Area of Curve – JEE Main 24 January 2025 Shift 1

Solution:

The area is bounded between the parabola $y = x^2+4x+2 = (x+2)^2 – 2$ and the modulus function $y = |x+2|$.
Let’s find the intersection points by setting $(x+2)^2 – 2 = |x+2|$.
Let $u = |x+2|$ (where $u \ge 0$).
$$ u^2 – 2 = u \Rightarrow u^2 – u – 2 = 0 $$
$$ (u-2)(u+1) = 0 $$
Since $u \ge 0$, we get $u = 2$.
$$ |x+2| = 2 \Rightarrow x+2 = \pm 2 \Rightarrow x = 0, -4 $$


The region is symmetric about the line $x = -2$. We can integrate from $-2$ to $0$ and multiply by 2.
In the interval $[-2, 0]$, $|x+2| = x+2$.
$$ \text{Area} = 2 \int_{-2}^{0} (\text{Upper Curve} – \text{Lower Curve}) dx $$
$$ \text{Area} = 2 \int_{-2}^{0} \left( (x+2) – (x^2+4x+2) \right) dx $$
Let $t = x+2$, then $dx = dt$. Limits change from $0$ to $2$.
$$ \text{Area} = 2 \int_{0}^{2} \left( t – (t^2-2) \right) dt $$
$$ = 2 \int_{0}^{2} (t – t^2 + 2) dt $$

Integrate:
$$ = 2 \left[ \frac{t^2}{2} – \frac{t^3}{3} + 2t \right]_{0}^{2} $$
$$ = 2 \left[ \left( \frac{4}{2} – \frac{8}{3} + 4 \right) – 0 \right] $$
$$ = 2 \left[ 2 – \frac{8}{3} + 4 \right] = 2 \left[ 6 – \frac{8}{3} \right] $$
$$ = 2 \left[ \frac{18-8}{3} \right] = 2 \left( \frac{10}{3} \right) = \frac{20}{3} $$

Ans. (3)