Find the Area of the Region Bounded by S = {(x, y) : x² + 4x + 2 ≤ y ≤ |x + 2|}

A standard area-under-curves problem, not tied to a specific exam paper.

Problem

Find the area of the region bounded by $S=\{(x,y) : x^2+4x+2 \leq y \leq |x+2|\}$.

Key insight. Both curves are functions of $x+2$ in disguise — the parabola completes to $(x+2)^2-2$, and the boundary is literally $|x+2|$. Substituting $u=x+2$ turns the problem into finding the area between $y=u^2-2$ and $y=|u|$, which is symmetric about $u=0$ since both are even functions of $u$.

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Approach

Complete the square on the parabola to reveal it as a shifted curve in $u=x+2$. Rewrite the boundary as $y=|u|$ too, and note both curves are even functions of $u$, so the enclosed region is symmetric about $u=0$. Find where the curves intersect, then integrate the difference (upper minus lower) using that symmetry to halve the work.

Solution

Step 1 — Complete the square on the parabola

$$x^2+4x+2 = (x+2)^2-2$$

Step 2 — Substitute u = x + 2

The region becomes $u^2-2 \leq y \leq |u|$, where the lower boundary is $y=u^2-2$ and the upper boundary is $y=|u|$ — both even functions of $u$, so the region is symmetric about $u=0$.

Step 3 — Find the intersection points for u ≥ 0

For $u\geq0$, $|u|=u$, so set $u=u^2-2$:

$$u^2-u-2=0 \implies (u-2)(u+1)=0 \implies u=2 \text{ or } u=-1$$

Since this case assumes $u\geq0$, only $u=2$ is valid here.

Step 4 — Confirm the intersection at u = −2 by symmetry

Since both curves are even in $u$, the mirror intersection at $u=-2$ follows automatically (or can be verified directly using $|u|=-u$ for $u<0$, giving $-u=u^2-2 \implies u^2+u-2=0 \implies (u+2)(u-1)=0$, valid root $u=-2$).

Step 5 — Set up the area integral using symmetry

$$\text{Area} = \int_{-2}^{2}\left[|u|-(u^2-2)\right]du = 2\int_{0}^{2}\left[u-(u^2-2)\right]du$$

Step 6 — Integrate

$$= 2\int_0^2\left(u-u^2+2\right)du = 2\left[\frac{u^2}{2}-\frac{u^3}{3}+2u\right]_0^2$$

Step 7 — Evaluate at the limits

$$= 2\left[\left(\frac{4}{2}-\frac{8}{3}+4\right)-0\right] = 2\left[2-\frac83+4\right] = 2\left[6-\frac83\right] = 2\times\frac{10}{3} = \frac{20}{3}$$

Answer

Common mistakes

• Trying to sketch and integrate the original curves in terms of $x$ directly, without the $u=x+2$ substitution. This works but requires tracking a vertex at $x=-2$ and a shifted absolute-value corner at the same point, which is easy to get wrong — the substitution makes both curves centre on the origin, exposing the symmetry immediately.
• Integrating over the full range $[-2,2]$ without using symmetry. Since $|u|-(u^2-2)$ is an even function of $u$, computing $2\int_0^2[\ldots]du$ is both faster and less error-prone than integrating $\int_{-2}^{2}$ directly (though both give the same answer).

Practise next

• Find the area of the region bounded by $S=\{(x,y) : x^2-2x-1 \leq y \leq |x-1|\}$, using the same complete-the-square substitution.