JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region enclosed by the parabola \(y=4 x-x^2\) and \(3 y=(x-4)^2\) is equal to
- A \(\frac{32}{9}\)
- B \(4\)
- C \(6\)
- D \(\frac{14}{3}\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\( \text { Area }=\left\lvert\, \int_1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \) \( \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \)…
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