§-Syllabus dot point

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What does it mean to reverse differentiation, and how do we find the family of antiderivatives of a function?

Find antiderivatives of standard functions, include the constant of integration, and determine a particular antiderivative from an initial condition

WACE Year 12 Mathematics Methods Unit 4 antidifferentiation: reversing differentiation, the constant of integration, antiderivatives of powers, exponentials, trig and one over x, and finding a particular solution from a condition, with worked examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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What this dot point is asking
Reversing differentiation
Preparing an integrand before integrating
The constant of integration
Finding a particular antiderivative

What this dot point is asking

SCSA Unit 4 begins integral calculus by reversing the differentiation of Unit 3. This dot point asks you to find antiderivatives of standard functions, always include the constant of integration, and use a condition to determine a particular solution. It is examined in both the calculator-free and calculator-assumed sections.

Reversing differentiation

If $F'(x)=f(x)$ , then $F$ is an antiderivative of $f$ . Because the derivative of a constant is zero, adding any constant gives another valid antiderivative, so the general antiderivative includes $+c$ . Geometrically, the antiderivatives form a stack of parallel curves, each a vertical translate of the others, all sharing the same gradient function $f$ at every $x$ .

The $\dfrac{1}{k}$ factors reverse the chain-rule factor that differentiation of $e^{kx}$ , $\sin(kx)$ or $\cos(kx)$ would have produced. A reliable check on any antiderivative is to differentiate your answer: if it returns the original integrand, the antiderivative is correct. This self-check catches the most common errors, such as a missing $\dfrac{1}{k}$ factor or a sign slip on the cosine, and costs only a few seconds in the exam.

Preparing an integrand before integrating

Many integrals do not match a standard form until you rewrite them. The power rule needs a single power of $x$ , so expand brackets, split fractions, and convert roots and reciprocals to powers first. For example $\sqrt{x}=x^{1/2}$ , $\dfrac{1}{x^{2}}=x^{-2}$ , and $\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}$ . Once each term is a clean power, exponential, or trig form, the standard antiderivatives apply directly.

The constant of integration

The constant $c$ is not optional: the indefinite integral represents an entire family of parallel curves differing only by a vertical shift. Omitting it both loses a mark and makes it impossible to apply an initial condition.

Finding a particular antiderivative

When extra information is given, such as a point on the curve, substitute it to solve for $c$ and select the one antiderivative that fits.

Using a condition to find c

A curve has gradient $\dfrac{dy}{dx}=3x^{2}-2$ and passes through $(1,4)$ . Find $y$ .

Step 1: Antidifferentiate to find the general solution

Because the gradient function is given, antidifferentiation produces the family of all curves with that gradient. Each member of the family differs from the others only by a vertical shift, captured by the constant $c$ .

y = x^{3}-2x+c.

Step 2: Substitute the given point to find $c$

The point $(1,4)$ lies on the curve, so substituting $x=1$ and $y=4$ gives an equation in $c$ alone that can be solved. This selects the single curve that passes through the given point.

4 = 1^{3}-2(1)+c = 1-2+c = -1+c,

so $c=5$ .

Step 3: Write the particular solution

With $c$ determined, replace it to obtain the unique antiderivative that satisfies the condition.

y = x^{3}-2x+5.

Final answer: $y = x^{3}-2x+5$ .

Markers reward the general antiderivative, substitution of the point, and the solved constant.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20215 marksCalculator-free. Find the following indefinite integrals: (a)

\displaystyle\int\left(4x^{3}-\dfrac{2}{x}\right)dx

; (b)

\displaystyle\int\left(e^{-2x}+\cos(3x)\right)dx

Show worked answer →

Calculator-free antidifferentiation across standard forms.

(a) $\displaystyle\int 4x^{3}\,dx=x^{4}$ and $\displaystyle\int\dfrac{2}{x}\,dx=2\ln|x|$ , so the integral is $x^{4}-2\ln|x|+c$ .

(b) $\displaystyle\int e^{-2x}\,dx=-\dfrac{1}{2}e^{-2x}$ and $\displaystyle\int\cos(3x)\,dx=\dfrac{1}{3}\sin(3x)$ , so the integral is $-\dfrac{1}{2}e^{-2x}+\dfrac{1}{3}\sin(3x)+c$ .

Markers reward the power rule, the $\ln|x|$ form for $\dfrac{1}{x}$ , the $\dfrac{1}{k}$ factors, and a single constant.

WACE 20235 marksCalculator-assumed. The gradient of a curve is

\dfrac{dy}{dx}=6x-4e^{2x}

and the curve passes through

(0,3)

. Find the equation of the curve.

Show worked answer →

A particular-solution question.

Antidifferentiate: $y=3x^{2}-4\cdot\dfrac{1}{2}e^{2x}+c=3x^{2}-2e^{2x}+c$ .

Apply the condition $(0,3)$ : $3=3(0)-2e^{0}+c=-2+c$ , so $c=5$ .

The curve is $y=3x^{2}-2e^{2x}+5$ .

Markers reward the general antiderivative with the $\dfrac{1}{2}$ factor, substituting the point, and the solved constant.