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SAMath MethodsSyllabus dot point

How does the definite integral connect antidifferentiation to accumulated change?

The Fundamental Theorem of Calculus evaluates a definite integral as the difference of an antiderivative at the two limits.

How to evaluate definite integrals using the Fundamental Theorem of Calculus, the key properties of definite integrals, and how they represent accumulated change.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. The Fundamental Theorem of Calculus
  3. Properties of definite integrals
  4. Net accumulated change
  5. Common errors
  6. Why it matters

What this dot point is asking

A definite integral ∫abf(x) dx\int_a^b f(x)\,dx has lower limit aa and upper limit bb. Unlike an indefinite integral (a family of functions), a definite integral evaluates to a single number.

The Fundamental Theorem of Calculus

The theorem links the two halves of calculus: to accumulate a rate of change ff over [a,b][a,b], find an antiderivative FF and subtract its values at the endpoints. The constant CC cancels in the subtraction, so it is omitted.

Properties of definite integrals

These properties simplify many problems:

Swapping the limits flips the sign; splitting the interval at an intermediate point cc lets you integrate piecewise.

Net accumulated change

If f(t)f(t) is a rate, the definite integral gives the net change in the quantity:

∫abf′(t) dt=F(b)−F(a).\int_a^b f'(t)\,dt=F(b)-F(a).

For example, if v(t)v(t) is velocity, ∫abv(t) dt\int_a^b v(t)\,dt is the displacement (net, signed) over the interval - distances travelled backward count as negative.

Common errors

Why it matters

The Fundamental Theorem is the central result of Stage 2 integral calculus. It converts every area-under-a-curve and accumulated-change problem into a quick antidifferentiate-and-subtract, and it is examined directly in both the Skills and Applications Tasks and the external exam.

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.

2017 SACE Stage 23 marksGiven that the integral of x e^x dx is x e^x - e^x + c, find the exact value of the area bounded by f(x) = x e^x, the x-axis, and the vertical lines x = 1 and x = 5.
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Since f(x) = x e^x is positive on [1, 5], the area equals the definite integral, which we evaluate with the Fundamental Theorem of Calculus using the given antiderivative F(x) = x e^x - e^x.

Area = integral from 1 to 5 of x e^x dx = [x e^x - e^x] from 1 to 5 = F(5) - F(1).

F(5) = 5e^5 - e^5 = 4e^5.
F(1) = 1 e^1 - e^1 = e - e = 0.

Area = 4e^5 - 0 = 4e^5 (exact value).

Marks: one for applying the Fundamental Theorem F(b) - F(a), one for correctly substituting each limit, and one for the exact simplified answer 4e^5. Leaving the answer as 4e^5 (not a decimal) is essential because the question asks for the exact value.

2024 SACE Stage 21 marksVehicles enter a car park at a rate E(t) vehicles per hour, with the car park empty at t = 0. A table gives the value of the integral from 0 to t of E(t) dt at various times, including the value 471 at t = 24. State how many vehicles entered the car park during the 24-hour period.
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The integral of a rate of flow accumulates the total quantity. Here E(t) is the rate at which vehicles enter (vehicles per hour), so the integral from 0 to 24 of E(t) dt is the total number of vehicles that entered over the whole day.

From the table, the integral from 0 to 24 of E(t) dt = 471.

So 471 vehicles entered the car park during the 24-hour period. The single mark is for reading the correct accumulated total from the table and recognising that integrating the entry rate over [0, 24] gives the total number that entered. This is the Fundamental Theorem idea that a definite integral of a rate gives net accumulated change.