What is properties worth quoting?

Two properties simplify many definite integrals: _a^b f(x)\,dx = -_b^a f(x)\,dx (reversing the limits flips the sign), and _a^b f + _b^c f = _a^c f (adjacent intervals combine). For an even function _-a^a f = 2_0^a f, and for an odd function _-a^a f = 0. Quoting the odd-function rule on a symmetric interval can turn a page of working into a one-line "the integral is 0 by symmetry", which markers accept and which removes the chance of an arithmetic slip.

NESA 2022 HSC Q13-style practice: Evaluate _0^2 (3 x^2 + 2 x) \, dx.

Find the antiderivative. int (3 x^2 + 2 x) \, dx = x^3 + x^2 + C. Apply the Fundamental Theorem of Calculus. _0^2 (3 x^2 + 2 x) \, dx = [x^3 + x^2]_0^2 = (8 + 4) - (0 + 0) = 12. Markers reward the explicit antiderivative, correct bracket notation, and the evaluated answer.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do we find antiderivatives and use the Fundamental Theorem of Calculus to evaluate definite integrals?

Find antiderivatives of standard functions, apply integration by substitution and evaluate definite integrals using the Fundamental Theorem of Calculus

A focused answer to the HSC Maths Advanced dot point on integration. Antiderivatives of standard functions, integration by substitution, definite integrals and the Fundamental Theorem of Calculus, with worked examples.

Generated by Claude Opus 4.816 min answerUpdated 2026-06-20

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What this dot point is asking

NESA wants you to find antiderivatives of standard functions, apply the reverse chain rule via substitution, and evaluate definite integrals using the Fundamental Theorem of Calculus (FTC). Integration underlies areas, volumes, motion problems and growth models.

Note

Differentiating tells you how fast something is changing; integrating runs the film backwards, adding up all those tiny changes to recover the total. Picture a car's speedometer: if you know how fast the car is going at every instant, integrating that speed over a trip gives the total distance travelled, because you are summing up all the little bits of distance covered second by second. Antiderivatives are just "what function had this rate of change?", so integrating is undoing a derivative. The Fundamental Theorem of Calculus is the shortcut that saves you from adding up millions of thin slices by hand: find the total-so-far function, read it at the start and the end, and subtract. A definite integral is really the area trapped between a curve and the horizontal axis, where anything below the axis counts as a negative amount.

The answer

Standard antiderivatives

Memorise these. The constant $C$ is omitted in definite integrals.

\int x^n \, dx = \frac{x^{n + 1}}{n + 1} + C \quad (n \neq -1)

\int \frac{1}{x} \, dx = \ln |x| + C

\int e^x \, dx = e^x + C, \qquad \int e^{kx} \, dx = \frac{e^{kx}}{k} + C

\int \sin x \, dx = -\cos x + C, \qquad \int \cos x \, dx = \sin x + C

\int \sec^2 x \, dx = \tan x + C

Linear inside argument

If the argument is linear, divide by the coefficient.

\int \sin(k x) \, dx = -\frac{1}{k} \cos(k x) + C

\int (a x + b)^n \, dx = \frac{(a x + b)^{n + 1}}{a (n + 1)} + C \quad (n \neq -1)

Integration by substitution

The substitution rule reverses the chain rule. To evaluate $\int f(g(x)) g'(x) \, dx$ , set $u = g(x)$ , so $du = g'(x) \, dx$ . The integral becomes $\int f(u) \, du$ , which you evaluate, then substitute back.

For a definite integral, you can either substitute back to $x$ and use the original limits, or change the limits to values of $u$ and skip the back-substitution.

The Fundamental Theorem of Calculus

If $F$ is any antiderivative of $f$ (so $F'(x) = f(x)$ ), then

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

The word any is the part textbooks rush past, and it is what makes the theorem usable. Two antiderivatives of the same function differ only by a constant, and that constant cancels in $F(b) - F(a)$ , so you never need the "right" antiderivative, just a convenient one, and you never need the $+ C$ for a definite integral. Notice too what the theorem buys you: an integral is defined as a limit of sums of thin strips, an object that looks impossible to compute, yet the FTC says you can get it exactly by evaluating a single function at two points. That shortcut is the whole reason calculus is practical.

A second statement: the function $G(x) = \int_a^x f(t) \, dt$ , the "area so far" as the upper limit slides, satisfies $G'(x) = f(x)$ . In short, differentiation and integration are inverse operations: integrating then differentiating gets you back where you started.

Choosing a substitution

The whole skill of substitution is recognising that the integrand contains a function and (a multiple of) its derivative. Scan the integrand for an inner function $u = g(x)$ whose derivative $g'(x)$ also appears, possibly up to a constant factor. Good signals include a power of a bracket multiplied by the bracket's derivative, a function inside a root, or a fraction whose numerator is the derivative of the denominator. If no such pair appears, substitution will not help and you should look for a standard form instead.

A clean substitution layout, written out in full, is what markers reward: state $u = g(x)$ , compute $\frac{du}{dx}$ , rearrange to replace $dx$ , rewrite the entire integral in terms of $u$ (including the limits for a definite integral), integrate, and either substitute back or evaluate the $u$ -limits.

The definite integral as signed area

Geometrically, $\int_a^b f(x)\,dx$ is the signed area between the curve and the $x$ -axis: regions above the axis count positive, regions below count negative. This is why a definite integral can be zero even when the curve is non-zero, as when an odd function is integrated over a symmetric interval. To find a genuine geometric area where the curve crosses the axis, split the integral at the crossing points and add the magnitudes, exactly as you split a motion problem at the times when velocity is zero.

The definite integral as signed area, stage by stage

The single most useful picture in this whole topic is the definite integral as a signed area. Below it is built up one stage at a time on a curve that is above the axis on the left part of the interval and below it on the right, so you can see exactly where the sign comes from.

Stage 1, mark the curve and the limits. Draw $y = f(x)$ and the interval you are integrating over, from $x = a$ to $x = b$ . Nothing is shaded yet; this is just the geometry the integral measures.

Stage 2, shade the part above the axis. Where the curve lies above the $x$ -axis (here from $a$ to the crossing point), the region between the curve and the axis contributes a positive amount to the integral, equal to its ordinary geometric area.

Stage 3, shade the part below the axis. Where the curve dips below the axis (from the crossing point to $b$ ), the region contributes a negative amount: the integral subtracts that area rather than adding it. It is drawn in a second tone to keep the two contributions distinct.

Stage 4, add the signed pieces. The definite integral $\int_a^b f(x)\,dx$ is the positive area minus the negative area. The crossing point, where $f = 0$ , is marked because it is exactly where the sign of the contribution flips, and it is where you must split the integral if a question asks for the geometric (always-positive) area instead.

Properties worth quoting

Two properties simplify many definite integrals: $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ (reversing the limits flips the sign), and $\int_a^b f + \int_b^c f = \int_a^c f$ (adjacent intervals combine). For an even function $\int_{-a}^{a} f = 2\int_0^a f$ , and for an odd function $\int_{-a}^{a} f = 0$ . Quoting the odd-function rule on a symmetric interval can turn a page of working into a one-line "the integral is $0$ by symmetry", which markers accept and which removes the chance of an arithmetic slip.

How exam questions ask about integration

The wording tells you which tool to reach for. Learn to translate the phrasing into a method:

"Find / evaluate $\int \ldots \, dx$ " (no limits). An indefinite integral: produce the antiderivative and you must write $+ C$ . Drop it and you lose the final mark.
"Evaluate $\int_a^b \ldots \, dx$ " (with limits). A definite integral: find the antiderivative, write it in square brackets $[\,]_a^b$ , substitute the top limit minus the bottom limit. No $+ C$ .
"Using the substitution $u = \ldots$ ". The substitution is handed to you; show $\frac{du}{dx}$ , replace $dx$ and (for limits) convert them to $u$ , integrate, then convert back or evaluate. The marks are for the visible transformation, not just the answer.
"Find the area bounded by / enclosed by / between." This is a geometric area, so it must be positive. Sketch first, find where the curve meets the axis or where two curves intersect, split at those points, and integrate "upper minus lower" (or take $\int |f|$ ), taking the magnitude of any piece below the axis.
"Show that $\frac{d}{dx}(\ldots) = \ldots$ , hence find $\int \ldots \, dx$ ". The "hence" is a gift: the differentiation result you just proved is the antiderivative you need, so quote it and reverse it rather than integrating from scratch.
"Find $\frac{d}{dx} \int_a^x f(t)\,dt$ ". A direct test of the second form of the FTC: the answer is just $f(x)$ , with no integration required.

The single most common silent error is treating "evaluate the integral" and "find the area" as the same task. They agree only when the curve never dips below the axis on the interval; the moment it does, the integral and the area differ, and the question wording decides which the marker wants.

Worked example

This example moves from the most direct integral to the most subtle: a standard antiderivative, a linear-inside-argument case, substitution, a definite integral with changed limits, and finally a direct use of the Fundamental Theorem.

Integrate a standard antiderivative

The integrand is a sum of powers of $x$ , so we integrate term by term with the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ , raising each index by one and dividing, and add a single constant at the end:

\int (5 x^3 + 4 x - 7) \, dx = \frac{5 x^4}{4} + 2 x^2 - 7 x + C.

Integrate functions with a linear inside argument

When the inside of the function is linear, integrating introduces a factor of the inner coefficient, so we divide by that coefficient to compensate. For $e^{3x}$ we divide by $3$ , and for $(2x+1)^4$ we divide by the coefficient $2$ as well as the new power $5$ :

\int e^{3 x} \, dx = \frac{e^{3 x}}{3} + C, \qquad \int (2 x + 1)^{4} \, dx = \frac{(2 x + 1)^{5}}{10} + C.

Evaluate $\int x \sqrt{x^2 + 4} \, dx$ by substitution

Evaluate $\int x \sqrt{x^2 + 4} \, dx$ .

The integrand contains the inner function $x^2 + 4$ together with (a multiple of) its derivative $2x$ , which is the signal to substitute. Setting $u = x^2 + 4$ turns the messy product into a plain power of $u$ :

u = x^2 + 4 \;\Rightarrow\; du = 2 x \, dx \;\Rightarrow\; x \, dx = \tfrac{1}{2} \, du.

Rewriting in $u$ , integrating, and substituting back gives the answer in terms of $x$ :

\int x \sqrt{x^2 + 4} \, dx = \int \sqrt{u} \cdot \tfrac{1}{2} \, du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 4)^{3/2} + C.

Evaluate $\int_0^1 2 x e^{x^2} \, dx$ with changed limits

Evaluate $\int_0^1 2 x e^{x^2} \, dx$ .

We substitute $u = x^2$ , and because this is a definite integral we convert the limits to $u$ values instead of substituting back, which avoids rewriting the antiderivative in $x$ . The $x$ -limits $0$ and $1$ become the $u$ -limits $0$ and $1$ :

u = x^2 \;\Rightarrow\; du = 2x\,dx, \qquad x = 0 \Rightarrow u = 0, \quad x = 1 \Rightarrow u = 1.

The integral collapses to a standard exponential, evaluated with the new limits:

\int_0^1 2 x e^{x^2} \, dx = \int_0^1 e^u \, du = [e^u]_0^1 = e - 1.

Differentiate an integral using the FTC in reverse

The second form of the Fundamental Theorem says differentiating an integral with a variable upper limit just returns the integrand evaluated at that limit. So no antiderivative of $\cos(t^2)$ is needed; we read the answer straight off:

G(x) = \int_1^x \cos(t^2) \, dt \;\Rightarrow\; G'(x) = \cos(x^2).

Signed integral versus geometric area

Here is the trap the stage-by-stage diagram warns about. Take $\int_0^2 (x^2 - 1)\,dx$ , where the curve is below the axis on $[0, 1]$ and above it on $[1, 2]$ . The plain integral mixes the signs:

\int_0^2 (x^2 - 1)\,dx = \left[\frac{x^3}{3} - x\right]_0^2 = \left(\frac{8}{3} - 2\right) - 0 = \frac{2}{3}.

If the question instead asks for the area enclosed by the curve and the axis on $[0, 2]$ , split at the crossing point $x = 1$ and take the magnitude of each piece. The piece on $[0, 1]$ is $\left[\frac{x^3}{3} - x\right]_0^1 = \frac{1}{3} - 1 = -\frac{2}{3}$ , so its area is $\frac{2}{3}$ ; the piece on $[1, 2]$ is $\left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$ .

\text{Area} = \left|-\tfrac{2}{3}\right| + \tfrac{4}{3} = \tfrac{2}{3} + \tfrac{4}{3} = 2 \text{ square units}.

The integral was $\frac{2}{3}$ but the geometric area is $2$ . They differ precisely because the negative piece partly cancelled the positive piece in the plain integral, which is the whole point of stages 2 to 4 above.

Common traps

Forgetting the $+ C$: An indefinite integral must include the constant of integration.
Dividing by the wrong coefficient: $\int e^{2 x} \, dx = \frac{1}{2} e^{2 x} + C$ , not $2 e^{2 x} + C$ .
Substituting without changing $dx$: When you substitute $u = g(x)$ , the $dx$ must become $\frac{du}{g'(x)}$ (or the integrand must already contain a multiple of $g'(x) \, dx$ ).
Mixing limits and the variable: After a substitution, either change the limits to $u$ values or substitute back to $x$ before evaluating. Do not mix the two.
Using the FTC with $|x|$ inside a logarithm wrongly: $\int \frac{1}{x} \, dx = \ln |x| + C$ . For a definite integral, the interval must not cross $x = 0$ .

Exam technique

Write the antiderivative explicitly inside square brackets, $[F(x)]_a^b$ , before you substitute; markers award a method mark for the correct antiderivative even if the final arithmetic slips. For an indefinite integral, the $+ C$ is worth a mark on its own, so never omit it. When a substitution is named in the question, show every transformed piece (the $\frac{du}{dx}$ line, the new $dx$ , the converted limits) because that visible working is what is being marked, not the number. For an "area" question, sketch the curve, mark where it crosses the axis, and split there before integrating; the difference between the signed integral and the geometric area is the single most common place marks are dropped. Leave volume and area answers in exact form (keep $\pi$ , fractions and surds) unless a decimal is requested.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC Q133 marksEvaluate

\int_0^2 (3 x^2 + 2 x) \, dx

Show worked answer →

Find the antiderivative.

$\int (3 x^2 + 2 x) \, dx = x^3 + x^2 + C$ .

Apply the Fundamental Theorem of Calculus.

$\int_0^2 (3 x^2 + 2 x) \, dx = [x^3 + x^2]_0^2 = (8 + 4) - (0 + 0) = 12$ .

Markers reward the explicit antiderivative, correct bracket notation, and the evaluated answer.

2020 HSC Q143 marksUse the substitution

u = x^2 + 1

to evaluate

\int 2 x (x^2 + 1)^4 \, dx

Show worked answer →

With $u = x^2 + 1$ , $\frac{du}{dx} = 2 x$ , so $du = 2 x \, dx$ .

The integral becomes $\int u^4 \, du = \frac{u^5}{5} + C$ .

Substitute back: $\int 2 x (x^2 + 1)^4 \, dx = \frac{(x^2 + 1)^5}{5} + C$ .

Markers expect a clear statement of the substitution, transformation of both the integrand and the differential, the antiderivative in $u$ , and the substitution back into $x$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind

\int (6x^2 - 4x + 3) \, dx

Show worked solution →

Integrate term by term with the power rule $\int x^n \, dx = \dfrac{x^{n+1}}{n+1}$ :

\int (6x^2 - 4x + 3) \, dx = \frac{6x^3}{3} - \frac{4x^2}{2} + 3x + C = 2x^3 - 2x^2 + 3x + C.

Marker's note: one mark for correctly integrating each power, one for the fully simplified expression with the $+ C$ . Omitting the constant of integration on an indefinite integral loses the second mark.

foundation3 marksEvaluate

\int_1^3 \dfrac{4}{x} \, dx

, giving your answer in exact form.

Show worked solution →

Use the standard antiderivative $\int \dfrac{1}{x} \, dx = \ln|x| + C$ , so the $4$ comes out as a factor:

\int_1^3 \frac{4}{x} \, dx = \left[\, 4\ln|x| \,\right]_1^3 = 4\ln 3 - 4\ln 1.

Simplify using $\ln 1 = 0$ :

4\ln 3 - 4(0) = 4\ln 3.

Marker's note: one mark for the antiderivative $4\ln|x|$ , one for correct substitution of the limits in $[\;]_1^3$ form, one for the exact answer $4\ln 3$ (using $\ln 1 = 0$ ). A decimal without the exact form is penalised when exact form is asked for.

foundation2 marksFind

\int (2\cos x - e^{2x}) \, dx

Show worked solution →

Integrate each term. The antiderivative of $\cos x$ is $\sin x$ , and for $e^{2x}$ the linear inner coefficient $2$ means we divide by $2$ :

\int (2\cos x - e^{2x}) \, dx = 2\sin x - \frac{1}{2}e^{2x} + C.

Marker's note: one mark for $2\sin x$ , one for $-\tfrac{1}{2}e^{2x} + C$ (dividing by the coefficient $2$ ). Writing $-2e^{2x}$ instead of $-\tfrac{1}{2}e^{2x}$ is the standard slip.

core3 marksUse the substitution

u = x^3 + 1

to find

\int 3x^2 (x^3 + 1)^5 \, dx

Show worked solution →

Set up the substitution. With $u = x^3 + 1$ ,

\frac{du}{dx} = 3x^2 \quad\Rightarrow\quad du = 3x^2 \, dx.

Rewrite the whole integral in $u$ . The factor $3x^2 \, dx$ is exactly $du$ :

\int 3x^2 (x^3 + 1)^5 \, dx = \int u^5 \, du = \frac{u^6}{6} + C.

Substitute back to $x$ :

\int 3x^2 (x^3 + 1)^5 \, dx = \frac{(x^3 + 1)^6}{6} + C.

Marker's note: one mark for $du = 3x^2 \, dx$ , one for the integral $\dfrac{u^6}{6} + C$ in $u$ , one for the back-substitution to $\dfrac{(x^3+1)^6}{6} + C$ . Leaving the answer in $u$ , or dropping the $+ C$ , forfeits the final mark.

core4 marksA particle moves in a straight line so that its velocity, in metres per second, is

v(t) = 3t^2 - 12t + 9

for

0 \le t \le 4

seconds. A sketch of

v

against

t

shows a parabola that opens upwards, cuts the time axis at

t = 1

and

t = 3

, is positive on

0 \le t < 1

and

3 < t \le 4

, and is negative on

1 < t < 3

. (a) Find the displacement of the particle over the interval

0 \le t \le 4

. (b) Find the total distance travelled over the same interval.

Show worked solution →

Displacement is the signed integral of velocity; distance is the integral of speed, so the sign changes at $t = 1$ and $t = 3$ matter for part (b).

Part (a): displacement is $\int_0^4 v \, dt$ .

\int_0^4 (3t^2 - 12t + 9) \, dt = \left[\, t^3 - 6t^2 + 9t \,\right]_0^4 = (64 - 96 + 36) - 0 = 4 \text{ m}.

Part (b): split at the sign changes and take magnitudes. Let $s(t) = t^3 - 6t^2 + 9t$ .

\int_0^1 v \, dt = s(1) - s(0) = 4 - 0 = 4, \qquad \int_1^3 v \, dt = s(3) - s(1) = 0 - 4 = -4,

\int_3^4 v \, dt = s(4) - s(3) = 4 - 0 = 4.

Adding the magnitudes:

\text{distance} = |4| + |-4| + |4| = 12 \text{ m}.

Marker's note: one mark for the antiderivative $t^3 - 6t^2 + 9t$ , one for the displacement $4$ m in (a); one for splitting at $t = 1$ and $t = 3$ and one for the total distance $12$ m in (b). Reporting the displacement $4$ m as the distance (not splitting at the sign changes) is the trap the graph is drawn to expose.

exam5 marksConsider the curves

y = x^2

and

y = 2x

. (a) Find the

x

-coordinates of the points where the curves intersect. (b) Sketch is not required: state which curve is the upper boundary on the interval between the intersection points, justifying your choice with a test value. (c) Hence find the exact area of the region enclosed between the two curves.

Show worked solution →

Part (a): set the curves equal.

x^2 = 2x \quad\Rightarrow\quad x^2 - 2x = 0 \quad\Rightarrow\quad x(x - 2) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 2.

Part (b): decide the upper curve with a test point. Take $x = 1$ (between $0$ and $2$ ):

y = 2x = 2(1) = 2, \qquad y = x^2 = 1^2 = 1.

Since $2 > 1$ , the line $y = 2x$ lies above the parabola $y = x^2$ on $0 < x < 2$ .

Part (c): integrate (upper minus lower) across the interval.

A = \int_0^2 \big(2x - x^2\big) \, dx = \left[\, x^2 - \frac{x^3}{3} \,\right]_0^2.

Substitute the limits:

A = \left(4 - \frac{8}{3}\right) - 0 = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \text{ square units}.

Marker's note: one mark for both intersection points in (a), one for identifying $y = 2x$ as the upper curve with a test value in (b), one for the "upper minus lower" integrand $2x - x^2$ , one for the antiderivative with the limits substituted, one for the exact area $\tfrac{4}{3}$ square units. Integrating "lower minus upper" gives $-\tfrac{4}{3}$ ; an area must be positive, so the sign matters.

exam5 marksLet

f(x) = x \ln x

for

x > 0

. (a) Show that

\dfrac{d}{dx}\big(x \ln x\big) = \ln x + 1

. (b) Hence find

\int \ln x \, dx

. (c) Hence evaluate

\int_1^e \ln x \, dx

, giving an exact answer.

Show worked solution →

Part (a): differentiate the product $x \ln x$ . Using the product rule with $u = x$ and $v = \ln x$ :

\frac{d}{dx}\big(x \ln x\big) = (1)(\ln x) + (x)\left(\frac{1}{x}\right) = \ln x + 1.

Part (b): reverse the result to integrate $\ln x$ . Integrating both sides of $\dfrac{d}{dx}(x\ln x) = \ln x + 1$ gives

\int (\ln x + 1) \, dx = x \ln x + C.

Split the left side and move the $\int 1 \, dx = x$ term across:

\int \ln x \, dx + x = x \ln x + C \quad\Rightarrow\quad \int \ln x \, dx = x \ln x - x + C.

Part (c): apply the Fundamental Theorem of Calculus.

\int_1^e \ln x \, dx = \left[\, x \ln x - x \,\right]_1^e = \big(e \ln e - e\big) - \big(1 \cdot \ln 1 - 1\big).

Using $\ln e = 1$ and $\ln 1 = 0$ :

= (e - e) - (0 - 1) = 0 - (-1) = 1.

Marker's note: one mark for the product-rule differentiation in (a), one for reversing it and one for the rearranged $\int \ln x \, dx = x\ln x - x + C$ in (b), one for substituting the limits in $[\;]_1^e$ form and one for the exact value $1$ in (c). Forgetting the $-x$ term in the antiderivative gives the wrong final value of $e - e = 0$ .