JEE Mains · Maths · STD 12 - 8. Application and integration
Let \(S\) be the region bounded by the curves \(y=x^{3}\) and \(y ^{2}= x\). The curve \(y =2| x |\) divides \(S\) into two regions of areas \(R_{1}\) and \(R_{2}\). If \(\max \left\{R_{1}, R_{2}\right\}=R_{2}\), then \(\frac{R_{2}}{R_{1}}\) is equal to
- A \(18\)
- B \(19\)
- C \(20\)
- D \(22\)
Answer & Solution
Correct Answer
(A) \(18\)
Step-by-step Solution
Detailed explanation
\(S=\int_{0}^{1} \sqrt{x}-x^{3}\) \(=\left[\frac{2 x^{3 / 2}}{3}-\frac{x^{4}}{4}\right]_{1}^{0}\) \(=\frac{5}{12}\) \(R_{1}=\int_{0}^{1 / 4}(\sqrt{x}-2 x) d x\) \(=\left[\frac{2 x^{3 / 2}}{3}-x^{2}\right]_{0}^{1 / 4}=\frac{1}{48}\) \(\therefore R_{2}=\frac{19}{48}\) So,…
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