JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region \(\left\{(x, y): x^2+4 x+2 \leq y \leq|x+2|\right\}\) is equal to
- A \(7\)
- B \(5\)
- C \(24 / 5\)
- D \(20 / 3\)
Answer & Solution
Correct Answer
(D) \(20 / 3\)
Step-by-step Solution
Detailed explanation
\(x^2+4 x+2 \leq y \leq|x+2|\) \(\begin{aligned} & A_1=\int_{-4}^0\left[2-\left(x^2+4 x+2\right)\right] d x-\frac{1}{2} \times 4 \times 2 \\ & =\left(\frac{-x^3}{3}-2 x^2\right)_{-4}^0-4 \\ & =0-\left(\frac{64}{3}-32\right)-4 \\ & =32-\frac{64}{3}-4=\frac{20}{3}\end{aligned}\)
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