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Antidifferentiation and indefinite integrals: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

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

8 min read

Quick answer

Antidifferentiation reverses differentiation. The indefinite integral ∫f(x) dx=F(x)+c\int f(x)\,dx=F(x)+c is the family of functions whose derivative is ff, and the constant cc is essential. A given condition fixes cc to select one particular antiderivative.

Jump to a section
  1. What this dot point is asking
  2. Reversing differentiation
  3. Preparing an integrand before integrating
  4. The constant of integration
  5. Finding a particular antiderivative
  6. Exam-style questions

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, integrals are Unit 3 (Topic 3.2), not Unit 4. This page sits in our Unit 4 folder for historical reasons; the content is Unit 3.

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

Standard antiderivatives

∫xn dx=xn+1n+1+c(n≠−1)\int x^{n}\,dx = \frac{x^{n+1}}{n+1}+c \quad (n\neq -1)

∫ekx dx=1kekx+c,∫1x dx=ln⁡∣x∣+c\int e^{kx}\,dx = \frac{1}{k}e^{kx}+c, \qquad \int \frac{1}{x}\,dx = \ln|x|+c

∫sin⁡(kx) dx=−1kcos⁡(kx)+c,∫cos⁡(kx) dx=1ksin⁡(kx)+c\int \sin(kx)\,dx = -\frac{1}{k}\cos(kx)+c, \qquad \int \cos(kx)\,dx = \frac{1}{k}\sin(kx)+c

The 1k\dfrac{1}{k} factors reverse the chain-rule factor that differentiation of ekxe^{kx}, sin⁡(kx)\sin(kx) or cos⁡(kx)\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 1k\dfrac{1}{k} factor or a sign slip on the cosine, and costs only a few seconds in the exam.

An indefinite integral

Find ∫(6x2−4x+e2x)dx\displaystyle\int\left(6x^{2}-4x+e^{2x}\right)dx

Antidifferentiate term by term, raising each power by one and dividing.

∫6x2 dx=2x3,∫−4x dx=−2x2,∫e2x dx=12e2x.\int 6x^{2}\,dx = 2x^{3}, \qquad \int -4x\,dx = -2x^{2}, \qquad \int e^{2x}\,dx = \tfrac{1}{2}e^{2x}.

Combine and add a single constant.

∫(6x2−4x+e2x)dx=2x3−2x2+12e2x+c.\int\left(6x^{2}-4x+e^{2x}\right)dx = 2x^{3}-2x^{2}+\tfrac{1}{2}e^{2x}+c.

Markers reward correct term-by-term reversal, the 12\tfrac{1}{2} factor on the exponential, and the constant cc.

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 xx, so expand brackets, split fractions, and convert roots and reciprocals to powers first. For example x=x1/2\sqrt{x}=x^{1/2}, 1x2=x−2\dfrac{1}{x^{2}}=x^{-2}, and x2+1x=x+1x\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.

Rewriting before integrating

Find ∫x2+3x dx\displaystyle\int\dfrac{x^{2}+3}{\sqrt{x}}\,dx

Split the fraction and convert to powers:

x2+3x=x2x1/2+3x1/2=x3/2+3x−1/2.\frac{x^{2}+3}{\sqrt{x}}=\frac{x^{2}}{x^{1/2}}+\frac{3}{x^{1/2}}=x^{3/2}+3x^{-1/2}.

Now apply the power rule to each term:
∫x3/2 dx=x5/25/2=25x5/2,∫3x−1/2 dx=3⋅x1/21/2=6x1/2.\int x^{3/2}\,dx=\frac{x^{5/2}}{5/2}=\frac{2}{5}x^{5/2},\qquad \int 3x^{-1/2}\,dx=3\cdot\frac{x^{1/2}}{1/2}=6x^{1/2}.

So the integral is 25x5/2+6x+c\dfrac{2}{5}x^{5/2}+6\sqrt{x}+c. Markers reward the rewrite into powers and the correct division by the new index.

The constant of integration

The constant cc 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 cc and select the one antiderivative that fits.

Using a condition to find c

A curve has gradient dydx=3x2−2\dfrac{dy}{dx}=3x^{2}-2 and passes through (1,4)(1,4). Find yy.

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 cc.

y=x3−2x+c.y = x^{3}-2x+c.

Step 2: Substitute the given point to find cc

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

4=13−2(1)+c=1−2+c=−1+c,4 = 1^{3}-2(1)+c = 1-2+c = -1+c,

so c=5c=5.

Step 3: Write the particular solution

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

y=x3−2x+5.y = x^{3}-2x+5.

Final answer: y=x3−2x+5y = x^{3}-2x+5.

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

Common errors
Omitting the constant of integration
Every indefinite integral needs +c+c; without it you cannot apply a condition.
Using the power rule when n=−1n=-1
∫x−1 dx=ln⁡∣x∣+c\int x^{-1}\,dx=\ln|x|+c, not x00\dfrac{x^{0}}{0}, which is undefined.
Forgetting the 1k\dfrac{1}{k} factor
∫e3x dx=13e3x+c\int e^{3x}\,dx=\dfrac{1}{3}e^{3x}+c, not e3x+ce^{3x}+c.
Solving for cc before antidifferentiating
Find the general antiderivative first, then substitute the condition.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style5 marks
Calculator-free. Find the following indefinite integrals: (a) ∫(4x3−2x)dx\displaystyle\int\left(4x^{3}-\dfrac{2}{x}\right)dx; (b) ∫(e−2x+cos⁡(3x))dx\displaystyle\int\left(e^{-2x}+\cos(3x)\right)dx.
Show worked answer →

Calculator-free antidifferentiation across standard forms.

(a) ∫4x3 dx=x4\displaystyle\int 4x^{3}\,dx=x^{4} and ∫2x dx=2ln⁡∣x∣\displaystyle\int\dfrac{2}{x}\,dx=2\ln|x|, so the integral is x4−2ln⁡∣x∣+cx^{4}-2\ln|x|+c.

(b) ∫e−2x dx=−12e−2x\displaystyle\int e^{-2x}\,dx=-\dfrac{1}{2}e^{-2x} and ∫cos⁡(3x) dx=13sin⁡(3x)\displaystyle\int\cos(3x)\,dx=\dfrac{1}{3}\sin(3x), so the integral is −12e−2x+13sin⁡(3x)+c-\dfrac{1}{2}e^{-2x}+\dfrac{1}{3}\sin(3x)+c.

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

WACE 2023-style5 marks
Calculator-assumed. The gradient of a curve is dydx=6x−4e2x\dfrac{dy}{dx}=6x-4e^{2x} and the curve passes through (0,3)(0,3). Find the equation of the curve.
Show worked answer →

A particular-solution question.

Antidifferentiate: y=3x2−4⋅12e2x+c=3x2−2e2x+cy=3x^{2}-4\cdot\dfrac{1}{2}e^{2x}+c=3x^{2}-2e^{2x}+c.

Apply the condition (0,3)(0,3): 3=3(0)−2e0+c=−2+c3=3(0)-2e^{0}+c=-2+c, so c=5c=5.

The curve is y=3x2−2e2x+5y=3x^{2}-2e^{2x}+5.

Markers reward the general antiderivative with the 12\dfrac{1}{2} factor, substituting the point, and the solved constant.

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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