JEE Mains · Maths · STD 12 - 8. Application and integration
If the area of the region \(\left\{(x, y): 0 \leq y \leq \min \left\{2 x, 6 x-x^2\right\}\right\}\) is \(A\), then \(12 A\) is equal to
- A \(421\)
- B \(304\)
- C \(321\)
- D \(123\)
Answer & Solution
Correct Answer
(B) \(304\)
Step-by-step Solution
Detailed explanation
We have \(A=\frac{1}{2} \times 4 \times 8+\int_4^6\left(6 x-x^2\right) d x \) \(A=\frac{76}{3} \) \(12 A=304\)
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