Application of Integrals: Exam-Pattern Drill (4)

Question

Find the area enclosed between the curves $y = x^2$ and $y = 2x$.

Solution — Step by Step

Why This Works

The “area between curves” formula treats the region as infinitely many vertical strips of width $dx$ and height (top - bottom). Integrating sums them all up.

The key prerequisite: identify which curve is on top, and over what interval. If curves cross multiple times, split the integral.

Alternative Method — Horizontal Strips

We could integrate with respect to $y$ instead:

$x_\text{right} = \sqrt{y}$ (from parabola), $x_\text{left} = y/2$ (from line). Range $0 \le y \le 4$.

$A = \int_0^4 (\sqrt{y} - y/2) \, dy = \left[\frac{2y^{3/2}}{3} - \frac{y^2}{4}\right]_0^4 = \frac{16}{3} - 4 = 4/3$. Same answer.

Use horizontal strips when curves are easier to express as $x = f(y)$.

Common Mistake

Students often forget to identify which curve is on top and end up with a negative answer (which they may write as positive without explanation, losing rigour marks in boards).