Revise Definite integrals: areas and volumes of revolution for IB Mathematics: Analysis and Approaches (5.6)

Approximate the area under a curve by drawing many thin vertical rectangles and summing their areas; make the rectangles infinitely thin and the sum becomes exact. That exact value is the definite integral $\int_a^b f(x)\,dx$, where $a$ and $b$ are the lower and upper limits. The remarkable fact — the fundamental theorem of calculus — is that you never have to add up slices by hand: if $F$ is any antiderivative of $f$, the whole sum collapses to a single subtraction.

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a), \quad \text{where } F'(x)=f(x)$.

When $f(x)\ge 0$ on $[a,b]$, the definite integral is exactly the area between the curve, the x-axis and the lines $x=a$ and $x=b$. The subtlety is sign. The integral measures 'signed' area: any part of the region lying below the x-axis contributes a NEGATIVE amount. So if a curve dips below the axis, integrating straight across lets the positive and negative parts cancel, and you get less than the true geometric area.

$\text{Area } A = \int_a^b f(x)\,dx \quad (f(x)\ge 0), \qquad \text{Total area} = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right| \text{ if the curve crosses at } x=c$.

The area of a region trapped between two curves $y=f(x)$ and $y=g(x)$ is the area under the top curve minus the area under the bottom curve. This collapses to a single integral of the difference. The crucial step is to identify which function is greater (the 'top' function) over the interval; a sketch or a test value settles it. This formula works even when both curves dip below the x-axis, because subtracting the lower from the upper always leaves a positive height.

$\text{Area } A = \int_a^b \big(\underbrace{f(x)}_{\text{top}} - \underbrace{g(x)}_{\text{bottom}}\big)\,dx, \qquad f(x)\ge g(x) \text{ on } [a,b]$.

Take the region under $y=f(x)$ between $x=a$ and $x=b$ and rotate it a full $360^\circ$ about the x-axis: it sweeps out a 3D solid of revolution. Slice the solid into thin discs perpendicular to the x-axis. Each disc is a short cylinder of radius $r=y=f(x)$ and thickness $dx$, with volume $\pi r^2\,dx = \pi y^2\,dx$. Integrating sums every disc from $a$ to $b$.

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

On the calculator paper the GDC can evaluate a definite integral directly and find intersection points numerically — ideal when the antiderivative is awkward. Still name the integral you are evaluating and quote the result to three significant figures. On the non-calculator Paper 1, however, you must integrate by hand and give an EXACT answer: a fraction such as $\tfrac{4}{3}$, a surd, or a multiple of $\pi$. A rounded decimal on Paper 1 typically forfeits the final accuracy mark, so keep everything exact until the very end.

On Paper 1 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark is dependent on the method mark it follows. Follow-through (FT) means a correct step built on an earlier slip can still score, and 'ignore subsequent working' (ISW) means a correct answer is not un-awarded if you then write something harmless afterwards. But that protection only exists if the method is written down. Study how each mark below is earned by a specific line in a classic 'area between two curves' question.

The definite integral is the bridge from antiderivatives to measurable quantities. The same $F(b)-F(a)$ idea reappears whenever you accumulate a rate over an interval — a distance from a velocity, a change from a rate of change — which is why kinematics problems in this course lean on exactly these integrals. Master the habit here: find the limits, integrate the right expression, watch the sign of regions below the axis, square before you integrate for volumes, and keep answers exact on Paper 1. The applications that follow are variations on a method you already own.