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The definite integral and Fundamental Theorem (SACE Stage 2 Mathematical Methods)

Syllabus dot point

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

8 min read

Quick answer

The Fundamental Theorem of Calculus says ∫abf(x) dx=F(b)−F(a)\displaystyle\int_a^b f(x)\,dx=F(b)-F(a), where FF is any antiderivative of ff. The definite integral is a single number - the net accumulated change of FF between x=ax=a and x=bx=b - and no constant of integration is needed.

Jump to a section
  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. The two forms of the theorem
  7. A worked accumulated-change problem
  8. Definite integrals on the calculator
  9. Why it matters
  10. Exam-style questions

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

Fundamental Theorem of Calculus

∫abf(x) dx=[F(x)]ab=F(b)−F(a),where F′(x)=f(x).\int_a^b f(x)\,dx=\Big[F(x)\Big]_a^b=F(b)-F(a), \qquad \text{where } F'(x)=f(x).

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.

Evaluating a polynomial integral

Evaluate ∫13(2x+1) dx\displaystyle\int_1^3 (2x+1)\,dx.

Step 1: Find an antiderivative

The Fundamental Theorem requires an antiderivative F(x)F(x) such that F′(x)=2x+1F'(x) = 2x + 1. Applying the power rule term by term gives:

F(x)=x2+x.F(x) = x^2 + x.

No constant of integration is needed here because it cancels when we subtract F(a)F(a) from F(b)F(b).

Step 2: Apply the limits using F(b)−F(a)F(b) - F(a)

Evaluate FF at the upper limit b=3b = 3 and the lower limit a=1a = 1, then subtract.

∫13(2x+1) dx=[x2+x]13=(32+3)−(12+1)=12−2=10.\int_1^3 (2x+1)\,dx = \Big[x^2 + x\Big]_1^3 = \big(3^2 + 3\big) - \big(1^2 + 1\big) = 12 - 2 = 10.

Final answer: ∫13(2x+1) dx=10\displaystyle\int_1^3 (2x+1)\,dx = 10.

Properties of definite integrals

These properties simplify many problems:

Properties of definite integrals

∫aaf(x) dx=0,∫baf(x) dx=−∫abf(x) dx,\int_a^a f(x)\,dx=0, \qquad \int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,

∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx,∫abk f(x) dx=k∫abf(x) dx.\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx, \qquad \int_a^b k\,f(x)\,dx=k\int_a^b f(x)\,dx.

Swapping the limits flips the sign; splitting the interval at an intermediate point cc lets you integrate piecewise. The additivity property is especially useful when a function is defined in pieces, or when a curve crosses the xx-axis and you want to handle the positive and negative parts separately.

An exponential definite integral

Evaluate ∫02e0.5x dx\displaystyle\int_0^2 e^{0.5x}\,dx.

Step 1: Find the antiderivative of e0.5xe^{0.5x}

For an exponential of the form ekxe^{kx}, the antiderivative is 1kekx\dfrac{1}{k}e^{kx}, because differentiating ekxe^{kx} multiplies by kk, so the antiderivative must divide by kk to cancel that factor. Here k=0.5k = 0.5, so:

F(x)=10.5e0.5x=2e0.5x.F(x) = \frac{1}{0.5}e^{0.5x} = 2e^{0.5x}.

Step 2: Evaluate at each limit and subtract

Apply the Fundamental Theorem by substituting the upper limit x=2x = 2 and the lower limit x=0x = 0.

∫02e0.5x dx=[2e0.5x]02=2e0.5(2)−2e0.5(0)=2e1−2e0=2e−2.\int_0^2 e^{0.5x}\,dx = \Big[2e^{0.5x}\Big]_0^2 = 2e^{0.5(2)} - 2e^{0.5(0)} = 2e^{1} - 2e^{0} = 2e - 2.

Step 3: Compute the numerical value

2e−2≈2(2.7183)−2=5.4366−2=3.437.2e - 2 \approx 2(2.7183) - 2 = 5.4366 - 2 = 3.437.

Final answer: ∫02e0.5x dx=2e−2≈3.437\displaystyle\int_0^2 e^{0.5x}\,dx = 2e - 2 \approx 3.437.

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.

Net vs total when the function changes sign

A definite integral measures net (signed) accumulation. Where f(x)f(x) dips below the axis it contributes negatively. If a question asks for total distance or area, split the integral at the zeros and take absolute values of the negative pieces.

Common errors

Subtracting the limits in the wrong order

The Fundamental Theorem requires upper minus lower: F(b)−F(a)F(b)-F(a). Reversing it negates the answer. Be especially careful with negative limits and brackets - write (F(b))−(F(a))\big(F(b)\big)-\big(F(a)\big) with full brackets so that subtracting a negative value of F(a)F(a) becomes an addition rather than a sign slip.

The two forms of the theorem

The version above evaluates an integral, but the Fundamental Theorem also has a differentiation form: if G(x)=∫axf(t) dtG(x)=\displaystyle\int_a^x f(t)\,dt, then G′(x)=f(x)G'(x)=f(x). In words, differentiating an integral with a variable upper limit just returns the integrand. This says that integration and differentiation are exact inverses. SACE questions occasionally use this to ask for the rate of change of an accumulated quantity, in which case the answer is simply the original rate function evaluated at the limit.

A worked accumulated-change problem

Total water from a varying flow

Worked example

Water flows into a tank at a rate r(t)=4+2tr(t)=4+2t litres per minute for 0≤t≤50\le t\le 5. Find the total volume added in the first five minutes.

The volume is the integral of the rate:

V=∫05(4+2t) dt=[4t+t2]05=(20+25)−0=45 litres.V=\int_0^5 (4+2t)\,dt=\Big[4t+t^2\Big]_0^5=(20+25)-0=45\text{ litres}.

The Fundamental Theorem turns the flow rate into accumulated volume in one antidifferentiate-and-subtract step.

Definite integrals on the calculator

In calculator-assumed sections SACE permits evaluating ∫abf(x) dx\int_a^b f(x)\,dx directly with a graphics or CAS calculator. The marks then sit in setting up the correct integral - the right integrand and the right limits - and in interpreting the number in context. For exact-value questions, though, you must show the antiderivative and the substitution by hand, because the calculator returns a decimal and an exact surd or logarithmic answer cannot be recovered from it. Read each question for the words "exact" or "to … decimal places" to decide which approach is required.

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 questions

Questions in the style of SACE Board 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.

SACE 2023-style3 marks
Calculator-assumed. Given that ∫xex dx=xex−ex+c\int x e^x\,dx = x e^x - e^x + c, find the exact value of the area bounded by f(x)=xexf(x) = x e^x, the xx-axis, and the vertical lines x=1x = 1 and x=5x = 5.
Show worked answer →

Since f(x)=xexf(x) = x e^x is positive on [1,5][1, 5], the area equals the definite integral, evaluated with the Fundamental Theorem using F(x)=xex−exF(x) = x e^x - e^x.

Area=∫15xex dx=[xex−ex]15=F(5)−F(1).\text{Area} = \int_1^5 x e^x\,dx = \Big[x e^x - e^x\Big]_1^5 = F(5) - F(1).

F(5)=5e5−e5=4e5F(5) = 5e^5 - e^5 = 4e^5 and F(1)=e−e=0F(1) = e - e = 0, so

Area=4e5−0=4e5.\text{Area} = 4e^5 - 0 = 4e^5.

Marks: one for applying F(b)−F(a)F(b) - F(a), one for substituting each limit, one for the exact answer 4e54e^5. Leaving it exact (not a decimal) is essential.

SACE 2024-style2 marks
Calculator-assumed. Vehicles enter a car park at a rate E(t)E(t) vehicles per hour, with the car park empty at t=0t = 0. A table gives ∫0tE(t) dt\int_0^t E(t)\,dt at various times, including the value 471471 at t=24t = 24. (a) State how many vehicles entered during the 24-hour period. (b) Explain what ∫824E(t) dt\int_8^{24} E(t)\,dt represents.
Show worked answer →

(a) The integral of a rate accumulates the total quantity. Since E(t)E(t) is the entry rate (vehicles per hour), ∫024E(t) dt\int_0^{24} E(t)\,dt is the total number that entered over the day. From the table this is 471471 vehicles. (1 mark)

(b) ∫824E(t) dt\int_8^{24} E(t)\,dt is the number of vehicles that entered between t=8t = 8 and t=24t = 24 hours, found by F(24)−F(8)F(24) - F(8) from the table. (1 mark)

This is the Fundamental Theorem idea that the definite integral of a rate gives net accumulated change over the interval.

SACE 2021-style2 marks
Calculator-free. Evaluate ∫02(3x2−4x+1) dx\int_0^2 (3x^2 - 4x + 1)\,dx.
Show worked answer →

Antidifferentiate to F(x)=x3−2x2+xF(x) = x^3 - 2x^2 + x, then apply the limits:

∫02(3x2−4x+1) dx=[x3−2x2+x]02=(8−8+2)−0=2.\int_0^2 (3x^2 - 4x + 1)\,dx = \Big[x^3 - 2x^2 + x\Big]_0^2 = (8 - 8 + 2) - 0 = 2.

Marks: one for the antiderivative, one for the correct evaluation F(2)−F(0)=2F(2) - F(0) = 2.

Practise this

Sources & how we know this

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

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