Application of Integrals — chapter notes

If $f$ is continuous and $f(x) \ge 0$ on $[a, b]$, the area of the region bounded by $y = f(x)$, the $x$-axis and the lines $x = a$, $x = b$ is

Think of the region as many thin vertical strips of height $y$ and width $dx$. Example: under $y = 2x + 1$ from $x = 1$ to $x = 3$ the area is $\left[x^2 + x\right]_1^3 = 10$ sq units, which is also the area of the trapezium with parallel sides $3$ and $7$ and width $2$.

If $f(x) \le 0$ on $[a, b]$, the integral is negative and the area is $\left|\displaystyle\int_a^b f(x)\,dx\right|$.

If the curve crosses the axis at $x = c$ inside $[a, b]$, split there and add absolute values:

Example: $y = 2x - 4$ on $[0, 4]$ crosses at $x = 2$; the parts are $4$ and $4$, so the area is $8$, although $\displaystyle\int_0^4 (2x - 4)\,dx = 0$.

For the region between a curve $x = g(y) \ge 0$, the $y$-axis and the lines $y = c$, $y = d$:

Example: the region in the first quadrant beside $y = x^2$ from $y = 1$ to $y = 4$ has $x = \sqrt{y}$ and area $\dfrac23(8 - 1) = \dfrac{14}{3}$ sq units. The limits are $y$-values.

• $x^2 = ky$ ($k > 0$): $y = \dfrac{x^2}{k}$, a U symmetric about the $y$-axis. Area under it from $0$ to $h$: $\dfrac{h^3}{3k}$.
• $y^2 = kx$ ($k > 0$): branches $y = \pm\sqrt{kx}$, symmetric about the $x$-axis. The region cut off by $x = h$ has area $2\displaystyle\int_0^h \sqrt{kx}\,dx = \dfrac43\sqrt{k}\,h^{3/2}$.

Example: $y^2 = 9x$ and $x = 4$: $2\displaystyle\int_0^4 3\sqrt{x}\,dx = 32$ sq units. This equals two-thirds of the enclosing rectangle ($4 \times 12$), a useful check.

Upper half of $x^2 + y^2 = a^2$: $y = \sqrt{a^2 - x^2}$. The key integral is

Quarter circle: $\displaystyle\int_0^a \sqrt{a^2 - x^2}\,dx = \dfrac{\pi a^2}{4}$, so the circle encloses $\pi a^2$. A piece cut off by $x = h$ ($0 < h < a$) has area $2\displaystyle\int_h^a \sqrt{a^2 - x^2}\,dx$; for $a = 4$, $h = 2$ this is $\dfrac{16\pi}{3} - 4\sqrt{3}$.

From $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$: upper half $y = \dfrac ba\sqrt{a^2 - x^2}$. By symmetry,

Convert to standard form first: $9x^2 + 25y^2 = 225$ becomes $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$, so the area is $15\pi$. A strip of an ellipse uses the same standard integral, e.g. $\dfrac34\displaystyle\int_0^2 \sqrt{16 - x^2}\,dx = \dfrac{3\sqrt{3}}{2} + \pi$.

1. Sketch and shade the region; mark where the curve meets the bounding lines.
2. Choose vertical strips ($dx$) or horizontal strips ($dy$).
3. Read the limits; note any axis crossing.
4. Use symmetry (factor $2$ or $4$).
5. Integrate, take absolute values, write sq units.

Not in the 2026–27 syllabus: the area between two curves (for example between a parabola and a slanting line, or between two circles). Every region in this chapter is bounded by one curve together with the axes or lines parallel to them.