JEE Mains · Maths · STD 12 - 8. Application and integration
If the area of the region \(\left\{(\mathrm{x}, \mathrm{y}): 1+\mathrm{x}^2 \leq \mathrm{y} \leq \min \{\mathrm{x}+7,11-3 \mathrm{x}\}\right\}\) is A , then 3 A is equal to
- A 50
- B 49
- C 46
- D 47
Answer & Solution
Correct Answer
(A) 50
Step-by-step Solution
Detailed explanation
\(A=\int_{-2}^1\left(x+7-x^2-1\right) d x+\int_1^2\left(11+3 x-x^2-1\right) d x \) \( =\left[\frac{x^2}{2}+6 x-\frac{x^3}{3}\right]_{-2}^1+\left[10 x-\frac{3 x^2}{2}-\frac{x^3}{3}\right]_1^2 \) \( =\frac{50}{3} \Rightarrow 3 A=50 \quad \text { Option (1) }\)
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