Application of Integrals: Edge Cases and Subtle Traps (5)

Question

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

The trap: figuring out which curve is “on top” between the intersection points. Switch them by accident, and you’ll get a negative or zero answer.

Solution — Step by Step

Why This Works

The integral $\int_a^b [f(x) - g(x)] dx$ gives the signed area between $f$ and $g$. If $f \geq g$ throughout $[a, b]$, the result is the actual geometric area.

If the curves cross within $[a, b]$, you must split the integral at the crossing points and take absolute values of each piece — otherwise positive and negative parts cancel and you get a wrong answer.

Alternative Method

Switch to integrating with respect to $y$. The curves become $x = y$ and $x = \sqrt{y}$ (taking the positive branch). For $0 \leq y \leq 1$, $\sqrt{y} \geq y$, so:

$A = \int_0^1 (\sqrt{y} - y) \, dy = \left[\frac{2y^{3/2}}{3} - \frac{y^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \checkmark$

Common Mistake

The five most common traps in area-by-integration problems:

1. Wrong order: $\int (x^2 - x)\,dx$ instead of $\int (x - x^2)\,dx$. Gives a negative; students take the absolute value at the end and might luck out, but the cleaner habit is to identify “top minus bottom” first.

2. Forgetting to split at crossings: if the question is $\int_{-1}^{2}(x^3)\,dx$ “as area,” you must split at $x = 0$ since $x^3$ changes sign.

3. Using the curve equation instead of $|f - g|$: $\int_0^1 x^2\,dx$ is the area under $y = x^2$, not between $y = x^2$ and $y = x$.

4. Wrong integration limits: forgetting to find the intersection points first.

5. Mixing up $dx$ and $dy$ integrations: both work, but you have to set up the limits correctly for the chosen variable.

A solid sketch defeats all five.