What does the integration module cover in N(A)-Level A-Maths 4051?

It covers three outcomes in the Calculus strand of syllabus 4051: integration as the reverse of differentiation, including the constant of integration; evaluating definite integrals by substituting upper and lower limits; and finding the area of a region bounded by a curve and the $x$-axis, handling regions below the axis. They build from the indefinite integral to its main application, area.

How do you integrate a power of $x$?

Reverse the power rule: $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + c$ for $n \neq -1$. You add one to the index, divide by the new index, and add the constant of integration $c$ for an indefinite integral. Integrate a sum term by term.

Why does an indefinite integral need a constant of integration?

Differentiating a constant gives zero, so many functions share the same derivative, differing only by a constant. Integration cannot recover that lost constant, so we write $+ c$ to represent every possible original function. A definite integral does not need $c$ because the constant cancels when you subtract the limits.

How do you find the area under a curve?

The area between a curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is the definite integral $\displaystyle\int_a^b f(x)\,dx$, provided the curve stays above the axis. Where the curve dips below the axis the integral is negative, so you compute that part separately and take its magnitude before adding.

How is the Calculus strand assessed in syllabus 4051?

Both Paper 1 and Paper 2 (each 1 hour 45 minutes, 70 marks, weighted 50 percent) draw on the whole syllabus, and candidates answer all questions. Integration questions range from short indefinite integrals to definite integrals and area problems, so practise reversing the power rule cleanly and handling limits with care.

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Integration in N(A)-Level Additional Mathematics (4051): integration as the reverse of differentiation, the constant of integration, evaluating definite integrals using limits, and finding the area under a curve

An N(A)-Level Additional Mathematics (4051) overview of integration in the Calculus strand. How integration reverses the power rule, why the constant of integration matters, how to evaluate a definite integral by substituting limits, and how to find the area under a curve including regions below the axis, with links to every dot point.

Generated by Claude Opus 4.87 min readSEAB-4051 CalculusUpdated 2026-06-13

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

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Differentiation in reverse
Integration as the reverse of differentiation
Definite integrals
Area under a curve
How this module is examined
Check your knowledge

Differentiation in reverse

Integration is the second half of the Calculus strand of N(A)-Level Additional Mathematics (SEAB 4051). Where differentiation finds a gradient, integration undoes it: from a derivative it recovers the original function, and from a curve it measures the area beneath it. The whole module rests on one move, reversing the power rule, and then applies it to definite integrals and areas.

This overview links three dot points: integration as the reverse of differentiation, definite integrals, and the area under a curve. Each has its own worked answers and practice.

Integration as the reverse of differentiation

The integration as the reverse of differentiation outcome establishes the indefinite integral. To integrate a power, add one to the index and divide by the new index:

\int x^n\,dx = \frac{x^{n+1}}{n+1} + c, \quad n \neq -1.

The constant of integration

c

stands for the constant that differentiation destroys, and you integrate a sum term by term.

Definite integrals

The definite integrals outcome evaluates an integral between limits. You integrate as usual, then substitute the upper limit and the lower limit and subtract:

\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a),

where

F

is an integral of

f

. No constant of integration is needed because it cancels in the subtraction. The answer is a number.

Area under a curve

The area under a curve outcome applies the definite integral to measure area. Between $x = a$ and $x = b$ , the area between the curve and the $x$ -axis is $\displaystyle\int_a^b f(x)\,dx$ when the curve lies above the axis. Where the curve drops below the axis the integral is negative, so you split the region at the $x$ -intercepts and take the magnitude of each piece before adding.

How this module is examined

Both papers, all questions. Paper 1 and Paper 2 (each 1 hour 45 minutes, 70 marks, 50 percent) cover the full syllabus, and you answer every question.
Add $+ c$ for indefinite integrals only. Include the constant of integration for an indefinite integral; omit it for a definite integral, where it cancels.
Watch for area below the axis. If part of the region lies below the $x$ -axis, integrate that part separately and take its magnitude; do not let a negative integral cancel a positive area.

Check your knowledge

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

Find $\displaystyle\int (4x^3)\,dx$ . (2 marks)
Find $\displaystyle\int (2x + 1)\,dx$ . (2 marks)
Evaluate $\displaystyle\int_0^2 3x^2\,dx$ . (3 marks)
Find the area between $y = 2x$ and the $x$ -axis from $x = 0$ to $x = 3$ . (3 marks)

Solutions

The four questions cover indefinite integration of a power, indefinite integration of a sum, a definite integral, and area under a curve.

Step 1: Integrate $4x^3$ with respect to $x$

Reverse the power rule: add one to the index and divide by the new index. The index rises from $3$ to $4$ , and dividing the coefficient $4$ by $4$ gives $1$ . Add the constant of integration because this is an indefinite integral:

\int 4x^3\,dx = x^4 + c.

Final answer (Q1): $x^4 + c$ .

Step 2: Integrate $(2x + 1)$ with respect to $x$

Integrate each term separately by the same reverse power rule. The term $2x$ integrates to $x^2$ , and the constant $1$ integrates to $x$ :

\int (2x + 1)\,dx = x^2 + x + c.

Final answer (Q2): $x^2 + x + c$ .

Step 3: Evaluate $\displaystyle\int_0^2 3x^2\,dx$

Integrate first to get $[x^3]_0^2$ , then substitute the upper limit and subtract the lower limit. The constant of integration cancels in the subtraction, so it is omitted:

\big[x^3\big]_0^2 = 2^3 - 0^3 = 8 - 0 = 8.

Final answer (Q3): $8$ .

Step 4: Find the area between $y = 2x$ and the $x$ -axis from $x = 0$ to $x = 3$

The line $y = 2x$ is above the axis on $[0, 3]$ (it is zero only at the origin), so the area equals the definite integral directly. No sign adjustment is needed:

\int_0^3 2x\,dx = \big[x^2\big]_0^3 = 9 - 0 = 9.

Final answer (Q4): The area is $9$ square units.

Sources & how we know this

Singapore-Cambridge GCE N(A)-Level Additional Mathematics (Syllabus 4051) — Singapore Examinations and Assessment Board (2026)