JEE Mains · Maths · STD 12 - 8. Application and integration
Let \( A_1 \) be the bounded area enclosed by the curves \( y=x^2+2 \), \( x+y=8 \) and y-axis that lies in the first quadrant. Let \( A_{2} \) be the bounded area enclosed by the curves \( y=x^2+2 \), \( y^{2}=x \), \( x=2 \), and y-axis that lies in the first quadrant. Then \( A_{1}-A_2 \) is equal to
- A \(\frac{2}{3}(2\sqrt{2}+1)\)
- B \(\frac{2}{3}(4\sqrt{2}+1)\)
- C \(\frac{2}{3}(\sqrt{2}+1)\)
- D \(\frac{2}{3}(3\sqrt{2}+1)\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}(2\sqrt{2}+1)\)
Step-by-step Solution
Detailed explanation
\(A_1=\int_0^2\left((8-x)-\left(x^2+2\right)\right) d x\) \(=A_1=\int_0^2\left(6-x-x^2\right) d x\) \(A_1\left(6 x-\frac{x^2}{2}-\frac{x^3}{3}\right)_0^2=12-2-\frac{8}{3}=10-\frac{8}{3}=\frac{22}{3}\) \(A_2=\int_0^2\left(x^2+2\right) d x-\frac{2}{3}(2 \sqrt{2})\)…
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