The area of the region \(x, y): x^2+4 x+2 \leq y \leq|x+2|\ is equal to

Area Under the Curves PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q40 with Solution

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Correct answer: C

We need the area of $\{(x,y): x^2+4x+2 \le y \le |x+2|\}.$

So the curve below is $y=x^2+4x+2=(x+2)^2-2,$ and the curve above is $y=|x+2|.$

Hence the required area is the area between these two curves where $x^2+4x+2 \le |x+2|.$

Let $t=x+2.$ Then $y=t^2-2$ and $y=|t|.$

So we need the area between $y=t^2-2 \quad \text{and} \quad y=|t|,$ where $t^2-2 \le |t|.$

Solve $t^2-2=|t|.$

Let $u=|t| \ge 0.$ Then $u^2-2=u$ $u^2-u-2=0$ $(u-2)(u+1)=0.$ Since $u\ge 0$ , we get $u=2.$ Thus $|t|=2 \implies t=\pm 2.$

So the bounded region lies for $-2 \le t \le 2.$

Area

=\int_{-2}^{2}(2+|t|-t^2)\,dt.$$ Since the integrand is even, $$\text{Area}=2\int_0^2 (2+t-t^2)\,dt.$$ --- 5. **Evaluate the integral** $$\int_0^2 (2+t-t^2)\,dt =\left[2t+\frac{t^2}{2}-\frac{t^3}{3}\right]_0^2.$$ At $t=2$, $$2(2)+\frac{2^2}{2}-\frac{2^3}{3} =4+2-\frac{8}{3} =6-\frac{8}{3} =\frac{10}{3}.$$ Therefore, $$\text{Area}=2\cdot \frac{10}{3}=\frac{20}{3}.$$ --- 6. **Match with options** $$\frac{20}{3}$$ corresponds to **Option C**. --- 7. **Compare with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So they agree.