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

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.

The answer

Standard antiderivatives

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

Linear inside argument

If the argument is linear, divide by the coefficient.

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

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.