The area of the region enclosed by the curve y = x 3 and its tangent at the point ( - 1 , - 1 ) is [2023]

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Solution

Q.
The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is [2023]

1 $\frac{27}{4}$

2 $\frac{31}{4}$

3 $\frac{23}{4}$

4 $\frac{19}{4}$

Ans.
(1)

$Given, y = x^{3}$

$Equation of tangent to the curve y = x^{3} at point (- 1, - 1) is$

$y + 1 = 3 (x + 1) \Rightarrow y = 3 x + 2$

$Point of intersection with the curve is (2, 8)$

$So, area = \int_{- 1}^{2} [(3 x + 2) - x^{3}] d x$

$= {[3 \times \frac{x^{2}}{2} + 2 x - \frac{x^{4}}{4}]}_{- 1}^{2}$

$= \frac{27}{4} sq. units$

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