Loav · VCE Mathematical Methods · Units 3 & 4
M34-INT-01 · Integration · Current study design
1. Integration starts with a derivative
An antiderivative of f is a function F whose derivative is f on the interval being considered. Differentiation asks for the rate belonging to a function; antidifferentiation asks which functions have a stated rate. This reverses a relationship, but it does not recover every piece of information. The derivative of x2 is 2x, and so are the derivatives of x2 + 7 and x2 − 100. A rate alone cannot identify the original vertical position.
The notation ∫f(x) dx represents the family F(x) + C. The integral sign signals antidifferentiation, f(x) is the integrand, and dx identifies x as the variable. C is an arbitrary constant. It is part of the answer to an indefinite integral, not a decoration that becomes optional when the algebra is difficult. On a connected interval, any two antiderivatives of the same continuous function differ by a constant.
A derivative graph also predicts the shape of every antiderivative. Where the derivative is positive, each antiderivative increases; where it is negative, each decreases. If the derivative changes from positive to negative at a zero, every antiderivative has a local maximum at that x-value. Their maximum heights can differ because of C, but their turning-point x-coordinates agree. This gives a graphical check on a recovered function before you calculate an initial-condition constant. A vertical shift cannot repair a wrong gradient or a turning point at the wrong x-value. A definite integral has numerical bounds and produces a signed accumulation. An indefinite integral produces a family of functions. Keep those tasks distinct. Here the focus is finding and checking antiderivatives; the next guide explains how bounds turn an antiderivative into accumulated change. If x has units and f measures a rate per unit x, the antiderivative has the units of f multiplied by the units of x.
2. Reverse the power rule carefully
For rational n ≠ −1, ∫xn dx = xn+1/(n + 1) + C, wherever the functions are real and differentiable. Increase the power by one, then divide by the new power. The division is essential: differentiating xn+1 creates a factor n + 1. Constant multiples can be taken outside an integral, and sums can be integrated term by term. Rewrite roots and reciprocal powers first when this makes the pattern clearer.
For example, x−2 becomes −x−1, whereas √x = x1/2 becomes (2/3)x3/2. These formulas do not remove domain restrictions. The reciprocal example excludes zero; the square-root example uses x ≥ 0 as a real-valued integrand, with derivative checks made in its interval interior. Never extend a rule across a point where the original function is undefined.
The exponent −1 is the exception because n + 1 would be zero. Instead, ∫(1/x) dx = ln x + C on x > 0. On x < 0, an antiderivative is ln(−x) + C, whose logarithm also has a positive argument. Work on one interval avoiding zero; neither expression claims a value at zero. A logarithm is the inverse of an exponential, so its derivative, rather than the power rule, provides the required reversal.
3. Compensate for a linear inner function
The chain rule explains the factors in standard composite antiderivatives. Differentiating (ax + b)n+1 multiplies by a as well as n + 1. Therefore, for a ≠ 0 and n ≠ −1, ∫(ax + b)n dx = (ax + b)n+1/[a(n + 1)] + C. Divide by the coefficient of x, not by the whole expression ax + b. If a = 0, the integrand is constant and should be treated directly.
The same reasoning gives ∫eax+b dx = eax+b/a + C, ∫sin(ax + b) dx = −cos(ax + b)/a + C, and ∫cos(ax + b) dx = sin(ax + b)/a + C. Angles are in radians. The negative sign in the sine formula comes from the derivative of cosine; memorising both as positive creates a predictable error.
For a reciprocal linear expression, ∫1/(ax + b) dx = ln(ax + b)/a + C when ax + b > 0. When ax + b < 0, use ln(−ax − b)/a + C. In both cases a ≠ 0, and the logarithm argument is positive. Check the stated interval before choosing a branch. The convenient divide-by-a pattern applies because the inside derivative is constant. It is not permission to divide by an arbitrary variable expression inside every composite function.
Worked example 1: Recover a function from its gradient and a point
Let f′(x) = 6x2 − 4x + 3 and f(1) = 8. Antidifferentiate term by term: f(x) = 2x3 − 2x2 + 3x + C. Substituting the point gives 8 = 2 − 2 + 3 + C, so C = 5.
Thus f(x) = 2x3 − 2x2 + 3x + 5. Differentiate the result to recover the given gradient, then check f(1) = 8. Both checks matter: every value of C passes the derivative check, but only one passes the point condition. A gradient value alone would not determine C.
Worked example 2: Keep three different reversal factors separate
Find ∫[3(2x + 1)4 + 4e2x − 6sin(3x)] dx. The power term becomes 3(2x + 1)5/(2 × 5); the exponential term becomes 2e2x; the sine term becomes 2cos(3x).
The answer is (3/10)(2x + 1)5 + 2e2x + 2cos(3x) + C. Its derivative is respectively 3(2x + 1)4, 4e2x and −6sin(3x). The two negative signs in the sine integration combine to make a positive cosine term. Only one arbitrary constant is needed for the sum.
4. Use recognition as a reasoned reversal
Recognition means identifying a derivative relationship and reading it backwards. It is especially useful when a question first asks you to differentiate a product or composite function and then says “hence” find an integral. Treat the earlier result as evidence. Compare the whole target integrand with the whole derivative, then adjust a constant multiple or separate an extra term.
This does not mean that the integral of a product equals the product of integrals. Nor does it mean that the integral of a quotient is the quotient of integrals. Differentiation has product and quotient rules, so reversing a single factor without considering the other factor usually fails. For this course, use the permitted standard forms, algebraic rearrangement and a derivative relationship that you can establish; do not invent an unsupported shortcut.
An efficient self-check starts before any technology. State a candidate, differentiate it and compare every factor with the integrand. If the result differs by a constant factor, repair that factor. If it contains an extra term, consider whether the supplied derivative can be rearranged to isolate the desired expression. A derivative that merely looks similar is not yet a valid justification.
Worked example 3: Reverse a supplied product-rule result
Let H(x) = x2ex. The product rule gives H′(x) = 2xex + x2ex = (x2 + 2x)ex. Hence find ∫(3x2 + 6x)ex dx.
The target integrand is 3H′(x), so its antiderivative is 3H(x) + C = 3x2ex + C. The word “hence” links the two parts; it does not authorise integrating x2 and ex separately and multiplying. Differentiating the final expression verifies the exact coefficient match for every real x.
5. Build an exam-ready checking routine
First simplify the integrand only as far as useful: expand a short polynomial, rewrite a reciprocal power, or factor a constant to expose a known pattern. Then identify the rule and its restrictions. After integrating, differentiate your full answer before applying an initial condition. This sequence separates a rule error from an arithmetic error when solving for C.
When technology is permitted, enter the entire integrand with clear brackets and the correct variable. A computer algebra system may return an equivalent expression, including a differently absorbed constant or logarithm form. Check equivalence on the stated domain rather than judging by appearance. Numerical substitution can detect some mistakes, but agreement at one point cannot prove a derivative identity.
If a question asks for an exact answer, preserve fractions, e, logarithms and π. A decimal antiderivative coefficient can create avoidable inaccuracy later in a definite integral. Show enough structure that a reader can see the reverse chain-rule factor. A final line with a correct-looking expression but no domain awareness is incomplete when the integrand has a restricted interval.
Practise, then check
Show the mathematical reason for each answer. Decide whether the result is a function, a signed quantity or a non-negative area before calculating.
- Find ∫(5x4 − 6x + 2) dx and verify your answer.
- Find ∫8/(4x + 3) dx for x > −3/4. Explain the logarithm coefficient.
- Find ∫[5cos(2x) − 3e−x] dx.
- Find ∫(3x − 2)−2 dx. State an excluded value.
- A function has f′(x) = 4x − 6 and f(2) = 1. Find f.
- A student writes ∫x−1 dx = x0/0 and omits C from every indefinite integral. Explain both problems.
- For x > −1, differentiate K(x) = (x + 1)ln(x + 1) − x. Hence find ∫ln(x + 1) dx.
Reasoned answers
- x5 − 3x2 + 2x + C. Differentiation restores 5x4 − 6x + 2, while the derivative of C is zero.
- 2ln(4x + 3) + C. The derivative of the logarithm is 4/(4x + 3), so multiplication by 2 gives the numerator 8. The supplied domain makes the argument positive.
- (5/2)sin(2x) + 3e−x + C. The sine differentiates with a factor 2; the exponential differentiates with a factor −1, producing the required negative term.
- −1/[3(3x − 2)] + C, on an interval excluding x = 2/3. Differentiation gives (3x − 2)−2; there is no antiderivative defined through the original singularity.
- f(x) = 2x2 − 6x + C. Since 1 = 8 − 12 + C, C = 5. Therefore f(x) = 2x2 − 6x + 5.
- The power rule explicitly excludes n = −1; division by zero is undefined. Use ln x + C for x > 0 or ln(−x) + C for x < 0. Omitting C gives one antiderivative when the indefinite integral asks for the family.
- K′(x) = ln(x + 1) + (x + 1)/(x + 1) − 1 = ln(x + 1). Hence the integral is (x + 1)ln(x + 1) − x + C on x > −1. This answer follows from the supplied derivative relationship.
Curriculum and sources
Prepared for the current VCE Mathematical Methods Units 3 and 4 course. Scope follows the VCAA Mathematics Study Design 2023–2027, Mathematical Methods Units 3 and 4, Area of Study 3: Calculus, with the current VCAA clarification. Curriculum checked 15 October 2026. Independent teaching material, not a VCAA publication. Teaching reference: Cambridge Senior Mathematics VCE Mathematical Methods 3&4, second edition (2023), Chapter 11; algorithm conventions also draw on the VCAA pseudocode guidance. All teaching prose, questions, numerical datasets and diagrams are original.
