Skip to content
spmaddmath.com.my
Tuition

Study

SyllabusFormulasMethodsExam & PapersTools
LocationsPricingBlogOur TeachersContact
EN

Chapters · Integration techniques

Integration techniques in SPM Add Math

Most integration questions in SPM Add Math come down to three moves: reversing the power rule (including the (ax+b)n(ax+b)^n version), using a given point to pin down the constant of integration, and evaluating a definite integral to find an area. Knowing which move a question is asking for, and applying it in the right order, matters more than memorising the notation.

What an integration question is actually asking for

Most integration questions in SPM Add Math fall into one of three shapes: find a general result and leave a constant in it, find a specific function using a given point on its graph, or evaluate a definite integral to get an actual number, usually an area. Before reaching for a rule, it helps to work out which of these three a question is asking for, because that decision shapes everything that follows, whether you need a +c+c, whether you substitute limits, and whether the final answer is an expression or a value.

The mechanics of integration in Add Math build directly on the reverse power rule, so if that step is solid the rest of the chapter becomes a matter of recognising which small variation applies, a linear term inside the brackets, a given point to substitute, or a pair of limits to evaluate between.

Read the wording before you integrate

Phrases like 'passes through the point', 'the curve has gradient function', and 'the area bounded by' are signals for which of the three question types you're facing. Underlining them for a few seconds before picking up your pen saves far more time than it costs.

The reverse power rule, and its (ax+b)n(ax+b)^n version

Every integral in this chapter reduces, eventually, to the reverse power rule. It undoes the ordinary power rule from differentiation: instead of bringing the power down and reducing it by one, integration raises the power by one and divides by the new power.

Reverse power ruleMust memorise
xndx=xn+1n+1+c,n1\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + c, \quad n \neq -1

When the term inside the brackets is linear rather than just xx, something of the form (ax+b)n(ax+b)^n, the same idea applies, but with one extra step: you also divide by the coefficient aa.

Linear composite versionMust memorise
(ax+b)ndx=(ax+b)n+1a(n+1)+c\int (ax+b)^{n}\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + c

That extra division by aa is the single most common thing students forget, especially when aa is something other than 1. It's worth building the habit of writing it into the formula every time, rather than trying to remember it only when a1a \neq 1.

  • Rewrite roots and fractions in index form first, x=x1/2\sqrt{x} = x^{1/2}, 1x2=x2\dfrac{1}{x^{2}} = x^{-2}, before applying either version of the rule.
  • If the integrand is a sum of terms, integrate each term separately; the reverse power rule applies term by term just as differentiation does.
  • Where an expression is a straightforward product that can be expanded first, like (x+1)(x2)(x+1)(x-2), expanding into separate power terms is usually faster and safer than trying to force the (ax+b)n(ax+b)^n form onto it.

Finding the constant of integration from a given point

When a question gives you a gradient function together with one point the curve passes through, it is asking you to find the specific curve out of the infinite family described by +c+c. The method is always the same three steps: integrate the gradient function normally, leaving +c+c at the end; substitute the given point's xx- and yy-values into that result; solve the resulting equation for cc.

Q1[4 marks]

The gradient function of a curve is dydx=6x24x+1\dfrac{dy}{dx} = 6x^{2} - 4x + 1. The curve passes through the point (1,5)(1, 5).

Find the equation of the curve.

Show worked solution

Step 1, integrate the gradient function. Integrating term by term with the reverse power rule gives y=2x32x2+x+cy = 2x^{3} - 2x^{2} + x + c.

Step 2, substitute the given point. At x=1x=1, y=5y=5, so 5=2(1)32(1)2+1+c=1+c5 = 2(1)^{3} - 2(1)^{2} + 1 + c = 1 + c.

Step 3, solve for cc. c=4c = 4, so the equation of the curve is y=2x32x2+x+4y = 2x^{3} - 2x^{2} + x + 4.

Each of those three steps is a separate opportunity to earn method marks under analytic marking, even if the final numerical answer is wrong, correctly integrating and correctly substituting are marked in their own right.

Definite integrals and the area under a curve

A definite integral asks for a number, not an expression, which is why the constant of integration disappears, it's added and then subtracted away when you evaluate between the two limits.

Definite integralMust memorise
abf(x)dx=F(b)F(a)\int_{a}^{b} f(x)\,dx = F(b) - F(a)

Geometrically, this value is the area between the curve and the xx-axis over that interval, with one important exception: where the curve dips below the xx-axis, the definite integral over that stretch comes out negative, even though an area itself cannot be negative.

Sketch before you integrate

A rough sketch of the curve between the two limits tells you immediately whether any part of the region lies below the xx-axis. If it does, integrate that part separately and take its absolute value before adding it to the rest, otherwise the negative and positive regions cancel each other out in the total.

The same reverse power rule from earlier does the actual integrating; a definite integral only adds the final step of substituting both limits and subtracting.

Where marks are typically lost in this chapter

Almost none of the marks lost in integration questions come from not knowing the rule. They come from small slips in applying it.

  • Forgetting to divide by aa when integrating (ax+b)n(ax+b)^n, which leaves an answer that looks right but is off by a constant factor.
  • Dropping the +c+c on an indefinite integral, even when no point is given to find its value, the +c+c itself is usually a scored step.
  • Not expanding a straightforward product first, and instead trying to force it into the (ax+b)n(ax+b)^n pattern where it doesn't actually fit.
  • Sign errors when evaluating F(b)F(a)F(b) - F(a), particularly when F(a)F(a) itself is negative and the subtraction becomes an addition.

Check your answer by differentiating it

If you differentiate your integrated answer and get back the original gradient function, you know the integration step itself was correct, even before you check the arithmetic around it. This takes seconds and catches a large share of avoidable mistakes.

How one-to-one teaching can help

Integration questions tend to fail in the small connecting steps, a missed division by aa, a constant left off, a sign dropped between two limits, rather than in understanding the topic itself. Watching a student's full working, line by line, is usually the fastest way to catch exactly where that happens, which is difficult to do from a printed mark scheme alone.

Our teachers are experienced, and lessons are taught online, in English, while the SPM papers themselves are set bilingually in Bahasa Melayu and English.

If it would help to work through your own integration questions with a teacher watching your working step by step, a one-hour paid trial class, at the teacher's own rate, from RM50 per hour depending on experience, is a reasonable way to find out whether the fit is right. We won't promise a particular grade, but a close look at where your working actually breaks down is often enough to turn near-misses into full marks.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

What's the difference between an indefinite and a definite integral in Add Math?

An indefinite integral gives a general expression with a +c+c at the end, because many different curves share the same gradient function. A definite integral is evaluated between two limits, abf(x)dx=F(b)F(a)\int_{a}^{b} f(x)\,dx = F(b)-F(a), and gives an actual number, the constant cancels out in the subtraction.

Do I always need to write +c+c?

Yes, on every indefinite integral, whether or not the question gives you a point to find its value. If a point is given, you go on to solve for cc; if it isn't, the +c+c still needs to appear in the final answer.

When do I use the (ax+b)n(ax+b)^n version of the power rule instead of expanding first?

Use the (ax+b)n(ax+b)^n version when expanding would create an unreasonably long polynomial, a high power, for instance. For something simple like (x+1)(x2)(x+1)(x-2), expanding into separate terms first is usually quicker and less error-prone.

What happens if the area I need to find dips below the x-axis?

The definite integral over that section evaluates to a negative number. Sketch the curve first, integrate any section below the axis separately, and take its absolute value before combining it with the rest of the area.

Source:SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Ready to get started?

Book a Trial Classfrom RM50/hr · One-hour paid trial · Same-day reply
Book a Trial ClassOne-hour paid trial · Same-day reply