JEE Mains · Maths · STD 12 - 8. Application and integration
Let the area of the region \(\{(x, y): 0 \leq x \leq 3,0 \leq y \leq\) \(\left.\min \left\{x^2+2,2 x+2\right\}\right\}\) be \(A\). Then \(12 \mathrm{~A}\) is equal to
- A \(164\)
- B \(145\)
- C \(165\)
- D \(146\)
Answer & Solution
Correct Answer
(A) \(164\)
Step-by-step Solution
Detailed explanation
\( A=\int_0^2\left(x^2+2\right) d x+\int_2^3(2 x+2) d x \) \( A=\frac{41}{3} \) \( 12 A=41 \times 4=164\)
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