Area Under a Curve — The Strip Method

Every area problem in this chapter is the same three-step routine: sketch the region, choose a strip, integrate the strip across the region.

Vertical strips (the default): for a curve $y = f(x)$ with $f(x) \geq 0$ between $x = a$ and $x = b$, a thin vertical strip at position $x$ has height $y$ and width $dx$, so $$\text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx$$

Horizontal strips (when the curve is given as $x$ in terms of $y$): for $x = g(y)$ between $y = c$ and $y = d$, the strip has length $x$ and thickness $dy$: $$\text{Area} = \int_c^d x\,dy = \int_c^d g(y)\,dy$$ Choose horizontal strips whenever the boundaries are lines $y = c$ and $y = d$ — fighting with vertical strips there costs time and marks.

Area Is Not the Same as the Integral

The definite integral counts region below the $x$-axis as negative; area is always positive. So when the curve crosses the axis inside $[a, b]$:

1. Find where $f(x) = 0$ inside the interval.
2. Split the integral at those points.
3. Take the modulus of each piece and add.

The classic warning: $\int_{-2}^{2} x^3\,dx = 0$, but the area between $y = x^3$ and the $x$-axis from $-2$ to $2$ is $2\int_0^2 x^3\,dx = 8$. Writing $0$ for an area is an instant zero for the question.

Symmetry saves work: even-symmetric regions can be computed on one side and doubled; a circle or ellipse is computed in the first quadrant and multiplied by $4$. State the symmetry in one line — examiners award the setup.

The Standard Regions (learn the results and their derivations)

Circle $x^2 + y^2 = a^2$: first-quadrant area $= \int_0^a \sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$, so the full circle has area $\pi a^2$.

Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: from $y = \frac{b}{a}\sqrt{a^2 - x^2}$, $$\text{Area} = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\,dx = 4\cdot\frac{b}{a}\cdot\frac{\pi a^2}{4} = \pi ab$$

Parabola $y^2 = 4ax$ up to its latus rectum $x = a$: by symmetry about the $x$-axis, $$\text{Area} = 2\int_0^a 2\sqrt{ax}\,dx = \frac{8a^2}{3}$$

The key antiderivative behind the circle and ellipse results is $\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ from the Integrals chapter — quote it, do not re-derive it in the exam.

Worked Examples — Strips and Sign Changes

Example 1 — A straight-line region

Find the area bounded by $y = 2x + 1$, the $x$-axis and the lines $x = 1$, $x = 3$.

Step 1 — check the sign: on $[1, 3]$, $y = 2x + 1 > 0$, so no splitting is needed.

Step 2 — integrate the strip: $\int_1^3 (2x + 1)\,dx = \left[x^2 + x\right]_1^3 = 12 - 2$.

Answer: $10$ square units — and since the region is a trapezium, the geometry formula $\frac{1}{2}(3 + 7)\cdot 2 = 10$ confirms it.

Example 2 — Under a parabola

Find the area under $y = x^2$ from $x = 0$ to $x = 2$.

Step 1 — the curve is non-negative: area $= \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2$.

Answer: $\dfrac{8}{3}$ square units.

Example 3 — A sideways parabola with vertical strips

Find the area of the region in the first quadrant bounded by $y^2 = x$, the $x$-axis and the lines $x = 1$ and $x = 4$.

Step 1 — solve for $y$ in the first quadrant: $y = \sqrt{x}$.

Step 2 — integrate: $\int_1^4 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{2}{3}(8 - 1)$.

Answer: $\dfrac{14}{3}$ square units.

Example 4 — Horizontal strips

Find the area of the region bounded by $x^2 = 4y$, the $y$-axis and the lines $y = 2$ and $y = 4$ in the first quadrant.

Step 1 — boundaries are horizontal lines, so use horizontal strips: $x = 2\sqrt{y}$.

Step 2 — integrate in $y$: $\int_2^4 2\sqrt{y}\,dy = \left[\frac{4}{3}y^{3/2}\right]_2^4 = \frac{4}{3}\left(8 - 2\sqrt{2}\right)$.

Answer: $\dfrac{32 - 8\sqrt{2}}{3}$ square units — choosing the strip to match the boundary lines is the whole trick.

Example 5 — The curve crosses the axis

Find the area bounded by $y = 3x + 2$, the $x$-axis and the lines $x = -1$ and $x = 1$.

Step 1 — find the crossing: $3x + 2 = 0$ at $x = -\frac{2}{3}$, which lies inside the interval — split there.

Step 2 — the negative piece: $\int_{-1}^{-2/3}(3x+2)\,dx = -\frac{1}{6}$, contributing area $\frac{1}{6}$.

Step 3 — the positive piece: $\int_{-2/3}^{1}(3x+2)\,dx = \frac{25}{6}$.

Answer: $\dfrac{1}{6} + \dfrac{25}{6} = \dfrac{13}{3}$ square units — integrating straight from $-1$ to $1$ gives $4$, which is the integral, not the area.

Example 6 — Area under a full sine arch and beyond

Find the area bounded by $y = \sin x$ and the $x$-axis from $x = 0$ to $x = 2\pi$.

Step 1 — locate the sign change: $\sin x \geq 0$ on $[0, \pi]$ and $\sin x \leq 0$ on $[\pi, 2\pi]$.

Step 2 — the two pieces: $\int_0^{\pi}\sin x\,dx = 2$ and $\left\vert\int_{\pi}^{2\pi}\sin x\,dx\right\vert = 2$.

Answer: $4$ square units — while $\int_0^{2\pi}\sin x\,dx = 0$. One curve, two different questions.

Worked Examples — Circles, Ellipses and Parabolas

Example 7 — Area of a circle, derived

Using integration, find the area of the circle $x^2 + y^2 = a^2$.

Step 1 — first quadrant + symmetry: area $= 4\int_0^a \sqrt{a^2 - x^2}\,dx$.

Step 2 — quote the standard antiderivative: $\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a = 0 + \frac{a^2}{2}\cdot\frac{\pi}{2}$.

Step 3 — multiply by 4: $4\cdot\frac{\pi a^2}{4}$.

Answer: $\pi a^2$ square units — a five-mark derivation that appears verbatim on board papers.

Example 8 — A quarter circle, numerically

Find the area of the region in the first quadrant enclosed by $x^2 + y^2 = 4$ and the coordinate axes.

Step 1 — set up: $\int_0^2\sqrt{4 - x^2}\,dx$ with $a = 2$.

Step 2 — evaluate: $\left[\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2}\right]_0^2 = 2\cdot\frac{\pi}{2}$.

Answer: $\pi$ square units — one quarter of the full circle's area $4\pi$, as it must be.

Example 9 — Area of an ellipse, derived

Using integration, find the area enclosed by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Step 1 — first-quadrant branch: $y = \dfrac{b}{a}\sqrt{a^2 - x^2}$.

Step 2 — symmetry and the standard integral: $$\text{Area} = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\,dx = \frac{4b}{a}\cdot\frac{\pi a^2}{4}$$

Answer: $\pi ab$ square units — setting $b = a$ recovers the circle's $\pi a^2$, a one-line sanity check worth writing.

Example 10 — Ellipse, numerically

Find the area of the region bounded by the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

Step 1 — identify $a = 4$, $b = 3$ and quote the result: area $= \pi ab$.

Answer: $12\pi$ square units. (If the derivation is demanded, reproduce Example 9 with these numbers.)

Example 11 — Parabola up to the latus rectum

Find the area bounded by the parabola $y^2 = 4ax$ and its latus rectum.

Step 1 — the latus rectum is the line $x = a$: by symmetry about the $x$-axis, area $= 2\int_0^a 2\sqrt{a}\sqrt{x}\,dx$.

Step 2 — integrate: $4\sqrt{a}\cdot\left[\frac{2}{3}x^{3/2}\right]_0^a = 4\sqrt{a}\cdot\frac{2}{3}a^{3/2}$.

Answer: $\dfrac{8a^2}{3}$ square units — for $y^2 = 8x$ (so $a = 2$) this is $\dfrac{32}{3}$.

Example 12 — Area versus integral, the cubic

Find the area bounded by $y = x^3$ and the $x$-axis between $x = -2$ and $x = 2$.

Step 1 — note the odd symmetry: the integral $\int_{-2}^{2}x^3\,dx = 0$, but the region has equal areas on both sides.

Step 2 — compute one side and double: $2\int_0^2 x^3\,dx = 2\cdot\frac{16}{4}$.

Answer: $8$ square units — parity kills the integral, never the area.

Example 13 — A cosine piece

Find the area under $y = \cos x$ from $x = 0$ to $x = \dfrac{\pi}{2}$.

Step 1 — cosine is non-negative on this interval: area $= \int_0^{\pi/2}\cos x\,dx = \left[\sin x\right]_0^{\pi/2}$.

Answer: $1$ square unit — always make the sign check before integrating; here it passes and the answer is one line.