve, negative below the axis. The area between two curves is the integral of the upper minus the lower function between their intersection points. Both papers draw on the whole syllabus.

Check your knowledge

Short questions across the four outcomes. Work them with full method, then check the solutions.

Find $\displaystyle\int (4x^3 - 2x + 1)\,dx$ . (2 marks)
Find $\displaystyle\int \cos x\,dx$ . (1 mark)
Evaluate $\displaystyle\int_0^2 3x^2\,dx$ . (2 marks)
Find $\displaystyle\int (3x + 1)^4\,dx$ . (2 marks)

Solutions

Step 1: Integrate term by term (Q1)

To integrate a polynomial you apply the power rule to each term separately. Add one to each exponent and divide by the new exponent, then attach the constant of integration.

\int (4x^3 - 2x + 1)\,dx = x^4 - x^2 + x + C.

Final answer: $x^4 - x^2 + x + C$

Step 2: Use the standard integral of cosine (Q2)

The integral of cosine is a standard result that mirrors the derivative of sine. Because differentiating $\sin x$ gives $\cos x$ , the reverse process gives:

\int \cos x\,dx = \sin x + C.

Final answer: $\sin x + C$

Step 3: Apply the power rule and evaluate between limits (Q3)

First find an antiderivative of $3x^2$ . Adding one to the power and dividing by the new power gives $x^3$ . For a definite integral you substitute the upper limit then subtract the value at the lower limit; the constant of integration cancels.

\int_0^2 3x^2\,dx = \Big[x^3\Big]_0^2 = 2^3 - 0^3 = 8 - 0 = 8.

Final answer: $8$

Step 4: Integrate a linear composite using the chain-rule adjustment (Q4)

When the base of a power is a linear expression rather than $x$ alone, you still add one to the power and divide by the new power, but you must also divide by the coefficient of $x$ inside the bracket. Here the coefficient is $3$ , so you divide by an additional factor of $3$ .

\int (3x + 1)^4\,dx = \frac{(3x + 1)^5}{5 \times 3} + C = \frac{(3x + 1)^5}{15} + C.

Final answer: $\dfrac{(3x + 1)^5}{15} + C$

Sources & how we know this

Singapore-Cambridge GCE O-Level Additional Mathematics (Syllabus 4049) — Singapore Examinations and Assessment Board (2026)