Question

Area of region enclosed by curve y=x3 and its tangent at (–1,–1)

• A. 4
• B. 27
• C. $\frac{4}{27}$
• D. $\frac{27}{4}$

Answer 1

Answer: (D)

$\frac{27}{4}$

Answer 2

The correct answer is option (D): $\frac{27}{4}$

y+1=3(x+1)

i.e y = 3x+2

Point of intersection with curve (2,8)

So area = $\int_{-1}^{2}((3x+2)-x^3)dx=\frac{27}{4}$