JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region enclosed by the curve \(y=x^3\) and its tangent at the point \((-1,-1)\) is
- A \(\frac{27}{4}\)
- B \(\frac{19}{4}\)
- C \(\frac{23}{4}\)
- D \(\frac{31}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{27}{4}\)
Step-by-step Solution
Detailed explanation
equation of tangent : \(y+1=3(x+1)\) i.e. \(y=3 x+2\) Point of intersection with curve \((2,8)\) So Area \(=\int \limits_{-1}^2\left((3 x+2)-x^3\right) d x=\frac{27}{4}\)
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