Loav · VCE Mathematical Methods · Units 3 & 4
M34-INT-02 · Integration · Current study design
1. From small pieces to a signed total
A definite integral combines the contributions of a function over an interval. Imagine dividing [a, b] into narrow strips. A strip with width Δx and height f(x) contributes approximately f(x)Δx. Adding many such contributions approaches ∫ab f(x) dx as the strip widths tend to zero, under the continuity conditions used here. This limit is the connection between local function values and a total over the interval.
When f is above the x-axis, a strip contributes positively; when f is below it, the contribution is negative. Therefore an integral is a signed accumulation, not automatically a geometric area. If f measures metres per second and x is time in seconds, the contributions have units of metres. If both graph axes measure metres, the geometric region has square-metre units. The multiplication of axis units is a useful interpretation check.
The lower and upper bounds specify an interval and an orientation. For a < b, we accumulate from left to right. Reversing them reverses the sign: ∫ba f(x) dx = −∫ab f(x) dx. Equal bounds give zero because the interval has zero width. Neither rule depends on whether the graph itself lies above or below the axis.
2. The fundamental theorem makes accumulation computable
For a continuous function f on [a, b], if F′(x) = f(x), then ∫ab f(x) dx = F(b) − F(a). This is the evaluation form of the fundamental theorem of calculus. It links the limiting sum of many small contributions to the change in one antiderivative. The notation [F(x)]ab is a compact instruction to substitute the upper bound and subtract the value at the lower bound.
Keep brackets around the complete lower-bound expression. If F(a) contains several terms or is negative, omitting brackets can reverse only part of it. Do not substitute bounds into the integrand instead of the antiderivative. That would give f(b) − f(a), which measures a change in function values, not accumulated area under the graph.
An arbitrary constant cancels: [F(b) + C] − [F(a) + C] = F(b) − F(a). You may use any one antiderivative, so the final definite-integral answer has no “+ C”. Exact values should remain exact unless a decimal is requested. For example, logarithmic and exponential expressions can be correct final answers even when they cannot be simplified to a rational number.
3. Use integral properties with their signs intact
Integrals add across adjacent intervals: ∫ac f = ∫ab f + ∫bc f. This lets you split a problem at a zero, a change of rule or an intersection. Constant multiples and sums also pass through integration. These properties explain why several simpler calculations can replace a complicated-looking region, provided every segment is counted once with the correct orientation.
Domain checks come before substitution. A function such as 1/x is undefined at zero, so you must not evaluate an ordinary integral across [−1, 1] by using separate logarithm branches and simply subtracting their endpoint values. The continuity requirement has failed inside the interval. The elementary fundamental-theorem procedure in this guide applies to functions continuous across the stated interval, or to a valid piecewise calculation with well-defined contributions.
A sketch is a decision tool, even if the final answer is algebraic. Mark the interval endpoints, relevant intercepts and intersections, and determine which expression is positive on each section. A graphing window may miss a small region or a repeated root, so solve the needed equations and use the graph to check the result. Never let the appearance of a screen replace a domain argument.
Worked example 1: Separate signed integral from geometric area
For f(x) = x2 − 1 on [−2, 2], the zeros are −1 and 1. An antiderivative is F(x) = x3/3 − x. The signed integral is F(2) − F(−2) = 2/3 − (−2/3) = 4/3.
The graph is positive on [−2, −1] and [1, 2], and negative between −1 and 1. The two outer signed integrals are each 4/3; the middle signed integral is −4/3. The geometric area is 4/3 + 4/3 + 4/3 = 4 square units. Its middle contribution is counted positively by integrating 1 − x2 there.
Taking the magnitude of the entire signed integral would still give 4/3, not 4. Cancellation has already occurred. To find geometric area, split at sign changes first and add non-negative contributions, using the appropriate positive function on each interval.
4. Area between curves means upper minus lower
At a fixed x, the vertical distance between two curves is the upper y-value minus the lower y-value. Hence a region between y = f(x) and y = g(x), with f ≥ g on [a, b], has area ∫ab[f(x) − g(x)] dx. Both curves may lie below the x-axis; their vertical distance is still determined by upper minus lower.
Find intersections by solving f(x) = g(x), then check which intersections bound the requested region. If the curves exchange order within the interval, split the calculation there. A single expression f − g over the whole interval may cancel one region against another. An intersection is not automatically a change of order: the curves can touch and remain in the same order on both sides.
Geometric boundaries may also include vertical lines or an axis. Translate every boundary into either a limit or an upper/lower function before integrating. For a region below the x-axis, zero is the upper function. If the question gives a diagram with shading, the shading identifies the intended region, but the equations are still needed for exact bounds and signs.
Worked example 2: Find an enclosed region between two curves
Find the area enclosed by y = 2x and y = x2. Intersections satisfy x2 = 2x, giving x = 0 and x = 2. At x = 1, the line has height 2 and the parabola height 1, so the line is above on the intervening interval.
The area is ∫02(2x − x2) dx = [x2 − x3/3]02 = 4 − 8/3 = 4/3 square units. The integrand is x(2 − x), which is non-negative throughout [0, 2]; this independently checks the ordering. Integrating only 2x would include the extra region underneath the parabola.
Worked example 3: Solve for an unknown endpoint
Let a ≥ 0 and suppose ∫0a(2x + 1) dx = 12. An antiderivative is x2 + x, so a2 + a = 12. Factoring gives (a + 4)(a − 3) = 0. The algebraic candidates are a = −4 and a = 3.
The condition a ≥ 0 selects a = 3. Substitution confirms ∫03(2x + 1) dx = 9 + 3 = 12. The discarded root solves the algebraic integral equation but violates the stated endpoint condition. On the allowed interval the integrand is positive, so the accumulated area increases with a; the answer also has a sensible geometric interpretation.
5. Write a solution that exposes its decisions
A complete area solution usually contains four visible decisions: relevant boundaries, sign or curve ordering, the integral setup, and evaluation with units. These lines are more valuable than a long calculator transcript. If a numerical solver supplies an intersection, retain unrounded precision during integration and round only the reported final result.
For functions symmetric about the y-axis, equal left and right geometric regions can reduce work, but explain the symmetry. A graph with rotational symmetry about the origin may have zero signed integral over symmetric bounds while enclosing non-zero geometric area. Zero net accumulation means positive and negative contributions balance; it does not imply that the function is zero throughout.
Use scale checks before accepting an answer. If a non-negative function never exceeds M on an interval of width w, its area cannot exceed Mw. If a curve lies between two constant heights, its integral lies between the corresponding rectangular areas. Such checks do not replace integration, but they can reveal a missing factor, mistaken limit or impossible negative area.
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.
- Evaluate ∫13(2x + 1) dx exactly.
- Given ∫−12 f(x) dx = 7 and ∫25 f(x) dx = −3, find ∫−15 f and ∫5−1 f.
- Find both the signed integral and geometric area for y = x − 2 on [0, 3].
- Find the area enclosed by y = 4 − x2 and the x-axis.
- Find the area between y = x and y = x2 on [0, 2]. Explain why one integral of x − x2 is insufficient.
- Why is an ordinary integral of 1/x across [−1, 1] not a valid application of the elementary fundamental-theorem formula?
- For a ≥ 0, solve ∫0a 3x2 dx = 8. Check the result and explain why no + C appears.
Reasoned answers
- [x2 + x]13 = 12 − 2 = 10. The bounds belong in an antiderivative, not in 2x + 1.
- Add adjacent intervals to obtain 7 + (−3) = 4. Reversing the complete interval changes the sign, so the second integral is −4. The negative second segment remains negative during addition.
- An antiderivative is x2/2 − 2x, giving signed integral −3/2. The zero is x = 2: the negative triangular region has area 2 and the positive one area 1/2. Total geometric area is 5/2 square units.
- The intercepts are −2 and 2, and the function is non-negative between them. Area = [4x − x3/3]−22 = 32/3 square units.
- The curves meet at 0 and 1; their order changes at 1. Area = ∫01(x − x2) dx + ∫12(x2 − x) dx = 1/6 + 5/6 = 1 square unit. One unsplit difference would permit cancellation.
- The integrand is undefined at zero, which lies inside the interval. The continuous-interval hypothesis for the elementary evaluation formula is not satisfied. Endpoint substitution conceals this failure; it does not justify the integral.
- [x3]0a = a3 = 8, so a = 2. Substitution gives 8. Constants cancel in the upper-minus-lower subtraction, and a definite integral is a number rather than an antiderivative family.
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.
