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SAMath MethodsSyllabus dot point

How do we reverse differentiation to recover a function from its rate of change?

Antidifferentiation reverses differentiation to find the family of functions whose derivative is a given function, always including a constant of integration.

How to antidifferentiate power, exponential and reciprocal functions, why the +C constant is essential, and how to find a particular antiderivative from a boundary condition.

Generated by Claude Opus 4.77 min answer

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  1. What this dot point is asking
  2. The standard antiderivatives
  3. Linearity
  4. Negative and fractional powers
  5. Finding a particular antiderivative
  6. Common errors
  7. Why it matters

What this dot point is asking

To antidifferentiate is to undo differentiation - to find a function F(x)F(x) whose derivative is the given function f(x)f(x). Because the derivative of any constant is zero, there is a whole family of antiderivatives differing only by a constant, written F(x)+CF(x)+C.

The standard antiderivatives

The power rule says "add one to the index, divide by the new index". It works for every power except n=−1n=-1, where it would divide by zero - that case gives the logarithm instead.

Linearity

Integration is linear, so you can integrate term by term and pull constants out:

∫(af(x)+bg(x)) dx=a∫f(x) dx+b∫g(x) dx.\int \big(af(x)+bg(x)\big)\,dx=a\int f(x)\,dx+b\int g(x)\,dx.

Negative and fractional powers

Rewrite roots and reciprocals as powers first, then apply the rule:

∫1x2 dx=∫x−2 dx=x−1−1+C=−1x+C,\int \frac{1}{x^2}\,dx=\int x^{-2}\,dx=\frac{x^{-1}}{-1}+C=-\frac{1}{x}+C,

∫x dx=∫x1/2 dx=x3/23/2+C=23x3/2+C.\int \sqrt{x}\,dx=\int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}x^{3/2}+C.

Finding a particular antiderivative

A boundary condition pins down the constant CC, selecting one curve from the family.

Common errors

Why it matters

Antidifferentiation is the engine of all of Topic 3. The definite integral, the Fundamental Theorem of Calculus, and every area calculation rely on first finding an antiderivative. Recovering a function from its rate of change is also the basis of every kinematics and growth model you will integrate.

Exam-style practice questions

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

2024 SACE Stage 22 marksCar A has velocity v(t) = 3 - 3e^(-1.5t) metres per second for 0 <= t <= 2. Given the displacement s(t) satisfies s(0) = 0, show that s(t) = 3t + 2e^(-1.5t) - 2.
Show worked answer →

Displacement is the antiderivative of velocity, so antidifferentiate v(t) = 3 - 3e^(-1.5t) term by term.

The antiderivative of 3 is 3t. For -3e^(-1.5t), recall that d/dt [e^(-1.5t)] = -1.5 e^(-1.5t), so the antiderivative of e^(-1.5t) is (1 / -1.5) e^(-1.5t) = -(2/3) e^(-1.5t). Therefore the antiderivative of -3 e^(-1.5t) is -3 times -(2/3) e^(-1.5t) = 2 e^(-1.5t).

So s(t) = 3t + 2e^(-1.5t) + C.

Apply the condition s(0) = 0: 0 = 0 + 2e^0 + C = 2 + C, so C = -2.

Hence s(t) = 3t + 2e^(-1.5t) - 2, as required. Marks: one for the correct antiderivative including the constant of integration, one for using s(0) = 0 to evaluate C and reach the given expression.

2018 SACE Stage 22 marksThe velocity of a braking vehicle is v(t) = -6t + k metres per second, where k is the initial velocity, and the vehicle stops when v = 0. Write an integral expression for the stopping distance d, and integrate it to find d in terms of k.
Show worked answer →

The vehicle stops when v(t) = 0: -6t + k = 0, giving t = k/6. The distance travelled is the integral of velocity from t = 0 to t = k/6.

d = integral from 0 to k/6 of (-6t + k) dt. (1 mark for this expression)

Antidifferentiate: integral of (-6t + k) dt = -3t^2 + kt. Evaluate between the limits:

d = [-3t^2 + kt] from 0 to k/6
= (-3 (k/6)^2 + k (k/6)) - 0
= -3 (k^2 / 36) + k^2 / 6
= -k^2 / 12 + 2k^2 / 12
= k^2 / 12.

So d = k^2 / 12. (1 mark for the correct integration and result.)