Revise Definite integrals: areas and volumes of revolution for IB Mathematics: Applications and Interpretation (5.6)

The definite integral $\int_a^b f(x)\,dx$ gives the signed area of the region bounded by $y=f(x)$, the x-axis and the vertical lines $x=a$ and $x=b$. 'Signed' means any part of the region lying below the x-axis contributes a negative amount. When the curve stays above the axis on $[a,b]$, the definite integral is exactly the geometric area. When it dips below, and you want the total physical area, integrate the absolute value of the function instead.

Signed area: $A=\int_a^b f(x)\,dx$

Total (physical) area, allowing for parts below the axis: $A=\int_a^b |f(x)|\,dx$

On the GDC, both are one-line calculations: the numerical integration feature evaluates $\int_a^b f(x)\,dx$ directly, and wrapping the integrand in an absolute value handles a curve that changes sign. This is the standard method in AI HL — you rarely need to find where the curve crosses the axis by hand.

To find the area enclosed between two curves $y=f(x)$ and $y=g(x)$, integrate the difference between the upper curve and the lower curve across the interval where they enclose the region. Subtracting the lower function strips away the area beneath it, leaving precisely the area trapped between the two. The limits are usually the x-coordinates where the curves intersect.

Area between curves (with $f\ge g$ on $[a,b]$):

$A=\int_a^b\big(\text{upper}-\text{lower}\big)\,dx = \int_a^b\big(f(x)-g(x)\big)\,dx$

A solid of revolution is formed by rotating a region through $360^\circ$ about an axis. Rotate the region under $y=f(x)$ about the x-axis and each vertical slice sweeps out a thin disc: its radius is the height of the curve, $y$, so its area is $\pi y^2$ and its volume is $\pi y^2\,dx$. Adding up all the discs — that is, integrating — gives the total volume. The single most important habit here is to square $y$ before integrating.

Volume of revolution about the x-axis, $x=a$ to $x=b$:

$V=\pi\int_a^b y^2\,dx = \pi\int_a^b \big[f(x)\big]^2\,dx$

Substitution reverses the chain rule. When the integrand is a composite function multiplied by (a constant times) the derivative of its inside, let $u$ be that inside function. Replacing $\mathrm{d}x$ using $\mathrm{d}u=\frac{du}{dx}\,dx$ turns the integral into a simpler one in $u$. For a definite integral, the cleanest approach is to convert the limits to $u$-values as you go — then you never have to substitute back.

If $u=g(x)$ then $\mathrm{d}u=g'(x)\,dx$, and

$\displaystyle\int_a^b f\big(g(x)\big)\,g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du$

Your GDC is the workhorse of this topic. Its numerical (definite) integration feature evaluates $\int_a^b f(x)\,dx$ for any function you can type — no algebra required — which is exactly what areas and volumes reduce to. For an area between curves, integrate the difference (or its absolute value). For a volume about the x-axis, integrate $y^2$ and multiply the result by $\pi$. Two accuracy habits matter: find intersection limits with the graph or solver and STORE them at full precision rather than retyping rounded values, and restrict any search to a sensible domain, since lengths, areas and volumes must be positive.

In an exam the marks are analytic: each is tied to a specific line of working, using the IB Mathematics conventions. A method mark (M) rewards a correct approach even if the arithmetic later slips; an accuracy mark (A) rewards a correct result and is normally dependent on the method mark being earned. Follow-through (FT) means a correct final step carried out on your own earlier (even wrong) value still scores, and equivalent correct forms — including an exact answer such as $8\pi$ or its decimal $25.1$ — are accepted. On this technology-active paper the GDC is allowed throughout. All of that protection exists only if your method is on the page. Study how each mark below is earned by a specific line.

Every calculation here is one idea — the definite integral adds up infinitely many thin slices — dressed for three jobs: area under a curve, area between curves, and volume of revolution. Substitution simply lets you reverse composite functions so those integrals can be evaluated exactly, while the GDC handles anything too awkward for hand-integration. The same set-up-then-evaluate discipline carries straight into the modelling and kinematics questions that dominate Paper 2, where an integral of velocity gives distance and an integral of a rate gives a total. Master the habit — sketch, choose the right integral, square where a volume needs it, evaluate, and state units — and the rest of integral calculus becomes variations on a method you already own.