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The Fundamental Theorem of Calculus: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“State and apply the Fundamental Theorem of Calculus to evaluate definite integrals and to differentiate integral functions”

8 min read

Quick answer

The Fundamental Theorem of Calculus links the two halves of calculus. To evaluate, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a) where F′=fF'=f. To differentiate an integral, ddx∫axf(t) dt=f(x)\dfrac{d}{dx}\int_a^x f(t)\,dt=f(x): differentiation and integration undo each other.

Jump to a section
  1. What this dot point is asking
  2. The two forms
  3. Evaluating a definite integral
  4. Differentiating an integral function
  5. Reverse problems with the evaluation form
  6. A note on notation
  7. Why the two processes are inverse
  8. Exam-style questions

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, integrals are Unit 3 (Topic 3.2), not Unit 4. This page sits in our Unit 4 folder for historical reasons; the content is Unit 3.

What this dot point is asking

SCSA Unit 4 emphasises the Fundamental Theorem of Calculus as the bridge between the definite integral (a limit of sums) and antidifferentiation. This dot point asks you to state and apply both forms, examined in both sections of the WACE written examination.

The two forms

The Fundamental Theorem of Calculus

Evaluation form. If FF is any antiderivative of ff, then

∫abf(x) dx=F(b)−F(a).\int_a^b f(x)\,dx = F(b)-F(a).

Derivative form. If g(x)=∫axf(t) dtg(x)=\displaystyle\int_a^x f(t)\,dt, then

g′(x)=f(x).g'(x)=f(x).

The evaluation form lets you compute an area without summing rectangles: find any antiderivative and subtract its values at the limits. The constant of integration cancels in F(b)−F(a)F(b)-F(a), so it is omitted in definite integrals. The derivative form says that integrating then differentiating returns the original function.

It does not matter which antiderivative you choose for the evaluation form. If F(x)F(x) and F(x)+5F(x)+5 are both antiderivatives of ff, then (F(b)+5)−(F(a)+5)=F(b)−F(a)\bigl(F(b)+5\bigr)-\bigl(F(a)+5\bigr)=F(b)-F(a): the added constant cancels in the subtraction. This is the formal reason the constant of integration is dropped in a definite integral, and it means you can always pick the simplest antiderivative without worrying about cc.

Evaluating a definite integral

Applying the evaluation form

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

Find an antiderivative.

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

Apply F(3)−F(1)F(3)-F(1).

∫13(2x+1) dx=(32+3)−(12+1)=12−2=10.\int_1^3 (2x+1)\,dx = (3^{2}+3) - (1^{2}+1) = 12 - 2 = 10.

Markers reward a correct antiderivative, clear substitution of both limits, and the final value. No constant of integration is needed because it cancels.

Differentiating an integral function

The derivative form is examined directly when a function is defined by an integral with a variable upper limit. Differentiating simply substitutes the upper limit into the integrand.

Applying the derivative form

Differentiate g(x)=∫2xt3+1 dtg(x)=\displaystyle\int_2^x \sqrt{t^{3}+1}\,dt

By the derivative form of the Fundamental Theorem, differentiating an integral with variable upper limit xx returns the integrand evaluated at xx.

g′(x)=x3+1.g'(x) = \sqrt{x^{3}+1}.

There is no need to evaluate the integral itself, which would be difficult. The lower limit 22 is constant and contributes nothing to the derivative. Markers reward recognising the theorem and substituting xx for tt.

Reverse problems with the evaluation form

The evaluation form also supports problems that give the value of a definite integral and ask for an unknown, such as a limit of integration or a constant in the integrand. The method is to carry out the integration in terms of the unknown, set the result equal to the given value, and solve the resulting equation. For instance, finding kk in ∫0k3x2 dx=8\int_0^k 3x^2\,dx=8 gives k3=8k^3=8, so k=2k=2. These reverse questions test whether you can apply the theorem flexibly rather than only computing a numerical area.

A note on notation

The vertical-bar notation [F(x)]ab\bigl[F(x)\bigr]_a^b is shorthand for F(b)−F(a)F(b)-F(a) and keeps the working tidy. Write the antiderivative inside the brackets, mark the limits, then substitute the upper limit first and subtract the lower. Keeping this order avoids the sign error of computing F(a)−F(b)F(a)-F(b), which negates the answer. When the integrand has several terms, antidifferentiate each term and apply the limits to the whole expression at once rather than term by term, which reduces arithmetic slips.

Why the two processes are inverse

The theorem shows that the area-accumulation function g(x)=∫axf(t) dtg(x)=\int_a^x f(t)\,dt has the original ff as its derivative, so accumulating area then measuring its rate of change recovers ff. Conversely, the evaluation form expresses the net accumulated area as the change in an antiderivative. This duality is the central result of single-variable calculus.

Common errors
Keeping the constant in a definite integral
The constant cancels in F(b)−F(a)F(b)-F(a); do not write +c+c in a definite integral.
Reversing F(b)−F(a)F(b)-F(a)
Substitute the upper limit first; computing F(a)−F(b)F(a)-F(b) negates the answer.
Trying to evaluate the integral in the derivative form
ddx∫axf(t) dt=f(x)\dfrac{d}{dx}\int_a^x f(t)\,dt=f(x) directly; you do not antidifferentiate first.
Mismatching the variable
In the derivative form replace the dummy variable tt with the upper limit xx in the integrand.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2021-style5 marks
Calculator-free. (a) Evaluate ∫0πsin⁡x dx\displaystyle\int_{0}^{\pi}\sin x\,dx. (b) Differentiate g(x)=∫1xln⁡(t2+1) dtg(x)=\displaystyle\int_{1}^{x}\ln(t^{2}+1)\,dt.
Show worked answer →

One part of each form of the theorem.

(a) An antiderivative of sin⁡x\sin x is −cos⁡x-\cos x. So [−cos⁡x]0π=−cos⁡π−(−cos⁡0)=−(−1)+1=2\bigl[-\cos x\bigr]_{0}^{\pi}=-\cos\pi-(-\cos 0)=-(-1)+1=2.

(b) By the derivative form, g′(x)=ln⁡(x2+1)g'(x)=\ln(x^{2}+1): substitute the upper limit xx for tt in the integrand; no integration is needed.

Markers reward the antiderivative with both limits in (a) and the direct application of the derivative form in (b).

WACE 2023-style5 marks
Calculator-assumed. (a) Evaluate ∫14(1x+2x)dx\displaystyle\int_{1}^{4}\left(\dfrac{1}{x}+2x\right)dx, giving an exact answer. (b) Find the value of kk for which ∫0k(2x−1) dx=6\displaystyle\int_{0}^{k}(2x-1)\,dx=6.
Show worked answer →

An evaluation and a reverse problem.

(a) Antiderivative ln⁡∣x∣+x2\ln|x|+x^{2}. So [ln⁡x+x2]14=(ln⁡4+16)−(ln⁡1+1)=ln⁡4+15\bigl[\ln x+x^{2}\bigr]_{1}^{4}=(\ln 4+16)-(\ln 1+1)=\ln 4+15.

(b) ∫0k(2x−1) dx=[x2−x]0k=k2−k\displaystyle\int_{0}^{k}(2x-1)\,dx=\bigl[x^{2}-x\bigr]_{0}^{k}=k^{2}-k. Set k2−k=6k^{2}-k=6, so k2−k−6=0k^{2}-k-6=0, (k−3)(k+2)=0(k-3)(k+2)=0. Taking the positive value, k=3k=3.

Markers reward the antiderivative including ln⁡x\ln x in (a), and forming and solving the quadratic in (b).

Practise this

Sources & how we know this

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

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