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

Question

Aicharya · Accepted Answer

$x^2+4 x+2 \leq y \leq|x+2|$

$\begin{aligned} & A_1=\int_{-4}^0\left[2-\left(x^2+4 x+2\right)\right] d x-\frac{1}{2} \times 4 \times 2 \\ & =\left(\frac{-x^3}{3}-2 x^2\right)_{-4}^0-4 \\ & =0-\left(\frac{64}{3}-32\right)-4 \\ & =32-\frac{64}{3}-4=\frac{20}{3}\end{aligned}$

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

Solution