JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in square units) of the region bounded by the curves \(y + 2x^2 = 0\) and \(y + 3x^2 = 1\) , is equal to
- A \(\frac {3}{5}\)
- B \(\frac {1}{3}\)
- C \(\frac {4}{3}\)
- D \(\frac {3}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac {4}{3}\)
Step-by-step Solution
Detailed explanation
Solving \(y=2 x^{2}=0\) \(y=3 x^{2}=1\) Point of intersection \((1,-2)\) and \((-1,-2)\) Area \( = 2\int\limits_0^1 {\left( {\left( {1 - 3{x^2}} \right) - \left( { - 2{x^2}} \right)dx} \right)} \)…
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